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Chemistry · Year 12 · Exceeding

Gas equilibrium: quotient, extent and physical solutions

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In this level

Model and critically evaluate a gas equilibrium using atom conservation, the reaction quotient and physically valid ICE solutions.

Learning objectives:

  • Construct the quotient from the balanced equation and justify net direction.
  • Solve forward and reverse ICE calculations, rejecting inadmissible roots.
  • Test rather than assume a small-x approximation.
  • Distinguish immediate changes from new equilibrium after volume or temperature changes.
  • Evaluate claims and transformed equations using conservation and equilibrium checks.

A compressed nitrogen-oxide mixture becomes more concentrated immediately, then reacts in the opposite direction. Can its final nitrogen dioxide concentration still exceed its original value?

Prerequisite review

  • Coefficients, concentration and reaction extent — Concentration equals amount divided by volume. For N₂O₄ ⇌ 2NO₂ at fixed volume, consuming x concentration units of N₂O₄ forms 2x of NO₂. Atom counts are conserved; molecule counts need not be.

    reaction extent x: The concentration change assigned to one N₂O₄ coefficient in this fixed-volume ICE calculation.

    coefficient: The number of formula units; it multiplies each atom count without changing the substance.

    concentration: Amount of a substance divided by its volume; here measured in moles per litre, mol L⁻¹.

    Watch the prerequisite 1

    Why does N₂O₄(g) ⇌ 2NO₂(g) use coefficient two?

    Step 1

    Before: One N₂O₄ molecule.
    Action: Count subscripts.
    After: Two nitrogen atoms and four oxygen atoms.
    Why: A subscript gives atom count in one molecule.
    Coefficients conserve atoms

    Step 2

    Before: One NO₂ contains one nitrogen and two oxygen atoms.
    Action: Use two NO₂ molecules.
    After: Nitrogen: 2 × 1 = 2. Oxygen: 2 × 2 = 4.
    Why: The coefficient multiplies every atom count in the formula; formulas remain unchanged.

    Both sides contain two nitrogen and four oxygen atoms. Changing NO₂ to another formula changes the substance.

    Watch the prerequisite 2

    A sample contains 0.100 mol in 0.500 L. Find its concentration, then find the immediate concentration if its volume doubles without reaction.

    Step 1

    Before: Concentration is amount per litre.
    Action: Write c = n/V.
    After: n = 0.100 mol; V = 0.500 L
    Why: The numerator is amount and the denominator is volume; use the stated litre units.

    Step 2

    Before: c = n/V
    Action: Substitute the amount and volume.
    After: c = 0.100/0.500 = 0.200 mol L⁻¹
    Why: Dividing by half a litre gives the amount in one litre.
    Amount divided by volume

    Step 3

    Before: Amount remains 0.100 mol.
    Action: Double the volume.
    After: New volume = 2 × 0.500 = 1.000 L
    Why: This immediate dilution changes volume, not the amount of reacting atoms.

    Step 4

    Before: c = n/V; n = 0.100; V = 1.000
    Action: Divide the unchanged amount by the new volume.
    After: c = 0.100/1.000 = 0.100 mol L⁻¹
    Why: Twice the volume gives half the concentration before any chemical adjustment.

    Step 5

    Before: Initial and new amounts should match.
    Action: Multiply each concentration by its own volume.
    After: 0.200 × 0.500 = 0.100; 0.100 × 1.000 = 0.100 mol
    Why: Both states contain the same amount.

    Initially 0.200 mol L⁻¹; immediately after doubling volume, 0.100 mol L⁻¹.

    Complete the missing step 1

    Complete the counts for 3N₂O₄ and 6NO₂. Do they contain matching totals?

    Complete the missing step 2

    Complete c = __/__: amount is 0.120 mol and volume is 0.600 L. What happens immediately to c if that volume halves without reaction?

    Try a fresh question 1

    An extent of 0.0200 mol L⁻¹ N₂O₄ is consumed at fixed volume. State both concentration changes and check atom conservation.

    Try a fresh question 2

    A sample contains 0.300 mol in 0.750 L. Find c, then its immediate concentration after volume triples without reaction.

  • Equal operations, bracket expansion and quadratic roots — Apply the same operation to both sides. Expand the square of a difference with its middle term. A quadratic may have two mathematical roots; concentrations and direction determine which are physical.

    physical root: A mathematical solution that also obeys nonnegative concentrations, stoichiometry and the stated direction of change.

    Watch the prerequisite 1

    Solve 2x − 10 = 6 using equal operations.

    Step 1

    Before: 2x − 10 = 6
    Action: Add ten to both sides.
    After: 2x − 10 + 10 = 6 + 10; 2x = 16
    Why: The marked opposite terms sum to zero.
    Apply +10 to both sides

    Step 2

    Before: 2x = 16
    Action: Divide both sides by two.
    After: 2x/2 = 16/2; x = 8
    Why: Matching factors two reduce to one. Check: 2 × 8 − 10 = 6.

    x = 8; each equal operation preserves equality.

    Complete the missing step 1

    Rearrange 0.100 − 0.200x = 4x² into a quadratic equal to zero. Show the operation on both sides.

    Try a fresh question 1

    Expand (0.400 − 2x)² and collect terms in (0.400 − 2x)² = 0.200(0.100 + x).

Readiness check

Count nitrogen and oxygen atoms in 2NO₂.
An amount 0.100 mol occupies 0.500 L. Find concentration.
What equal operation isolates y in y − 3 = 8?
Expand (u − v)².

Watch — I do

Gas equilibrium: quotient, extent and physical solutions

Count atoms, then construct the quotient

Use N₂O₄(g) ⇌ 2NO₂(g). Square the product concentration because its coefficient is two. In this lesson K means the classroom concentration equilibrium constant K with subscript c; Q is the corresponding reaction quotient at the current composition. Values are illustrative training data, not measured nitrogen-oxide constants.

K = [NO₂]²/[N₂O₄] Q < K: net forward change Q > K: net reverse change Q = K: equilibrium
Coefficients conserve atoms

Track the same extent through both species

At fixed volume, an ICE table lists initial, change and equilibrium concentrations. For forward change use −x and +2x; for reverse change use +x and −2x. Choose the direction by Q versus K before defining x.

Forward: [N₂O₄] = A − x; [NO₂] = B + 2x Reverse: [N₂O₄] = A + x; [NO₂] = B − 2x
Keep the coefficient in every change

Solve and test the root

For ar² + br + c = 0 with a nonzero, use the quadratic formula below. In an equilibrium calculation, reject roots that violate the concentration bounds or the defined direction. Substitute valid concentrations back into K and check atom conservation.

r = (−b ± √(b² − 4ac))/(2a)
A mathematical root must also be physical

Test a shortcut

Neglecting x in A − x is a conditional approximation. Here a 5% depletion screen is a stated modelling tolerance, not a universal accuracy guarantee. Test x/A after calculating the estimate, and use the complete equation if the screen fails.

Approximate only if justified: x/A × 100% < 5%
Compare estimated depletion

Distinguish the immediate state from equilibrium

A rapid volume change rescales concentrations before appreciable reaction. Q changes, then reaction adjusts composition until Q equals K. At fixed temperature K stays unchanged. A temperature change can change K. A catalyst does not change the equilibrium constant or equilibrium composition.

Halve volume: each concentration doubles Q after compression = 2Q before compression
Compression and later adjustment

Check claims using more than one condition

Closed fixed-volume atom conservation gives 2[N₂O₄] + [NO₂] equal to its initial value. Satisfying K alone does not prove that a proposed final composition came from that starting mixture. Reversing the balanced equation inverts K; doubling it squares K.

K for reverse equation = 1/K K for doubled equation = K²
Rewrite the expression before the constant

Worked examples

I do

At the supplied temperature K = 0.200. [N₂O₄] = 0.300; [NO₂] = 0.200. Predict net reaction direction.

Step 1

Before: Current concentrations may not be equilibrium values.
Action: Use Q to describe the present state.
After: Q = [NO₂]²/[N₂O₄]
Why: K is the equilibrium reference at this temperature.

Step 2

Before: Q = [NO₂]²/[N₂O₄]
Action: Square product concentration, then divide.
After: Q = (0.200)²/0.300 = 0.133
Why: The coefficient two becomes an exponent.

Step 3

Before: Q = 0.133; K = 0.200
Action: Compare the two values.
After: Net change is rightwards.
Why: Q below K needs a larger ratio; Q above K needs a smaller ratio.

Net rightwards; K remains 0.200.

I do

At the supplied temperature K = 0.200. [N₂O₄] = 0.200; [NO₂] = 0.400. Predict net reaction direction.

Step 1

Before: Current concentrations may not be equilibrium values.
Action: Use Q to describe the present state.
After: Q = [NO₂]²/[N₂O₄]
Why: K is the equilibrium reference at this temperature.

Step 2

Before: Q = [NO₂]²/[N₂O₄]
Action: Square product concentration, then divide.
After: Q = (0.400)²/0.200 = 0.800
Why: The coefficient two becomes an exponent.

Step 3

Before: Q = 0.800; K = 0.200
Action: Compare the two values.
After: Net change is leftwards.
Why: Q below K needs a larger ratio; Q above K needs a smaller ratio.

Net leftwards; K remains 0.200.

I do

Initially [N₂O₄] = 0.500 and [NO₂] = 0. K = 0.200. Find both equilibrium concentrations.

Step 1

Before: Initial Q = 0²/0.500 = 0.
Action: Compare with K = 0.200.
After: Q < K, so net forward change.
Why: Reactant is consumed; product forms.

Step 2

Before: Define x as concentration of N₂O₄ consumed.
Action: Apply the balanced coefficients.
After: [N₂O₄] = 0.500 − x; [NO₂] = 2x
Why: One reactant unit gives two product units.
Initial, change and equilibrium

Step 3

Before: Concentrations cannot be negative.
Action: Set the physical range.
After: 0 ≤ x ≤ 0.500
Why: The product concentration is 2x and reactant is 0.500 − x.

Step 4

Before: K = [NO₂]²/[N₂O₄]
Action: Substitute the equilibrium expressions.
After: 0.200 = (2x)²/(0.500 − x)
Why: Square the whole product expression, including coefficient two.

Step 5

Before: 0.200 = 4x²/(0.500 − x)
Action: Multiply both sides by 0.500 − x.
After: 0.200(0.500 − x) = 4x²
Why: The denominator factor reduces to one; x = 0.500 cannot satisfy finite K.
Multiply both sides by the denominator

Step 6

Before: 0.200(0.500 − x) = 4x²
Action: Distribute 0.200 across the bracket.
After: 0.100 − 0.200x = 4x²
Why: Multiply each term in the bracket.

Step 7

Before: 0.100 − 0.200x = 4x²
Action: Add 0.200x to both sides.
After: 0.100 = 4x² + 0.200x
Why: The opposite x terms sum to zero.
Apply +0.200x to both sides

Step 8

Before: 0.100 = 4x² + 0.200x
Action: Subtract 0.100 from both sides.
After: 4x² + 0.200x − 0.100 = 0
Why: Read the same equality with the quadratic on the left.
Apply −0.100 to both sides

Step 9

Before: 4x² + 0.200x − 0.100 = 0
Action: Identify quadratic coefficients.
After: a = 4; b = 0.200; c = −0.100
Why: Include the negative sign in c.

Step 10

Before: x = (−b ± √(b² − 4ac))/(2a)
Action: Substitute all coefficients.
After: x = (−0.200 ± √(0.0400 + 1.600))/8
Why: Subtracting the negative product adds 1.600.

Step 11

Before: Discriminant = 1.640.
Action: Calculate both signs.
After: x = 0.135078 or x = −0.185078
Why: Keep guard digits until final concentrations.

Step 12

Before: Physical range: 0 ≤ x ≤ 0.500.
Action: Reject the negative root.
After: x = 0.135078
Why: A negative extent would produce negative NO₂ from an initial zero amount.

Step 13

Before: [N₂O₄] = 0.500 − x; [NO₂] = 2x
Action: Substitute the valid extent.
After: [N₂O₄] = 0.364922; [NO₂] = 0.270156
Why: Report approximately 0.365 and 0.270 mol L⁻¹ to three significant figures.

Step 14

Before: Proposed concentrations 0.364922 and 0.270156.
Action: Check quotient and atoms.
After: 0.270156²/0.364922 ≈ 0.200; 2 × 0.364922 + 0.270156 = 1.000
Why: Both the equilibrium ratio and the initial nitrogen-atom inventory agree.

[N₂O₄] ≈ 0.365 mol L⁻¹; [NO₂] ≈ 0.270 mol L⁻¹.

I do

Initially [N₂O₄] = 0.100 and [NO₂] = 0.400. K = 0.200. Find equilibrium concentrations.

Step 1

Before: Q = 0.400²/0.100 = 1.60.
Action: Compare with K.
After: Q > 0.200; net reverse change.
Why: Define x as N₂O₄ formed.

Step 2

Before: One N₂O₄ forms from two NO₂.
Action: Write reverse changes.
After: [N₂O₄] = 0.100 + x; [NO₂] = 0.400 − 2x
Why: The range is 0 ≤ x ≤ 0.200.

Step 3

Before: K = [NO₂]²/[N₂O₄]
Action: Substitute and multiply both sides by 0.100 + x.
After: (0.400 − 2x)² = 0.200(0.100 + x)
Why: The whole squared bracket must be expanded.

Step 4

Before: (u − v)² = u² − 2uv + v²
Action: Expand both sides.
After: 0.160 − 1.600x + 4x² = 0.0200 + 0.200x
Why: The middle term includes twice 0.400 times 2x.
Expand all three square terms

Step 5

Before: 0.160 − 1.600x + 4x² = 0.0200 + 0.200x
Action: Subtract 0.0200 and 0.200x from both sides.
After: 4x² − 1.800x + 0.140 = 0
Why: Equal operations preserve the equation.

Step 6

Before: a = 4; b = −1.800; c = 0.140
Action: Use both signs of the quadratic formula.
After: x = (1.800 ± √(3.240 − 2.240))/8; x = 0.100 or 0.350
Why: The discriminant is 1.000.

Step 7

Before: x must be at most 0.200.
Action: Reject x = 0.350.
After: x = 0.100
Why: The other root gives [NO₂] = 0.400 − 0.700 = −0.300, impossible.

Step 8

Before: x = 0.100
Action: Calculate and check both concentrations.
After: [N₂O₄] = 0.200; [NO₂] = 0.200; Q = 0.200
Why: 2 × 0.200 + 0.200 = 0.600, matching 2 × 0.100 + 0.400.

Both equilibrium concentrations are 0.200 mol L⁻¹.

I do

Initially [N₂O₄] = 0.500 and [NO₂] = 0; supplied K = 0.2. Test neglecting x using a 5% depletion screen.

Step 1

Before: K = 4x²/(0.500 − x)
Action: Assume x is small relative to 0.500.
After: K ≈ 4x²/0.500
Why: This is a hypothesis to test, not an established equality.

Step 2

Before: K ≈ 4x²/0.500
Action: Multiply by 0.500, divide by four, then take the nonnegative square root.
After: x ≈ √(0.2 × 0.500/4) = 0.158114
Why: The defined forward extent is nonnegative.

Step 3

Before: x estimate 0.158114.
Action: Calculate the estimated depletion fraction.
After: x/0.500 × 100% = 31.62%
Why: Compare this with the stated 5% screen.

Step 4

Before: Depletion 31.62%.
Action: Decide whether the shortcut is suitable.
After: The screen fails; use the full quadratic.
Why: The full solution gives x = 0.13507811.

Estimated depletion 31.62%; reject the approximation.

I do

Initially [N₂O₄] = 0.500 and [NO₂] = 0; supplied K = 0.0002. Test neglecting x using a 5% depletion screen.

Step 1

Before: K = 4x²/(0.500 − x)
Action: Assume x is small relative to 0.500.
After: K ≈ 4x²/0.500
Why: This is a hypothesis to test, not an established equality.

Step 2

Before: K ≈ 4x²/0.500
Action: Multiply by 0.500, divide by four, then take the nonnegative square root.
After: x ≈ √(0.0002 × 0.500/4) = 0.005000
Why: The defined forward extent is nonnegative.

Step 3

Before: x estimate 0.005000.
Action: Calculate the estimated depletion fraction.
After: x/0.500 × 100% = 1.00%
Why: Compare this with the stated 5% screen.

Step 4

Before: Depletion 1.00%.
Action: Decide whether the shortcut is suitable.
After: The screen passes; assess the required numerical precision too.
Why: The full solution gives x = 0.00497506.

Step 5

Before: N₂O₄(g) ⇌ 2NO₂(g)
Action: Write the current concentration quotient.
After: Q = [NO₂]²/[N₂O₄]
Why: The coefficient two becomes an exponent.

Step 6

Before: [NO₂] = 0
Action: Square the product concentration first.
After: [NO₂]² = 0² = 0
Why: Square the whole concentration; do not multiply it by two.

Step 7

Before: Product square 0; reactant 0.5
Action: Divide the product square by reactant concentration.
After: Q = 0/0.5 = 0
Why: Use the current values to calculate Q.

Step 8

Before: Q = 0; K = 0.0002
Action: Compare Q with the reference at this temperature.
After: Q is below K: net forward change.
Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

Step 9

Before: The quotient determines the net reaction direction.
Action: Define nonnegative x as N₂O₄ consumed.
After: [N₂O₄] = 0.5 − x; [NO₂] = 0 + 2x
Why: The balanced equation gives one-to-two concentration changes.

Step 10

Before: Every final concentration must be nonnegative.
Action: Set the extent bounds.
After: 0 ≤ x ≤ 0.5
Why: Reject roots that violate these bounds or give zero denominator at finite K.

Step 11

Before: K = [NO₂]²/[N₂O₄]
Action: Substitute both equilibrium expressions.
After: 0.0002 = (0 + 2x)²/(0.5 − x)
Why: Square the entire numerator bracket.

Step 12

Before: 0.0002 = (0 + 2x)²/(0.5 − x)
Action: Multiply both sides by 0.5 − x.
After: 0.0002(0.5 − x) = (0 + 2x)²
Why: The denominator factor reduces to one.

Step 13

Before: (0 + 2x)²
Action: Expand the square, including its middle term.
After: 0 + 0x + 4x²
Why: Use u² ± 2uv + v²; here v is 2x.

Step 14

Before: 0.0002(0.5 − x)
Action: Distribute K to both terms.
After: 0.0001 − 0.0002x
Why: Both the constant term and the x term are multiplied by K.

Step 15

Before: 0 + 0x + 4x² = 0.0001 − 0.0002x
Action: Subtract 0.0001 from both sides.
After: −0.0001 + 0x + 4x² = − 0.0002x
Why: The matching constant on the right sums to zero.

Step 16

Before: The right side contains − 0.0002x.
Action: Add 0.0002x on both sides.
After: 4x² + 0.0002x − 0.0001 = 0
Why: Combine the x coefficients; the opposite right-side x terms sum to zero.

Step 17

Before: 4x² + 0.0002x − 0.0001 = 0
Action: Read the quadratic coefficients.
After: a = 4; b = 0.0002; c = −0.0001
Why: Retain each coefficient sign.

Step 18

Before: Discriminant = b² − 4ac
Action: Substitute and calculate.
After: (0.0002)² − 4 × 4 × (−0.0001) = 0.00160004
Why: Bracket a negative coefficient when substituting.

Step 19

Before: x = (−b ± √(b² − 4ac))/(2a)
Action: Substitute into the quadratic formula.
After: x = (−(0.0002) ± √0.00160004)/8
Why: Calculate both signs; the denominator is two times four.

Step 20

Before: Both signs of the formula are required.
Action: Calculate both numerical roots.
After: x = 0.00497506 or −0.00502506
Why: Use the physical constraints to choose the valid root.

Step 21

Before: 0 ≤ x ≤ 0.5
Action: Reject x = −0.00502506.
After: Use x = 0.00497506
Why: The rejected value violates the extent bounds and gives a negative species concentration.

Step 22

Before: Use the valid extent in both species.
Action: Calculate final concentrations.
After: [N₂O₄] = 0.495025; [NO₂] = 0.00995012
Why: Keep guard digits while checking the solution.

Step 23

Before: Proposed final concentrations are now known.
Action: Check the quotient and conserved atoms.
After: Q = 0.00995012²/0.495025 ≈ 0.0002; 2[N₂O₄] + [NO₂] = 1
Why: Both the equilibrium ratio and starting atom inventory agree.

Estimated depletion 1.00%; passes this stated screen.

I do

An equilibrium mixture has both concentrations 0.200. Halve its volume at unchanged temperature. Find the new equilibrium.

Step 1

Before: Amount is unchanged during rapid compression.
Action: Divide volume by two.
After: Both concentrations immediately become 0.400
Why: Concentration equals amount divided by volume.

Step 2

Before: Q = 0.400²/0.400
Action: Calculate and compare.
After: Q = 0.400 > K = 0.200
Why: K remains unchanged because temperature is unchanged.

Step 3

Before: Net change is reverse.
Action: Use +x and −2x.
After: [N₂O₄] = 0.400 + x; [NO₂] = 0.400 − 2x
Why: 0 ≤ x ≤ 0.200.

Step 4

Before: 0.200 = (0.400 − 2x)²/(0.400 + x)
Action: Multiply by the denominator and expand.
After: 0.160 − 1.600x + 4x² = 0.0800 + 0.200x
Why: The middle term of the square remains essential.

Step 5

Before: 0.160 − 1.600x + 4x² = 0.0800 + 0.200x
Action: Subtract the right expression from both sides.
After: 4x² − 1.800x + 0.0800 = 0
Why: Both sides undergo the same operations.

Step 6

Before: a = 4; b = −1.800; c = 0.0800
Action: Use the quadratic formula.
After: x = (1.800 ± √(3.240 − 1.280))/8 = 0.0500 or 0.400
Why: The second root exceeds 0.200 and gives negative NO₂.

Step 7

Before: Valid x = 0.0500
Action: Substitute and check.
After: [N₂O₄] = 0.450; [NO₂] = 0.300; Q = 0.300²/0.450 = 0.200
Why: The new NO₂ concentration exceeds its original 0.200 even though reaction adjusted leftwards.

New equilibrium: 0.450 mol L⁻¹ N₂O₄ and 0.300 mol L⁻¹ NO₂.

I do

At fixed volume, heat an equilibrium mixture with both concentrations 0.200. Supplied K changes from 0.200 to 0.600. Find the new equilibrium.

Step 1

Before: The volume and amounts have not yet changed.
Action: Compare current Q with the new K.
After: Q = 0.200 < new K = 0.600
Why: Temperature changed K; reaction then changes concentrations.

Step 2

Before: Net change is forward.
Action: Use −x and +2x.
After: [N₂O₄] = 0.200 − x; [NO₂] = 0.200 + 2x
Why: 0 ≤ x ≤ 0.200.

Step 3

Before: 0.600 = (0.200 + 2x)²/(0.200 − x)
Action: Multiply by the denominator and expand.
After: 0.120 − 0.600x = 0.0400 + 0.800x + 4x²
Why: The square has three terms.

Step 4

Before: 0.120 − 0.600x = 0.0400 + 0.800x + 4x²
Action: Add 0.600x and subtract 0.120 on both sides.
After: 4x² + 1.400x − 0.0800 = 0
Why: Equal operations collect the quadratic.

Step 5

Before: a = 4; b = 1.400; c = −0.0800
Action: Calculate both roots.
After: x = (−1.400 ± √(1.960 + 1.280))/8 = 0.0500 or −0.400
Why: Reject the negative forward extent.

Step 6

Before: x = 0.0500
Action: Substitute and check.
After: [N₂O₄] = 0.150; [NO₂] = 0.300; Q = 0.300²/0.150 = 0.600
Why: The new concentrations satisfy the new equilibrium constant.

New equilibrium: 0.150 mol L⁻¹ N₂O₄ and 0.300 mol L⁻¹ NO₂.

I do

K = 0.200 for N₂O₄ ⇌ 2NO₂. Find the constant for 2NO₂ ⇌ N₂O₄.

Step 1

Before: Original K = [NO₂]²/[N₂O₄]
Action: Write the expression for the reversed equation.
After: Reverse constant = [N₂O₄]/[NO₂]²
Why: The new reactant and product roles swap; their balanced exponents stay attached to each formula.

Step 2

Before: [N₂O₄]/[NO₂]²
Action: Recognise the reciprocal of the original ratio.
After: Reverse constant = 1/K
Why: Multiplying the two ratios cancels matching concentration factors and gives one.

Step 3

Before: Reverse constant = 1/K
Action: Substitute and calculate.
After: 1/0.200 = 5.00
Why: Check: 5.00 × 0.200 = 1.00.

The reversed constant is 5.00.

I do

K = 0.200 for N₂O₄ ⇌ 2NO₂. Find the constant for 2N₂O₄ ⇌ 4NO₂.

Step 1

Before: Balanced coefficients are now two and four.
Action: Apply each coefficient as an exponent.
After: New constant = [NO₂]⁴/[N₂O₄]²
Why: A coefficient changes an exponent, not a multiplier outside the expression.

Step 2

Before: [NO₂]⁴/[N₂O₄]²
Action: Rewrite as the square of the original ratio.
After: ([NO₂]²/[N₂O₄])² = K²
Why: Squaring a quotient squares its numerator and denominator.

Step 3

Before: New constant = K²
Action: Substitute and square the full numerical constant.
After: 0.200² = 0.0400
Why: Using the equilibrium pair 0.200 and 0.200: 0.200⁴/0.200² = 0.0400, confirming the new expression.

The doubled-equation constant is 0.0400.

Together — We do

Worked examples

We do

Complete the forward ICE equation after expansion from concentrations 0.225 N₂O₄ and 0.150 NO₂. Supplied K = 0.200 at this fixed temperature.

  1. Complete the counts for 3N₂O₄ and 6NO₂. Do they contain matching totals?

    Hint 1

    Use the coefficient to multiply each subscript.

    Hint 2

    For 3N₂O₄, nitrogen count is three groups of two.

    Hint 3

    Find both oxygen totals, then compare each element separately.

    Show the complete working

    3N₂O₄: N = 3 × 2 = 6; O = 3 × 4 = 12.

    6NO₂: N = 6 × 1 = 6; O = 6 × 2 = 12.

    Atom totals match, although molecule counts differ.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  2. Rearrange 0.100 − 0.200x = 4x² into a quadratic equal to zero. Show the operation on both sides.

    Hint 1

    Preserve equality while collecting the quadratic terms.

    Hint 2

    Add 0.200x to both sides; the opposite x terms sum to zero.

    Hint 3

    Now subtract 0.100 from both sides and write the quadratic expression on the left.

    Show the complete working

    Add 0.200x to both sides: 0.100 − 0.200x + 0.200x = 4x² + 0.200x, so 0.100 = 4x² + 0.200x.

    Subtract 0.100 from both sides: 0 = 4x² + 0.200x − 0.100.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  3. Complete Q = __²/__ for [NO₂] = 0.300 and [N₂O₄] = 0.450. Decide whether the mixture is at equilibrium. Use supplied K = 0.200 at the stated fixed temperature.

    Track the supplied mixtureN₂O₄(g) ⇌ 2NO₂(g) Keep coefficients in Q and ICE Separate immediate and later statesN₂O₄(g) 2NO₂(g)Keep coefficients in Q and ICESeparate immediate and later states
    Track the supplied mixture
    Hint 1

    Use the balanced coefficient as the exponent.

    Hint 2

    Square 0.300 before dividing by 0.450.

    Hint 3

    The quotient equals the supplied reference; explain what this means for net change.

    Show the complete working

    Step 1

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 2

    Before: [NO₂] = 0.3
    Action: Square the product concentration first.
    After: [NO₂]² = 0.3² = 0.09
    Why: Square the whole concentration; do not multiply it by two.

    Step 3

    Before: Product square 0.09; reactant 0.45
    Action: Divide the product square by reactant concentration.
    After: Q = 0.09/0.45 = 0.2
    Why: Use the current values to calculate Q.

    Step 4

    Before: Q = 0.2; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is equal to K: dynamic equilibrium.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Q = 0.300²/0.450 = 0.0900/0.450 = 0.200 = K. The mixture is at dynamic equilibrium.

    Q = 0.300²/0.450 = 0.0900/0.450 = 0.200 = K.

    The mixture is at dynamic equilibrium.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  4. A mixture starts with [N₂O₄] = 0.450 and [NO₂] = 0.300, then volume doubles at fixed temperature. Complete immediate concentrations and the ICE change signs. Use supplied K = 0.200 at the stated fixed temperature.

    Track the supplied mixtureN₂O₄(g) ⇌ 2NO₂(g) Keep coefficients in Q and ICE Separate immediate and later statesN₂O₄(g) 2NO₂(g)Keep coefficients in Q and ICESeparate immediate and later states
    Track the supplied mixture
    Hint 1

    Separate immediate dilution from reaction adjustment.

    Hint 2

    Doubling volume halves both concentrations.

    Hint 3

    Calculate Q from the halved values and use its comparison to K for the signs.

    Show the complete working

    Immediate concentrations are 0.225 and 0.150.

    Q = 0.150²/0.225 = 0.100 < K.

    Net change is forward, so changes are −x for N₂O₄ and +2x for NO₂.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  5. After that expansion, complete (0.150 + 2x)² = 0.200(0.225 − x), then find the physical root.

    Track the supplied mixtureN₂O₄(g) ⇌ 2NO₂(g) Keep coefficients in Q and ICE Separate immediate and later statesN₂O₄(g) 2NO₂(g)Keep coefficients in Q and ICESeparate immediate and later states
    Track the supplied mixture
    Hint 1

    Expand the square with its middle term.

    Hint 2

    The expanded equality is 0.0225 + 0.600x + 4x² = 0.0450 − 0.200x.

    Hint 3

    Collect 4x² + 0.800x − 0.0225 = 0. Use both signs, then test 0 ≤ x ≤ 0.225.

    Show the complete working

    Step 1

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 2

    Before: [NO₂] = 0.15
    Action: Square the product concentration first.
    After: [NO₂]² = 0.15² = 0.0225
    Why: Square the whole concentration; do not multiply it by two.

    Step 3

    Before: Product square 0.0225; reactant 0.225
    Action: Divide the product square by reactant concentration.
    After: Q = 0.0225/0.225 = 0.1
    Why: Use the current values to calculate Q.

    Step 4

    Before: Q = 0.1; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is below K: net forward change.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Step 5

    Before: The quotient determines the net reaction direction.
    Action: Define nonnegative x as N₂O₄ consumed.
    After: [N₂O₄] = 0.225 − x; [NO₂] = 0.15 + 2x
    Why: The balanced equation gives one-to-two concentration changes.

    Step 6

    Before: Every final concentration must be nonnegative.
    Action: Set the extent bounds.
    After: 0 ≤ x ≤ 0.225
    Why: Reject roots that violate these bounds or give zero denominator at finite K.

    Step 7

    Before: K = [NO₂]²/[N₂O₄]
    Action: Substitute both equilibrium expressions.
    After: 0.2 = (0.15 + 2x)²/(0.225 − x)
    Why: Square the entire numerator bracket.

    Step 8

    Before: 0.2 = (0.15 + 2x)²/(0.225 − x)
    Action: Multiply both sides by 0.225 − x.
    After: 0.2(0.225 − x) = (0.15 + 2x)²
    Why: The denominator factor reduces to one.

    Step 9

    Before: (0.15 + 2x)²
    Action: Expand the square, including its middle term.
    After: 0.0225 + 0.6x + 4x²
    Why: Use u² ± 2uv + v²; here v is 2x.

    Step 10

    Before: 0.2(0.225 − x)
    Action: Distribute K to both terms.
    After: 0.045 − 0.2x
    Why: Both the constant term and the x term are multiplied by K.

    Step 11

    Before: 0.0225 + 0.6x + 4x² = 0.045 − 0.2x
    Action: Subtract 0.045 from both sides.
    After: −0.0225 + 0.6x + 4x² = − 0.2x
    Why: The matching constant on the right sums to zero.

    Step 12

    Before: The right side contains − 0.2x.
    Action: Add 0.2x on both sides.
    After: 4x² + 0.8x − 0.0225 = 0
    Why: Combine the x coefficients; the opposite right-side x terms sum to zero.

    Step 13

    Before: 4x² + 0.8x − 0.0225 = 0
    Action: Read the quadratic coefficients.
    After: a = 4; b = 0.8; c = −0.0225
    Why: Retain each coefficient sign.

    Step 14

    Before: Discriminant = b² − 4ac
    Action: Substitute and calculate.
    After: (0.8)² − 4 × 4 × (−0.0225) = 1
    Why: Bracket a negative coefficient when substituting.

    Step 15

    Before: x = (−b ± √(b² − 4ac))/(2a)
    Action: Substitute into the quadratic formula.
    After: x = (−(0.8) ± √1)/8
    Why: Calculate both signs; the denominator is two times four.

    Step 16

    Before: Both signs of the formula are required.
    Action: Calculate both numerical roots.
    After: x = 0.025 or −0.225
    Why: Use the physical constraints to choose the valid root.

    Step 17

    Before: 0 ≤ x ≤ 0.225
    Action: Reject x = −0.225.
    After: Use x = 0.025
    Why: The rejected value violates the extent bounds and gives a negative species concentration.

    Step 18

    Before: Use the valid extent in both species.
    Action: Calculate final concentrations.
    After: [N₂O₄] = 0.2; [NO₂] = 0.2
    Why: Keep guard digits while checking the solution.

    Step 19

    Before: Proposed final concentrations are now known.
    Action: Check the quotient and conserved atoms.
    After: Q = 0.2²/0.2 ≈ 0.2; 2[N₂O₄] + [NO₂] = 0.6
    Why: Both the equilibrium ratio and starting atom inventory agree.

    Expand: 0.0225 + 0.600x + 4x² = 0.0450 − 0.200x. Add 0.200x and subtract 0.0450 on both sides: 4x² + 0.800x − 0.0225 = 0. Roots: (−0.800 ± √(0.640 + 0.360))/8 = 0.0250 or −0.225. Reject the negative forward extent.

    Expand: 0.0225 + 0.600x + 4x² = 0.0450 − 0.200x.

    Add 0.200x and subtract 0.0450 on both sides: 4x² + 0.800x − 0.0225 = 0.

    Roots: (−0.800 ± √(0.640 + 0.360))/8 = 0.0250 or −0.225.

    Reject the negative forward extent.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  6. At fixed volume, both concentrations remain 0.200 immediately after heating changes K from 0.200 to 0.600. Complete current Q and the net direction.

    Hint 1

    Separate a changed equilibrium reference from unchanged immediate concentrations.

    Hint 2

    Calculate Q from both current concentrations 0.200.

    Hint 3

    Compare your quotient with the new constant 0.600, then select the net direction.

    Show the complete working

    Q = 0.200²/0.200 = 0.200.

    Compare with the new K = 0.600: Q < K.

    Net forward change forms more NO₂ and consumes N₂O₄.

    Temperature changed K; amounts have not yet adjusted.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  7. At a different supplied temperature the original reaction has K = 0.500. Complete the reverse constant 1/___ and doubled-equation constant (___)².

    Hint 1

    Rewrite each balanced equation and its expression.

    Hint 2

    Reversing gives the reciprocal 1/K; doubling coefficients doubles exponents.

    Hint 3

    Use the supplied 0.500 in each expression; calculate the reciprocal and square separately.

    Show the complete working

    Reverse: 1/0.500 = 2.00.

    Doubled equation: 0.500² = 0.250.

    Reversal swaps numerator and denominator; doubled coefficients square the full original expression.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  8. A proposed shortcut gives estimated extent 0.00300 from initial reactant 0.300. Complete the depletion calculation ___/___ × 100%. Does it pass a stated 5% screen, and does that alone establish numerical accuracy?

    Hint 1

    Use the estimated extent as a fraction of initial reactant.

    Hint 2

    Write 0.00300/0.300 × 100%.

    Hint 3

    Calculate the percentage and compare with five; distinguish this screen from an exact-equation check.

    Show the complete working

    Depletion estimate = 0.00300/0.300 × 100% = 1.00%.

    It passes the stated depletion screen.

    This is only a consistency screen for the small-x assumption; the full equation or required numerical precision must still justify the final accuracy.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

With help

Worked examples

You do

A prepared mixture has N₂O₄ 0.400 and NO₂ 0.200. Supplied K = 0.200 at this fixed temperature. Choose a method to predict its change.

  1. An extent of 0.0200 mol L⁻¹ N₂O₄ is consumed at fixed volume. State both concentration changes and check atom conservation.

    One extent changes both speciesN₂O₄ change: −0.0200 NO₂ change: twice the opposite amount Check nitrogen and oxygen separatelyN₂O₄ change: −0.0200NO₂ change: twice the opposite amountCheck nitrogen and oxygen separately
    One extent changes both species
    Hint 1

    Use the balanced one-to-two relationship.

    Hint 2

    Set the N₂O₄ change to minus the stated extent.

    Hint 3

    NO₂ forms at twice that extent. Check two times the reactant change plus the product change.

    Show the complete working

    Δ[N₂O₄] = −0.0200 mol L⁻¹; Δ[NO₂] = +0.0400 mol L⁻¹.

    Nitrogen change: 2 × (−0.0200) + 0.0400 = 0; oxygen change is twice that, also zero.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  2. Expand (0.400 − 2x)² and collect terms in (0.400 − 2x)² = 0.200(0.100 + x).

    Hint 1

    Use the square of a difference, including its middle term.

    Hint 2

    Expand both sides: 0.160 − 1.600x + 4x² = 0.0200 + 0.200x.

    Hint 3

    Subtract the entire right expression from both sides, then combine like terms.

    Show the complete working

    (u − v)² = u² − 2uv + v².

    Here the left becomes 0.160 − 1.600x + 4x²; the right is 0.0200 + 0.200x.

    Subtract the right expression from both sides: 4x² − 1.800x + 0.140 = 0.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  3. For reverse change from [N₂O₄] = 0.100 and [NO₂] = 0.400, decide whether x = 0.350 is physical.

    Hint 1

    Use concentration bounds to test the proposed extent.

    Hint 2

    Substitute into [NO₂] = 0.400 − 2x.

    Hint 3

    The proposed product concentration is below zero; explain the physical consequence.

    Show the complete working

    [NO₂] = 0.400 − 2 × 0.350 = −0.300.

    A negative concentration is impossible; reject the root.

    The allowed reverse extent is at most 0.200.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  4. A learner estimates x = 0.158 from initial [N₂O₄] = 0.500. Assess the approximation using a stated 5% depletion screen.

    Hint 1

    Test the assumption after estimating x.

    Hint 2

    Divide the estimated extent by the initial reactant concentration.

    Hint 3

    Convert the fraction to per cent and compare with the stated tolerance.

    Show the complete working

    Depletion estimate = 0.158/0.500 × 100% = 31.6%.

    This exceeds the chosen 5% screen.

    Use the full quadratic.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  5. At fixed temperature, both equilibrium concentrations 0.200 are doubled by compression. Predict adjustment and state whether K changes.

    Hint 1

    Calculate Q at the immediate compressed state.

    Hint 2

    Both concentrations become 0.400 before reaction adjustment.

    Hint 3

    Compare the quotient with the unchanged temperature-dependent reference.

    Show the complete working

    Q = 0.400²/0.400 = 0.400 > K.

    Net change is leftwards.

    Temperature is unchanged, so K stays 0.200.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

On my own

  1. Initially [N₂O₄] = 0.400 and [NO₂] = 0.200. Predict direction and explain. Use supplied K = 0.200 at the stated fixed temperature.

    Hint 1

    Identify the requested comparison, calculation or claim.

    Hint 2

    Write the balanced equation and the quotient or conservation relationship relevant to the stated conditions.

    Hint 3

    Carry the coefficient two through the working; check the result against K, the physical bounds and the stated constraint.

    Show the complete working

    Step 1

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 2

    Before: [NO₂] = 0.2
    Action: Square the product concentration first.
    After: [NO₂]² = 0.2² = 0.04
    Why: Square the whole concentration; do not multiply it by two.

    Step 3

    Before: Product square 0.04; reactant 0.4
    Action: Divide the product square by reactant concentration.
    After: Q = 0.04/0.4 = 0.1
    Why: Use the current values to calculate Q.

    Step 4

    Before: Q = 0.1; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is below K: net forward change.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Q = 0.200²/0.400 = 0.100 < 0.200. Net change is forward; N₂O₄ decreases and NO₂ increases.

    Q = 0.200²/0.400 = 0.100 < 0.200.

    Net change is forward; N₂O₄ decreases and NO₂ increases.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  2. An equilibrium mixture has [N₂O₄] = 0.450 and [NO₂] = 0.300. Prove that it satisfies K = 0.200.

    Hint 1

    Identify the requested comparison, calculation or claim.

    Hint 2

    Write the balanced equation and the quotient or conservation relationship relevant to the stated conditions.

    Hint 3

    Carry the coefficient two through the working; check the result against K, the physical bounds and the stated constraint.

    Show the complete working

    Step 1

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 2

    Before: [NO₂] = 0.3
    Action: Square the product concentration first.
    After: [NO₂]² = 0.3² = 0.09
    Why: Square the whole concentration; do not multiply it by two.

    Step 3

    Before: Product square 0.09; reactant 0.45
    Action: Divide the product square by reactant concentration.
    After: Q = 0.09/0.45 = 0.2
    Why: Use the current values to calculate Q.

    Step 4

    Before: Q = 0.2; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is equal to K: dynamic equilibrium.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Q = 0.300²/0.450 = 0.0900/0.450 = 0.200. It satisfies the equilibrium condition at this temperature.

    Q = 0.300²/0.450 = 0.0900/0.450 = 0.200.

    It satisfies the equilibrium condition at this temperature.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  3. Initially [N₂O₄] = 0.100 and [NO₂] = 0.400. Construct and solve the reverse ICE equation, rejecting invalid roots. Use supplied K = 0.200 at the stated fixed temperature.

    Hint 1

    Identify the requested comparison, calculation or claim.

    Hint 2

    Write the balanced equation and the quotient or conservation relationship relevant to the stated conditions.

    Hint 3

    Carry the coefficient two through the working; check the result against K, the physical bounds and the stated constraint.

    Show the complete working

    Step 1

    Before: Q = 0.400²/0.100 = 1.60.
    Action: Compare with K.
    After: Q > 0.200; net reverse change.
    Why: Define x as N₂O₄ formed.

    Step 2

    Before: One N₂O₄ forms from two NO₂.
    Action: Write reverse changes.
    After: [N₂O₄] = 0.100 + x; [NO₂] = 0.400 − 2x
    Why: The range is 0 ≤ x ≤ 0.200.

    Step 3

    Before: K = [NO₂]²/[N₂O₄]
    Action: Substitute and multiply both sides by 0.100 + x.
    After: (0.400 − 2x)² = 0.200(0.100 + x)
    Why: The whole squared bracket must be expanded.

    Step 4

    Before: (u − v)² = u² − 2uv + v²
    Action: Expand both sides.
    After: 0.160 − 1.600x + 4x² = 0.0200 + 0.200x
    Why: The middle term includes twice 0.400 times 2x.
    Expand all three square terms

    Step 5

    Before: 0.160 − 1.600x + 4x² = 0.0200 + 0.200x
    Action: Subtract 0.0200 and 0.200x from both sides.
    After: 4x² − 1.800x + 0.140 = 0
    Why: Equal operations preserve the equation.

    Step 6

    Before: a = 4; b = −1.800; c = 0.140
    Action: Use both signs of the quadratic formula.
    After: x = (1.800 ± √(3.240 − 2.240))/8; x = 0.100 or 0.350
    Why: The discriminant is 1.000.

    Step 7

    Before: x must be at most 0.200.
    Action: Reject x = 0.350.
    After: x = 0.100
    Why: The other root gives [NO₂] = 0.400 − 0.700 = −0.300, impossible.

    Step 8

    Before: x = 0.100
    Action: Calculate and check both concentrations.
    After: [N₂O₄] = 0.200; [NO₂] = 0.200; Q = 0.200
    Why: 2 × 0.200 + 0.200 = 0.600, matching 2 × 0.100 + 0.400.

    Q = 1.60 > K. Let x be N₂O₄ formed. (0.400 − 2x)² = 0.200(0.100 + x). Expand and collect: 4x² − 1.800x + 0.140 = 0. Roots x = 0.100 and 0.350; reject the latter because it gives negative NO₂. Final concentrations both 0.200. Check Q = 0.200.

    Q = 1.60 > K.

    Let x be N₂O₄ formed.

    (0.400 − 2x)² = 0.200(0.100 + x).

    Expand and collect: 4x² − 1.800x + 0.140 = 0.

    Roots x = 0.100 and 0.350; reject the latter because it gives negative NO₂.

    Final concentrations both 0.200.

    Check Q = 0.200.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  4. A rapid expansion halves concentrations initially 0.450 N₂O₄ and 0.300 NO₂. Find the new equilibrium at K = 0.200.

    Hint 1

    Identify the requested comparison, calculation or claim.

    Hint 2

    Write the balanced equation and the quotient or conservation relationship relevant to the stated conditions.

    Hint 3

    Carry the coefficient two through the working; check the result against K, the physical bounds and the stated constraint.

    Show the complete working

    Step 1

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 2

    Before: [NO₂] = 0.15
    Action: Square the product concentration first.
    After: [NO₂]² = 0.15² = 0.0225
    Why: Square the whole concentration; do not multiply it by two.

    Step 3

    Before: Product square 0.0225; reactant 0.225
    Action: Divide the product square by reactant concentration.
    After: Q = 0.0225/0.225 = 0.1
    Why: Use the current values to calculate Q.

    Step 4

    Before: Q = 0.1; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is below K: net forward change.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Step 5

    Before: The quotient determines the net reaction direction.
    Action: Define nonnegative x as N₂O₄ consumed.
    After: [N₂O₄] = 0.225 − x; [NO₂] = 0.15 + 2x
    Why: The balanced equation gives one-to-two concentration changes.

    Step 6

    Before: Every final concentration must be nonnegative.
    Action: Set the extent bounds.
    After: 0 ≤ x ≤ 0.225
    Why: Reject roots that violate these bounds or give zero denominator at finite K.

    Step 7

    Before: K = [NO₂]²/[N₂O₄]
    Action: Substitute both equilibrium expressions.
    After: 0.2 = (0.15 + 2x)²/(0.225 − x)
    Why: Square the entire numerator bracket.

    Step 8

    Before: 0.2 = (0.15 + 2x)²/(0.225 − x)
    Action: Multiply both sides by 0.225 − x.
    After: 0.2(0.225 − x) = (0.15 + 2x)²
    Why: The denominator factor reduces to one.

    Step 9

    Before: (0.15 + 2x)²
    Action: Expand the square, including its middle term.
    After: 0.0225 + 0.6x + 4x²
    Why: Use u² ± 2uv + v²; here v is 2x.

    Step 10

    Before: 0.2(0.225 − x)
    Action: Distribute K to both terms.
    After: 0.045 − 0.2x
    Why: Both the constant term and the x term are multiplied by K.

    Step 11

    Before: 0.0225 + 0.6x + 4x² = 0.045 − 0.2x
    Action: Subtract 0.045 from both sides.
    After: −0.0225 + 0.6x + 4x² = − 0.2x
    Why: The matching constant on the right sums to zero.

    Step 12

    Before: The right side contains − 0.2x.
    Action: Add 0.2x on both sides.
    After: 4x² + 0.8x − 0.0225 = 0
    Why: Combine the x coefficients; the opposite right-side x terms sum to zero.

    Step 13

    Before: 4x² + 0.8x − 0.0225 = 0
    Action: Read the quadratic coefficients.
    After: a = 4; b = 0.8; c = −0.0225
    Why: Retain each coefficient sign.

    Step 14

    Before: Discriminant = b² − 4ac
    Action: Substitute and calculate.
    After: (0.8)² − 4 × 4 × (−0.0225) = 1
    Why: Bracket a negative coefficient when substituting.

    Step 15

    Before: x = (−b ± √(b² − 4ac))/(2a)
    Action: Substitute into the quadratic formula.
    After: x = (−(0.8) ± √1)/8
    Why: Calculate both signs; the denominator is two times four.

    Step 16

    Before: Both signs of the formula are required.
    Action: Calculate both numerical roots.
    After: x = 0.025 or −0.225
    Why: Use the physical constraints to choose the valid root.

    Step 17

    Before: 0 ≤ x ≤ 0.225
    Action: Reject x = −0.225.
    After: Use x = 0.025
    Why: The rejected value violates the extent bounds and gives a negative species concentration.

    Step 18

    Before: Use the valid extent in both species.
    Action: Calculate final concentrations.
    After: [N₂O₄] = 0.2; [NO₂] = 0.2
    Why: Keep guard digits while checking the solution.

    Step 19

    Before: Proposed final concentrations are now known.
    Action: Check the quotient and conserved atoms.
    After: Q = 0.2²/0.2 ≈ 0.2; 2[N₂O₄] + [NO₂] = 0.6
    Why: Both the equilibrium ratio and starting atom inventory agree.

    Immediate values 0.225 and 0.150 give Q = 0.100 < K. (0.150 + 2x)² = 0.200(0.225 − x). Expand and collect 4x² + 0.800x − 0.0225 = 0. Roots 0.0250 and −0.225; reject negative forward extent. Final concentrations: N₂O₄ = 0.200, NO₂ = 0.200. Check Q = 0.200.

    Immediate values 0.225 and 0.150 give Q = 0.100 < K.

    (0.150 + 2x)² = 0.200(0.225 − x).

    Expand and collect 4x² + 0.800x − 0.0225 = 0.

    Roots 0.0250 and −0.225; reject negative forward extent.

    Final concentrations: N₂O₄ = 0.200, NO₂ = 0.200.

    Check Q = 0.200.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  5. A suitable catalyst is added to an equilibrium mixture at unchanged temperature. Does equilibrium NO₂ concentration rise? Explain.

    Hint 1

    Identify the requested comparison, calculation or claim.

    Hint 2

    Write the balanced equation and the quotient or conservation relationship relevant to the stated conditions.

    Hint 3

    Carry the coefficient two through the working; check the result against K, the physical bounds and the stated constraint.

    Show the complete working

    No.

    A catalyst increases both reaction rates without changing K or equilibrium composition.

    At equilibrium rates remain equal; no net composition change occurs.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  6. At fixed volume, heating changes K from 0.200 to 0.600 while both concentrations initially remain 0.200. Predict the net change and explain why this differs from compression.

    Hint 1

    Identify the requested comparison, calculation or claim.

    Hint 2

    Write the balanced equation and the quotient or conservation relationship relevant to the stated conditions.

    Hint 3

    Carry the coefficient two through the working; check the result against K, the physical bounds and the stated constraint.

    Show the complete working

    Initial Q remains 0.200 but is below new K = 0.600, so net forward change occurs.

    Heating changed K; compression at fixed temperature changes Q while K stays unchanged.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  7. A proposed final pair 0.200 N₂O₄ and 0.200 NO₂ satisfies K. Could it come from initial 0.500 N₂O₄ and zero NO₂ in a closed fixed-volume vessel?

    Hint 1

    Identify the requested comparison, calculation or claim.

    Hint 2

    Write the balanced equation and the quotient or conservation relationship relevant to the stated conditions.

    Hint 3

    Carry the coefficient two through the working; check the result against K, the physical bounds and the stated constraint.

    Show the complete working

    No.

    Initial atom inventory 2[N₂O₄] + [NO₂] = 1.000.

    Proposed pair gives 2 × 0.200 + 0.200 = 0.600.

    It satisfies K but fails atom conservation.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  8. With original K = 0.200, find constants for the reversed and doubled balanced equations. Explain using exponents.

    Hint 1

    Identify the requested comparison, calculation or claim.

    Hint 2

    Write the balanced equation and the quotient or conservation relationship relevant to the stated conditions.

    Hint 3

    Carry the coefficient two through the working; check the result against K, the physical bounds and the stated constraint.

    Show the complete working

    Step 1

    Before: Original K = [NO₂]²/[N₂O₄]
    Action: Write the expression for the reversed equation.
    After: Reverse constant = [N₂O₄]/[NO₂]²
    Why: The new reactant and product roles swap; their balanced exponents stay attached to each formula.

    Step 2

    Before: [N₂O₄]/[NO₂]²
    Action: Recognise the reciprocal of the original ratio.
    After: Reverse constant = 1/K
    Why: Multiplying the two ratios cancels matching concentration factors and gives one.

    Step 3

    Before: Reverse constant = 1/K
    Action: Substitute and calculate.
    After: 1/0.200 = 5.00
    Why: Check: 5.00 × 0.200 = 1.00.

    Step 4

    Before: Balanced coefficients are now two and four.
    Action: Apply each coefficient as an exponent.
    After: New constant = [NO₂]⁴/[N₂O₄]²
    Why: A coefficient changes an exponent, not a multiplier outside the expression.

    Step 5

    Before: [NO₂]⁴/[N₂O₄]²
    Action: Rewrite as the square of the original ratio.
    After: ([NO₂]²/[N₂O₄])² = K²
    Why: Squaring a quotient squares its numerator and denominator.

    Step 6

    Before: New constant = K²
    Action: Substitute and square the full numerical constant.
    After: 0.200² = 0.0400
    Why: Using the equilibrium pair 0.200 and 0.200: 0.200⁴/0.200² = 0.0400, confirming the new expression.

    Reverse: [N₂O₄]/[NO₂]² = 1/K = 5.00. Doubled: [NO₂]⁴/[N₂O₄]² = K² = 0.0400. Doubling coefficients doubles exponents, so it squares rather than doubles K.

    Reverse: [N₂O₄]/[NO₂]² = 1/K = 5.00.

    Doubled: [NO₂]⁴/[N₂O₄]² = K² = 0.0400.

    Doubling coefficients doubles exponents, so it squares rather than doubles K.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  9. Which statement is justified at fixed temperature?

    • Compression changes Q immediately; K remains unchanged.
    • Compression doubles K.
    • A catalyst increases equilibrium product concentration.
    Hint 1

    Identify the relationship being tested.

    Hint 2

    Use the definitions before comparing the options.

    Hint 3

    Reject options that contradict the relationship; select the remaining option.

    Show the complete working

    K depends on temperature for the stated reaction convention.

    Compression changes concentrations and Q; a catalyst changes rates.

  10. A shortcut that neglects x gives estimated extent 0.0200 from initial reactant 0.200. Assess it using a stated 5% depletion screen and specify the next calculation.

    Hint 1

    Test the neglected term relative to the initial amount.

    Hint 2

    Use extent divided by initial reactant, then multiply by 100%.

    Hint 3

    Compare the computed percentage with the stated screen; identify which term must be restored.

    Show the complete working

    Depletion estimate = 0.0200/0.200 × 100% = 10.0%.

    It fails the stated 5% screen.

    Restore the extent in the denominator, solve the full ICE equation, test both quadratic roots and check the quotient.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

Come back

Return on another day. Try these fresh questions before revealing a hint or answer. Explain what changed in your approach.

  1. Sketch one N₂O₄ forming two NO₂. Count atoms and explain why changing product subscripts is invalid.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: One N₂O₄ molecule.
    Action: Count subscripts.
    After: Two nitrogen atoms and four oxygen atoms.
    Why: A subscript gives atom count in one molecule.
    Coefficients conserve atoms

    Step 2

    Before: One NO₂ contains one nitrogen and two oxygen atoms.
    Action: Use two NO₂ molecules.
    After: Nitrogen: 2 × 1 = 2. Oxygen: 2 × 2 = 4.
    Why: The coefficient multiplies every atom count in the formula; formulas remain unchanged.

    N₂O₄ has two N and four O atoms. Two NO₂ have 2 × 1 = 2 N and 2 × 2 = 4 O. Atom totals match. Changing a subscript changes the substance; coefficient changes preserve the formula.

    N₂O₄ has two N and four O atoms.

    Two NO₂ have 2 × 1 = 2 N and 2 × 2 = 4 O.

    Atom totals match.

    Changing a subscript changes the substance; coefficient changes preserve the formula.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  2. At equilibrium [NO₂] = 0.200 and [N₂O₄] = 0.200. Calculate K and justify the exponent.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 2

    Before: [NO₂] = 0.2
    Action: Square the product concentration first.
    After: [NO₂]² = 0.2² = 0.04
    Why: Square the whole concentration; do not multiply it by two.

    Step 3

    Before: Product square 0.04; reactant 0.2
    Action: Divide the product square by reactant concentration.
    After: Q = 0.04/0.2 = 0.2
    Why: Use the current values to calculate Q.

    Step 4

    Before: Q = 0.2; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is equal to K: dynamic equilibrium.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    K = 0.200²/0.200 = 0.0400/0.200 = 0.200. The NO₂ coefficient two becomes exponent two, not a multiplier outside the square.

    K = 0.200²/0.200 = 0.0400/0.200 = 0.200.

    The NO₂ coefficient two becomes exponent two, not a multiplier outside the square.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  3. A prepared mixture has [NO₂] = 0.200 and [N₂O₄] = 0.300. Predict its net reaction direction. Use supplied K = 0.200 at the stated fixed temperature.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: Current concentrations may not be equilibrium values.
    Action: Use Q to describe the present state.
    After: Q = [NO₂]²/[N₂O₄]
    Why: K is the equilibrium reference at this temperature.

    Step 2

    Before: Q = [NO₂]²/[N₂O₄]
    Action: Square product concentration, then divide.
    After: Q = (0.200)²/0.300 = 0.133
    Why: The coefficient two becomes an exponent.

    Step 3

    Before: Q = 0.133; K = 0.200
    Action: Compare the two values.
    After: Net change is rightwards.
    Why: Q below K needs a larger ratio; Q above K needs a smaller ratio.

    Q = 0.200²/0.300 = 0.133 < 0.200. Net forward change consumes N₂O₄ and forms NO₂ until Q reaches K.

    Q = 0.200²/0.300 = 0.133 < 0.200.

    Net forward change consumes N₂O₄ and forms NO₂ until Q reaches K.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  4. A prepared mixture has [NO₂] = 0.400 and [N₂O₄] = 0.200. Predict changes and express them using a nonnegative extent x. Use supplied K = 0.200 at the stated fixed temperature.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: Current concentrations may not be equilibrium values.
    Action: Use Q to describe the present state.
    After: Q = [NO₂]²/[N₂O₄]
    Why: K is the equilibrium reference at this temperature.

    Step 2

    Before: Q = [NO₂]²/[N₂O₄]
    Action: Square product concentration, then divide.
    After: Q = (0.400)²/0.200 = 0.800
    Why: The coefficient two becomes an exponent.

    Step 3

    Before: Q = 0.800; K = 0.200
    Action: Compare the two values.
    After: Net change is leftwards.
    Why: Q below K needs a larger ratio; Q above K needs a smaller ratio.

    Q = 0.400²/0.200 = 0.800 > K. Net reverse change: Δ[N₂O₄] = +x and Δ[NO₂] = −2x.

    Q = 0.400²/0.200 = 0.800 > K.

    Net reverse change: Δ[N₂O₄] = +x and Δ[NO₂] = −2x.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  5. A learner says the reaction stops when both concentrations are 0.200 because Q equals K. Evaluate.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Q = 0.200²/0.200 = 0.200 = K.

    Equilibrium is dynamic: forward and reverse reactions continue at equal rates.

    Concentrations have no net change.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  6. Initially [N₂O₄] = 0.500 and [NO₂] = 0. Find equilibrium concentrations using the complete quadratic and reject the invalid root. Use supplied K = 0.200 at the stated fixed temperature.

    Hint 1

    Use Q versus K to choose the direction before the ICE table.

    Hint 2

    Write equilibrium concentrations using one extent: reactant changes by x, product by twice x.

    Hint 3

    Multiply the quotient equation by its denominator on both sides, expand and collect the quadratic. Test both roots before reporting concentrations.

    Show the complete working

    Step 1

    Before: Initial Q = 0²/0.500 = 0.
    Action: Compare with K = 0.200.
    After: Q < K, so net forward change.
    Why: Reactant is consumed; product forms.

    Step 2

    Before: Define x as concentration of N₂O₄ consumed.
    Action: Apply the balanced coefficients.
    After: [N₂O₄] = 0.500 − x; [NO₂] = 2x
    Why: One reactant unit gives two product units.
    Initial, change and equilibrium

    Step 3

    Before: Concentrations cannot be negative.
    Action: Set the physical range.
    After: 0 ≤ x ≤ 0.500
    Why: The product concentration is 2x and reactant is 0.500 − x.

    Step 4

    Before: K = [NO₂]²/[N₂O₄]
    Action: Substitute the equilibrium expressions.
    After: 0.200 = (2x)²/(0.500 − x)
    Why: Square the whole product expression, including coefficient two.

    Step 5

    Before: 0.200 = 4x²/(0.500 − x)
    Action: Multiply both sides by 0.500 − x.
    After: 0.200(0.500 − x) = 4x²
    Why: The denominator factor reduces to one; x = 0.500 cannot satisfy finite K.
    Multiply both sides by the denominator

    Step 6

    Before: 0.200(0.500 − x) = 4x²
    Action: Distribute 0.200 across the bracket.
    After: 0.100 − 0.200x = 4x²
    Why: Multiply each term in the bracket.

    Step 7

    Before: 0.100 − 0.200x = 4x²
    Action: Add 0.200x to both sides.
    After: 0.100 = 4x² + 0.200x
    Why: The opposite x terms sum to zero.
    Apply +0.200x to both sides

    Step 8

    Before: 0.100 = 4x² + 0.200x
    Action: Subtract 0.100 from both sides.
    After: 4x² + 0.200x − 0.100 = 0
    Why: Read the same equality with the quadratic on the left.
    Apply −0.100 to both sides

    Step 9

    Before: 4x² + 0.200x − 0.100 = 0
    Action: Identify quadratic coefficients.
    After: a = 4; b = 0.200; c = −0.100
    Why: Include the negative sign in c.

    Step 10

    Before: x = (−b ± √(b² − 4ac))/(2a)
    Action: Substitute all coefficients.
    After: x = (−0.200 ± √(0.0400 + 1.600))/8
    Why: Subtracting the negative product adds 1.600.

    Step 11

    Before: Discriminant = 1.640.
    Action: Calculate both signs.
    After: x = 0.135078 or x = −0.185078
    Why: Keep guard digits until final concentrations.

    Step 12

    Before: Physical range: 0 ≤ x ≤ 0.500.
    Action: Reject the negative root.
    After: x = 0.135078
    Why: A negative extent would produce negative NO₂ from an initial zero amount.

    Step 13

    Before: [N₂O₄] = 0.500 − x; [NO₂] = 2x
    Action: Substitute the valid extent.
    After: [N₂O₄] = 0.364922; [NO₂] = 0.270156
    Why: Report approximately 0.365 and 0.270 mol L⁻¹ to three significant figures.

    Step 14

    Before: Proposed concentrations 0.364922 and 0.270156.
    Action: Check quotient and atoms.
    After: 0.270156²/0.364922 ≈ 0.200; 2 × 0.364922 + 0.270156 = 1.000
    Why: Both the equilibrium ratio and the initial nitrogen-atom inventory agree.

    Let x be N₂O₄ consumed: concentrations 0.500 − x and 2x. Multiply K = 4x²/(0.500 − x) by the denominator: 0.100 − 0.200x = 4x². Add 0.200x and subtract 0.100 on both sides: 4x² + 0.200x − 0.100 = 0. Roots (−0.200 ± √1.640)/8 = 0.135078 or −0.185078. Reject negative x. Final N₂O₄ = 0.364922 and NO₂ = 0.270156. Their quotient is 0.200 and 2[N₂O₄] + [NO₂] = 1.000.

    Let x be N₂O₄ consumed: concentrations 0.500 − x and 2x.

    Multiply K = 4x²/(0.500 − x) by the denominator: 0.100 − 0.200x = 4x².

    Add 0.200x and subtract 0.100 on both sides: 4x² + 0.200x − 0.100 = 0.

    Roots (−0.200 ± √1.640)/8 = 0.135078 or −0.185078.

    Reject negative x.

    Final N₂O₄ = 0.364922 and NO₂ = 0.270156.

    Their quotient is 0.200 and 2[N₂O₄] + [NO₂] = 1.000.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  7. Initially [N₂O₄] = 0.100 and [NO₂] = 0.400. Find equilibrium concentrations and reject the invalid root. Use supplied K = 0.200 at the stated fixed temperature.

    Hint 1

    Use Q versus K to choose the direction before the ICE table.

    Hint 2

    Write equilibrium concentrations using one extent: reactant changes by x, product by twice x.

    Hint 3

    Multiply the quotient equation by its denominator on both sides, expand and collect the quadratic. Test both roots before reporting concentrations.

    Show the complete working

    Step 1

    Before: Q = 0.400²/0.100 = 1.60.
    Action: Compare with K.
    After: Q > 0.200; net reverse change.
    Why: Define x as N₂O₄ formed.

    Step 2

    Before: One N₂O₄ forms from two NO₂.
    Action: Write reverse changes.
    After: [N₂O₄] = 0.100 + x; [NO₂] = 0.400 − 2x
    Why: The range is 0 ≤ x ≤ 0.200.

    Step 3

    Before: K = [NO₂]²/[N₂O₄]
    Action: Substitute and multiply both sides by 0.100 + x.
    After: (0.400 − 2x)² = 0.200(0.100 + x)
    Why: The whole squared bracket must be expanded.

    Step 4

    Before: (u − v)² = u² − 2uv + v²
    Action: Expand both sides.
    After: 0.160 − 1.600x + 4x² = 0.0200 + 0.200x
    Why: The middle term includes twice 0.400 times 2x.
    Expand all three square terms

    Step 5

    Before: 0.160 − 1.600x + 4x² = 0.0200 + 0.200x
    Action: Subtract 0.0200 and 0.200x from both sides.
    After: 4x² − 1.800x + 0.140 = 0
    Why: Equal operations preserve the equation.

    Step 6

    Before: a = 4; b = −1.800; c = 0.140
    Action: Use both signs of the quadratic formula.
    After: x = (1.800 ± √(3.240 − 2.240))/8; x = 0.100 or 0.350
    Why: The discriminant is 1.000.

    Step 7

    Before: x must be at most 0.200.
    Action: Reject x = 0.350.
    After: x = 0.100
    Why: The other root gives [NO₂] = 0.400 − 0.700 = −0.300, impossible.

    Step 8

    Before: x = 0.100
    Action: Calculate and check both concentrations.
    After: [N₂O₄] = 0.200; [NO₂] = 0.200; Q = 0.200
    Why: 2 × 0.200 + 0.200 = 0.600, matching 2 × 0.100 + 0.400.

    Q = 1.60 > K. Let x be N₂O₄ formed. (0.400 − 2x)² = 0.200(0.100 + x). Expand: 0.160 − 1.600x + 4x² = 0.0200 + 0.200x. Subtract the right expression on both sides: 4x² − 1.800x + 0.140 = 0. Roots (1.800 ± 1.000)/8 = 0.100 or 0.350. Reject 0.350, giving negative NO₂. Final concentrations both 0.200; quotient 0.200.

    Q = 1.60 > K.

    Let x be N₂O₄ formed.

    (0.400 − 2x)² = 0.200(0.100 + x).

    Expand: 0.160 − 1.600x + 4x² = 0.0200 + 0.200x.

    Subtract the right expression on both sides: 4x² − 1.800x + 0.140 = 0.

    Roots (1.800 ± 1.000)/8 = 0.100 or 0.350.

    Reject 0.350, giving negative NO₂.

    Final concentrations both 0.200; quotient 0.200.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  8. For initial [N₂O₄] = 0.500 and zero NO₂, test neglecting x when K = 0.200 using a stated 5% depletion screen.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: K = 4x²/(0.500 − x)
    Action: Assume x is small relative to 0.500.
    After: K ≈ 4x²/0.500
    Why: This is a hypothesis to test, not an established equality.

    Step 2

    Before: K ≈ 4x²/0.500
    Action: Multiply by 0.500, divide by four, then take the nonnegative square root.
    After: x ≈ √(0.2 × 0.500/4) = 0.158114
    Why: The defined forward extent is nonnegative.

    Step 3

    Before: x estimate 0.158114.
    Action: Calculate the estimated depletion fraction.
    After: x/0.500 × 100% = 31.62%
    Why: Compare this with the stated 5% screen.

    Step 4

    Before: Depletion 31.62%.
    Action: Decide whether the shortcut is suitable.
    After: The screen fails; use the full quadratic.
    Why: The full solution gives x = 0.13507811.

    K ≈ 4x²/0.500 gives x ≈ √(0.200 × 0.500/4) = 0.158114. Depletion = 0.158114/0.500 × 100% = 31.6%. The screen fails; use the complete quadratic. A small K alone is insufficient justification.

    K ≈ 4x²/0.500 gives x ≈ √(0.200 × 0.500/4) = 0.158114.

    Depletion = 0.158114/0.500 × 100% = 31.6%.

    The screen fails; use the complete quadratic.

    A small K alone is insufficient justification.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  9. At another supplied temperature K = 0.000200. Initial N₂O₄ is 0.500 and NO₂ zero. Test the same approximation and compare with the complete solution.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: K = 4x²/(0.500 − x)
    Action: Assume x is small relative to 0.500.
    After: K ≈ 4x²/0.500
    Why: This is a hypothesis to test, not an established equality.

    Step 2

    Before: K ≈ 4x²/0.500
    Action: Multiply by 0.500, divide by four, then take the nonnegative square root.
    After: x ≈ √(0.0002 × 0.500/4) = 0.005000
    Why: The defined forward extent is nonnegative.

    Step 3

    Before: x estimate 0.005000.
    Action: Calculate the estimated depletion fraction.
    After: x/0.500 × 100% = 1.00%
    Why: Compare this with the stated 5% screen.

    Step 4

    Before: Depletion 1.00%.
    Action: Decide whether the shortcut is suitable.
    After: The screen passes; assess the required numerical precision too.
    Why: The full solution gives x = 0.00497506.

    Step 5

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 6

    Before: [NO₂] = 0
    Action: Square the product concentration first.
    After: [NO₂]² = 0² = 0
    Why: Square the whole concentration; do not multiply it by two.

    Step 7

    Before: Product square 0; reactant 0.5
    Action: Divide the product square by reactant concentration.
    After: Q = 0/0.5 = 0
    Why: Use the current values to calculate Q.

    Step 8

    Before: Q = 0; K = 0.0002
    Action: Compare Q with the reference at this temperature.
    After: Q is below K: net forward change.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Step 9

    Before: The quotient determines the net reaction direction.
    Action: Define nonnegative x as N₂O₄ consumed.
    After: [N₂O₄] = 0.5 − x; [NO₂] = 0 + 2x
    Why: The balanced equation gives one-to-two concentration changes.

    Step 10

    Before: Every final concentration must be nonnegative.
    Action: Set the extent bounds.
    After: 0 ≤ x ≤ 0.5
    Why: Reject roots that violate these bounds or give zero denominator at finite K.

    Step 11

    Before: K = [NO₂]²/[N₂O₄]
    Action: Substitute both equilibrium expressions.
    After: 0.0002 = (0 + 2x)²/(0.5 − x)
    Why: Square the entire numerator bracket.

    Step 12

    Before: 0.0002 = (0 + 2x)²/(0.5 − x)
    Action: Multiply both sides by 0.5 − x.
    After: 0.0002(0.5 − x) = (0 + 2x)²
    Why: The denominator factor reduces to one.

    Step 13

    Before: (0 + 2x)²
    Action: Expand the square, including its middle term.
    After: 0 + 0x + 4x²
    Why: Use u² ± 2uv + v²; here v is 2x.

    Step 14

    Before: 0.0002(0.5 − x)
    Action: Distribute K to both terms.
    After: 0.0001 − 0.0002x
    Why: Both the constant term and the x term are multiplied by K.

    Step 15

    Before: 0 + 0x + 4x² = 0.0001 − 0.0002x
    Action: Subtract 0.0001 from both sides.
    After: −0.0001 + 0x + 4x² = − 0.0002x
    Why: The matching constant on the right sums to zero.

    Step 16

    Before: The right side contains − 0.0002x.
    Action: Add 0.0002x on both sides.
    After: 4x² + 0.0002x − 0.0001 = 0
    Why: Combine the x coefficients; the opposite right-side x terms sum to zero.

    Step 17

    Before: 4x² + 0.0002x − 0.0001 = 0
    Action: Read the quadratic coefficients.
    After: a = 4; b = 0.0002; c = −0.0001
    Why: Retain each coefficient sign.

    Step 18

    Before: Discriminant = b² − 4ac
    Action: Substitute and calculate.
    After: (0.0002)² − 4 × 4 × (−0.0001) = 0.00160004
    Why: Bracket a negative coefficient when substituting.

    Step 19

    Before: x = (−b ± √(b² − 4ac))/(2a)
    Action: Substitute into the quadratic formula.
    After: x = (−(0.0002) ± √0.00160004)/8
    Why: Calculate both signs; the denominator is two times four.

    Step 20

    Before: Both signs of the formula are required.
    Action: Calculate both numerical roots.
    After: x = 0.00497506 or −0.00502506
    Why: Use the physical constraints to choose the valid root.

    Step 21

    Before: 0 ≤ x ≤ 0.5
    Action: Reject x = −0.00502506.
    After: Use x = 0.00497506
    Why: The rejected value violates the extent bounds and gives a negative species concentration.

    Step 22

    Before: Use the valid extent in both species.
    Action: Calculate final concentrations.
    After: [N₂O₄] = 0.495025; [NO₂] = 0.00995012
    Why: Keep guard digits while checking the solution.

    Step 23

    Before: Proposed final concentrations are now known.
    Action: Check the quotient and conserved atoms.
    After: Q = 0.00995012²/0.495025 ≈ 0.0002; 2[N₂O₄] + [NO₂] = 1
    Why: Both the equilibrium ratio and starting atom inventory agree.

    x ≈ √(0.000200 × 0.500/4) = 0.00500. Depletion estimate 1.00% passes the stated screen. Full quadratic 4x² + 0.000200x − 0.000100 = 0 gives positive x = 0.00497506. Final N₂O₄ = 0.495025 and NO₂ = 0.00995012. Assess whether approximate accuracy suits the purpose.

    x ≈ √(0.000200 × 0.500/4) = 0.00500.

    Depletion estimate 1.00% passes the stated screen.

    Full quadratic 4x² + 0.000200x − 0.000100 = 0 gives positive x = 0.00497506.

    Final N₂O₄ = 0.495025 and NO₂ = 0.00995012.

    Assess whether approximate accuracy suits the purpose.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  10. A piston halves volume at fixed temperature from equilibrium concentrations both 0.200. Find immediate and final concentrations.

    Hint 1

    Use Q versus K to choose the direction before the ICE table.

    Hint 2

    Write equilibrium concentrations using one extent: reactant changes by x, product by twice x.

    Hint 3

    Multiply the quotient equation by its denominator on both sides, expand and collect the quadratic. Test both roots before reporting concentrations.

    Show the complete working

    Step 1

    Before: Amount is unchanged during rapid compression.
    Action: Divide volume by two.
    After: Both concentrations immediately become 0.400
    Why: Concentration equals amount divided by volume.

    Step 2

    Before: Q = 0.400²/0.400
    Action: Calculate and compare.
    After: Q = 0.400 > K = 0.200
    Why: K remains unchanged because temperature is unchanged.

    Step 3

    Before: Net change is reverse.
    Action: Use +x and −2x.
    After: [N₂O₄] = 0.400 + x; [NO₂] = 0.400 − 2x
    Why: 0 ≤ x ≤ 0.200.

    Step 4

    Before: 0.200 = (0.400 − 2x)²/(0.400 + x)
    Action: Multiply by the denominator and expand.
    After: 0.160 − 1.600x + 4x² = 0.0800 + 0.200x
    Why: The middle term of the square remains essential.

    Step 5

    Before: 0.160 − 1.600x + 4x² = 0.0800 + 0.200x
    Action: Subtract the right expression from both sides.
    After: 4x² − 1.800x + 0.0800 = 0
    Why: Both sides undergo the same operations.

    Step 6

    Before: a = 4; b = −1.800; c = 0.0800
    Action: Use the quadratic formula.
    After: x = (1.800 ± √(3.240 − 1.280))/8 = 0.0500 or 0.400
    Why: The second root exceeds 0.200 and gives negative NO₂.

    Step 7

    Before: Valid x = 0.0500
    Action: Substitute and check.
    After: [N₂O₄] = 0.450; [NO₂] = 0.300; Q = 0.300²/0.450 = 0.200
    Why: The new NO₂ concentration exceeds its original 0.200 even though reaction adjusted leftwards.

    Immediate concentrations both 0.400. Q = 0.400 > K, so let x be N₂O₄ formed. (0.400 − 2x)² = 0.200(0.400 + x). Collect 4x² − 1.800x + 0.0800 = 0. Roots 0.0500 and 0.400; reject the latter because NO₂ becomes negative. Final N₂O₄ = 0.450 and NO₂ = 0.300. Check 0.300²/0.450 = 0.200. K is unchanged.

    Immediate concentrations both 0.400.

    Q = 0.400 > K, so let x be N₂O₄ formed.

    (0.400 − 2x)² = 0.200(0.400 + x).

    Collect 4x² − 1.800x + 0.0800 = 0.

    Roots 0.0500 and 0.400; reject the latter because NO₂ becomes negative.

    Final N₂O₄ = 0.450 and NO₂ = 0.300.

    Check 0.300²/0.450 = 0.200.

    K is unchanged.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  11. A piston doubles volume from equilibrium N₂O₄ 0.450 and NO₂ 0.300 at fixed temperature. Find immediate and new equilibrium values.

    Hint 1

    Use Q versus K to choose the direction before the ICE table.

    Hint 2

    Write equilibrium concentrations using one extent: reactant changes by x, product by twice x.

    Hint 3

    Multiply the quotient equation by its denominator on both sides, expand and collect the quadratic. Test both roots before reporting concentrations.

    Show the complete working

    Step 1

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 2

    Before: [NO₂] = 0.15
    Action: Square the product concentration first.
    After: [NO₂]² = 0.15² = 0.0225
    Why: Square the whole concentration; do not multiply it by two.

    Step 3

    Before: Product square 0.0225; reactant 0.225
    Action: Divide the product square by reactant concentration.
    After: Q = 0.0225/0.225 = 0.1
    Why: Use the current values to calculate Q.

    Step 4

    Before: Q = 0.1; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is below K: net forward change.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Step 5

    Before: The quotient determines the net reaction direction.
    Action: Define nonnegative x as N₂O₄ consumed.
    After: [N₂O₄] = 0.225 − x; [NO₂] = 0.15 + 2x
    Why: The balanced equation gives one-to-two concentration changes.

    Step 6

    Before: Every final concentration must be nonnegative.
    Action: Set the extent bounds.
    After: 0 ≤ x ≤ 0.225
    Why: Reject roots that violate these bounds or give zero denominator at finite K.

    Step 7

    Before: K = [NO₂]²/[N₂O₄]
    Action: Substitute both equilibrium expressions.
    After: 0.2 = (0.15 + 2x)²/(0.225 − x)
    Why: Square the entire numerator bracket.

    Step 8

    Before: 0.2 = (0.15 + 2x)²/(0.225 − x)
    Action: Multiply both sides by 0.225 − x.
    After: 0.2(0.225 − x) = (0.15 + 2x)²
    Why: The denominator factor reduces to one.

    Step 9

    Before: (0.15 + 2x)²
    Action: Expand the square, including its middle term.
    After: 0.0225 + 0.6x + 4x²
    Why: Use u² ± 2uv + v²; here v is 2x.

    Step 10

    Before: 0.2(0.225 − x)
    Action: Distribute K to both terms.
    After: 0.045 − 0.2x
    Why: Both the constant term and the x term are multiplied by K.

    Step 11

    Before: 0.0225 + 0.6x + 4x² = 0.045 − 0.2x
    Action: Subtract 0.045 from both sides.
    After: −0.0225 + 0.6x + 4x² = − 0.2x
    Why: The matching constant on the right sums to zero.

    Step 12

    Before: The right side contains − 0.2x.
    Action: Add 0.2x on both sides.
    After: 4x² + 0.8x − 0.0225 = 0
    Why: Combine the x coefficients; the opposite right-side x terms sum to zero.

    Step 13

    Before: 4x² + 0.8x − 0.0225 = 0
    Action: Read the quadratic coefficients.
    After: a = 4; b = 0.8; c = −0.0225
    Why: Retain each coefficient sign.

    Step 14

    Before: Discriminant = b² − 4ac
    Action: Substitute and calculate.
    After: (0.8)² − 4 × 4 × (−0.0225) = 1
    Why: Bracket a negative coefficient when substituting.

    Step 15

    Before: x = (−b ± √(b² − 4ac))/(2a)
    Action: Substitute into the quadratic formula.
    After: x = (−(0.8) ± √1)/8
    Why: Calculate both signs; the denominator is two times four.

    Step 16

    Before: Both signs of the formula are required.
    Action: Calculate both numerical roots.
    After: x = 0.025 or −0.225
    Why: Use the physical constraints to choose the valid root.

    Step 17

    Before: 0 ≤ x ≤ 0.225
    Action: Reject x = −0.225.
    After: Use x = 0.025
    Why: The rejected value violates the extent bounds and gives a negative species concentration.

    Step 18

    Before: Use the valid extent in both species.
    Action: Calculate final concentrations.
    After: [N₂O₄] = 0.2; [NO₂] = 0.2
    Why: Keep guard digits while checking the solution.

    Step 19

    Before: Proposed final concentrations are now known.
    Action: Check the quotient and conserved atoms.
    After: Q = 0.2²/0.2 ≈ 0.2; 2[N₂O₄] + [NO₂] = 0.6
    Why: Both the equilibrium ratio and starting atom inventory agree.

    Immediate values 0.225 and 0.150 give Q = 0.100 < K. Forward changes −x and +2x give (0.150 + 2x)² = 0.200(0.225 − x). Collect 4x² + 0.800x − 0.0225 = 0. Roots 0.0250 and −0.225; reject negative forward extent. Final concentrations both 0.200.

    Immediate values 0.225 and 0.150 give Q = 0.100 < K.

    Forward changes −x and +2x give (0.150 + 2x)² = 0.200(0.225 − x).

    Collect 4x² + 0.800x − 0.0225 = 0.

    Roots 0.0250 and −0.225; reject negative forward extent.

    Final concentrations both 0.200.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  12. At fixed volume and temperature, a feed raises NO₂ from 0.200 to 0.400 while N₂O₄ stays 0.200. Must all added product disappear? Calculate the final composition. Use supplied K = 0.200 at the stated fixed temperature.

    Hint 1

    Use Q versus K to choose the direction before the ICE table.

    Hint 2

    Write equilibrium concentrations using one extent: reactant changes by x, product by twice x.

    Hint 3

    Multiply the quotient equation by its denominator on both sides, expand and collect the quadratic. Test both roots before reporting concentrations.

    Show the complete working

    Step 1

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 2

    Before: [NO₂] = 0.4
    Action: Square the product concentration first.
    After: [NO₂]² = 0.4² = 0.16
    Why: Square the whole concentration; do not multiply it by two.

    Step 3

    Before: Product square 0.16; reactant 0.2
    Action: Divide the product square by reactant concentration.
    After: Q = 0.16/0.2 = 0.8
    Why: Use the current values to calculate Q.

    Step 4

    Before: Q = 0.8; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is above K: net reverse change.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Step 5

    Before: The quotient determines the net reaction direction.
    Action: Define nonnegative x as N₂O₄ formed.
    After: [N₂O₄] = 0.2 + x; [NO₂] = 0.4 − 2x
    Why: The balanced equation gives one-to-two concentration changes.

    Step 6

    Before: Every final concentration must be nonnegative.
    Action: Set the extent bounds.
    After: 0 ≤ x ≤ 0.2
    Why: Reject roots that violate these bounds or give zero denominator at finite K.

    Step 7

    Before: K = [NO₂]²/[N₂O₄]
    Action: Substitute both equilibrium expressions.
    After: 0.2 = (0.4 − 2x)²/(0.2 + x)
    Why: Square the entire numerator bracket.

    Step 8

    Before: 0.2 = (0.4 − 2x)²/(0.2 + x)
    Action: Multiply both sides by 0.2 + x.
    After: 0.2(0.2 + x) = (0.4 − 2x)²
    Why: The denominator factor reduces to one.

    Step 9

    Before: (0.4 − 2x)²
    Action: Expand the square, including its middle term.
    After: 0.16 − 1.6x + 4x²
    Why: Use u² ± 2uv + v²; here v is 2x.

    Step 10

    Before: 0.2(0.2 + x)
    Action: Distribute K to both terms.
    After: 0.04 + 0.2x
    Why: Both the constant term and the x term are multiplied by K.

    Step 11

    Before: 0.16 − 1.6x + 4x² = 0.04 + 0.2x
    Action: Subtract 0.04 from both sides.
    After: 0.12 − 1.6x + 4x² = 0.2x
    Why: The matching constant on the right sums to zero.

    Step 12

    Before: The right side contains + 0.2x.
    Action: Subtract 0.2x on both sides.
    After: 4x² − 1.8x + 0.12 = 0
    Why: Combine the x coefficients; the opposite right-side x terms sum to zero.

    Step 13

    Before: 4x² − 1.8x + 0.12 = 0
    Action: Read the quadratic coefficients.
    After: a = 4; b = −1.8; c = 0.12
    Why: Retain each coefficient sign.

    Step 14

    Before: Discriminant = b² − 4ac
    Action: Substitute and calculate.
    After: (−1.8)² − 4 × 4 × (0.12) = 1.32
    Why: Bracket a negative coefficient when substituting.

    Step 15

    Before: x = (−b ± √(b² − 4ac))/(2a)
    Action: Substitute into the quadratic formula.
    After: x = (−(−1.8) ± √1.32)/8
    Why: Calculate both signs; the denominator is two times four.

    Step 16

    Before: Both signs of the formula are required.
    Action: Calculate both numerical roots.
    After: x = 0.368614 or 0.0813859
    Why: Use the physical constraints to choose the valid root.

    Step 17

    Before: 0 ≤ x ≤ 0.2
    Action: Reject x = 0.368614.
    After: Use x = 0.0813859
    Why: The rejected value violates the extent bounds and gives a negative species concentration.

    Step 18

    Before: Use the valid extent in both species.
    Action: Calculate final concentrations.
    After: [N₂O₄] = 0.281386; [NO₂] = 0.237228
    Why: Keep guard digits while checking the solution.

    Step 19

    Before: Proposed final concentrations are now known.
    Action: Check the quotient and conserved atoms.
    After: Q = 0.237228²/0.281386 ≈ 0.2; 2[N₂O₄] + [NO₂] = 0.8
    Why: Both the equilibrium ratio and starting atom inventory agree.

    Q = 0.400²/0.200 = 0.800 > K. Let x be N₂O₄ formed. (0.400 − 2x)² = 0.200(0.200 + x). Expand and collect 4x² − 1.800x + 0.120 = 0. Roots (1.800 ± √1.320)/8 = 0.0813859 or 0.368614; reject the latter because NO₂ becomes negative. Final N₂O₄ = 0.281386 and NO₂ = 0.237228. Added product is partly consumed; the original concentration need not return.

    Q = 0.400²/0.200 = 0.800 > K.

    Let x be N₂O₄ formed.

    (0.400 − 2x)² = 0.200(0.200 + x).

    Expand and collect 4x² − 1.800x + 0.120 = 0.

    Roots (1.800 ± √1.320)/8 = 0.0813859 or 0.368614; reject the latter because NO₂ becomes negative.

    Final N₂O₄ = 0.281386 and NO₂ = 0.237228.

    Added product is partly consumed; the original concentration need not return.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  13. A separation step halves N₂O₄ from 0.200 to 0.100 while NO₂ remains 0.200 at fixed volume and temperature. Predict adjustment; evaluate the rule that removing reactant always favours forward reaction. Use supplied K = 0.200 at the stated fixed temperature.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 2

    Before: [NO₂] = 0.2
    Action: Square the product concentration first.
    After: [NO₂]² = 0.2² = 0.04
    Why: Square the whole concentration; do not multiply it by two.

    Step 3

    Before: Product square 0.04; reactant 0.1
    Action: Divide the product square by reactant concentration.
    After: Q = 0.04/0.1 = 0.4
    Why: Use the current values to calculate Q.

    Step 4

    Before: Q = 0.4; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is above K: net reverse change.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Q = 0.200²/0.100 = 0.400 > K. Net reverse change consumes NO₂ and forms N₂O₄. Removing this reactant raised the quotient; the stated rule is false.

    Q = 0.200²/0.100 = 0.400 > K.

    Net reverse change consumes NO₂ and forms N₂O₄.

    Removing this reactant raised the quotient; the stated rule is false.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  14. Compare inert-gas addition at constant volume and at constant total pressure, each at fixed temperature.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    At fixed volume, reacting amounts and concentrations stay unchanged, so Q stays equal to K; no shift.

    At constant total pressure, a movable piston expands volume.

    If each reacting concentration is scaled by f < 1, new Q = f²[NO₂]²/(f[N₂O₄]) = fK < K.

    Net forward change occurs in this ideal-gas model.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  15. At fixed volume, heating changes supplied K from 0.200 to 0.600 from equilibrium concentrations both 0.200. Find the new equilibrium.

    Hint 1

    Use Q versus K to choose the direction before the ICE table.

    Hint 2

    Write equilibrium concentrations using one extent: reactant changes by x, product by twice x.

    Hint 3

    Multiply the quotient equation by its denominator on both sides, expand and collect the quadratic. Test both roots before reporting concentrations.

    Show the complete working

    Step 1

    Before: The volume and amounts have not yet changed.
    Action: Compare current Q with the new K.
    After: Q = 0.200 < new K = 0.600
    Why: Temperature changed K; reaction then changes concentrations.

    Step 2

    Before: Net change is forward.
    Action: Use −x and +2x.
    After: [N₂O₄] = 0.200 − x; [NO₂] = 0.200 + 2x
    Why: 0 ≤ x ≤ 0.200.

    Step 3

    Before: 0.600 = (0.200 + 2x)²/(0.200 − x)
    Action: Multiply by the denominator and expand.
    After: 0.120 − 0.600x = 0.0400 + 0.800x + 4x²
    Why: The square has three terms.

    Step 4

    Before: 0.120 − 0.600x = 0.0400 + 0.800x + 4x²
    Action: Add 0.600x and subtract 0.120 on both sides.
    After: 4x² + 1.400x − 0.0800 = 0
    Why: Equal operations collect the quadratic.

    Step 5

    Before: a = 4; b = 1.400; c = −0.0800
    Action: Calculate both roots.
    After: x = (−1.400 ± √(1.960 + 1.280))/8 = 0.0500 or −0.400
    Why: Reject the negative forward extent.

    Step 6

    Before: x = 0.0500
    Action: Substitute and check.
    After: [N₂O₄] = 0.150; [NO₂] = 0.300; Q = 0.300²/0.150 = 0.600
    Why: The new concentrations satisfy the new equilibrium constant.

    Immediately Q = 0.200 < new K = 0.600. Let x be reactant consumed. (0.200 + 2x)² = 0.600(0.200 − x). Collect 4x² + 1.400x − 0.0800 = 0. Roots 0.0500 and −0.400; reject negative extent. Final N₂O₄ = 0.150 and NO₂ = 0.300. Check 0.300²/0.150 = 0.600.

    Immediately Q = 0.200 < new K = 0.600.

    Let x be reactant consumed.

    (0.200 + 2x)² = 0.600(0.200 − x).

    Collect 4x² + 1.400x − 0.0800 = 0.

    Roots 0.0500 and −0.400; reject negative extent.

    Final N₂O₄ = 0.150 and NO₂ = 0.300.

    Check 0.300²/0.150 = 0.600.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  16. An engineer adds a suitable catalyst to improve equilibrium NO₂ yield at unchanged temperature. Evaluate.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    The catalyst changes rates, not K or equilibrium composition.

    A nonequilibrium mixture can approach equilibrium faster.

    At equilibrium both rates increase while remaining equal.

    Equilibrium product yield is unchanged under otherwise identical conditions.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  17. After compression, sensor values are: before (0.200, 0.200), immediate (0.400, 0.400), final (0.450, 0.300), ordered N₂O₄ then NO₂. Describe the curves and check the adjustment ratio.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Both curves jump upward when volume halves.

    During chemical adjustment N₂O₄ rises gradually by 0.0500 while NO₂ falls by 0.100.

    Change ratio is +x and −2x.

    Final plateaus show no net concentration change, not stopped reactions.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  18. At fixed calibrated optical path, absorbance is proportional to NO₂ concentration. It changes 0.200 to 0.400 immediately after compression, then to 0.300. Does the leftward adjustment make absorbance lower than originally?

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: Initial NO₂ concentration 0.200.
    Action: Compare immediate 0.400 with the initial value.
    After: 0.400/0.200 = 2.00: absorbance doubles.
    Why: The optical path and calibration are explicitly fixed.

    Step 2

    Before: Final NO₂ concentration 0.300.
    Action: Compare final concentration with the initial value.
    After: 0.300/0.200 = 1.50: final absorbance is 1.5 times original.
    Why: It decreased relative to the immediately compressed state but remains above the original.

    Step 3

    Before: A leftward chemical adjustment occurred.
    Action: State the reference state when comparing colour.
    After: Leftward adjustment does not imply a lower concentration than before compression.
    Why: Volume changed the concentration before reaction adjustment.

    Absorbance immediately doubles, then decreases to 1.5 times its original value. The leftward reaction reduces NO₂ relative to the compressed state, but final concentration exceeds the original. Do not confuse the reference states or assume an unchanged optical path in an uncalibrated vessel.

    Absorbance immediately doubles, then decreases to 1.5 times its original value.

    The leftward reaction reduces NO₂ relative to the compressed state, but final concentration exceeds the original.

    Do not confuse the reference states or assume an unchanged optical path in an uncalibrated vessel.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  19. From initial N₂O₄ 0.500 and NO₂ zero in a closed fixed-volume vessel, audit proposed final pairs A (0.200, 0.200), B (0.400, 0.200), C (0.364922, 0.270156). Check K and atoms independently. Use supplied K = 0.200 at the stated fixed temperature.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: Initial N₂O₄ 0.500; NO₂ zero.
    Action: Count conserved nitrogen atoms per litre.
    After: 2 × 0.500 + 0 = 1.000
    Why: At fixed volume the atom inventory concentration stays constant.

    Step 2

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 3

    Before: [NO₂] = 0.2
    Action: Square the product concentration first.
    After: [NO₂]² = 0.2² = 0.04
    Why: Square the whole concentration; do not multiply it by two.

    Step 4

    Before: Product square 0.04; reactant 0.2
    Action: Divide the product square by reactant concentration.
    After: Q = 0.04/0.2 = 0.2
    Why: Use the current values to calculate Q.

    Step 5

    Before: Q = 0.2; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is equal to K: dynamic equilibrium.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Step 6

    Before: Report A: 0.200 N₂O₄; 0.200 NO₂.
    Action: Check inventory independently.
    After: 2 × 0.200 + 0.200 = 0.600 ≠ 1.000
    Why: A satisfies the equilibrium ratio but cannot conserve the original atoms.

    Step 7

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 8

    Before: [NO₂] = 0.2
    Action: Square the product concentration first.
    After: [NO₂]² = 0.2² = 0.04
    Why: Square the whole concentration; do not multiply it by two.

    Step 9

    Before: Product square 0.04; reactant 0.4
    Action: Divide the product square by reactant concentration.
    After: Q = 0.04/0.4 = 0.1
    Why: Use the current values to calculate Q.

    Step 10

    Before: Q = 0.1; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is below K: net forward change.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Step 11

    Before: Report B: 0.400 N₂O₄; 0.200 NO₂.
    Action: Check inventory independently.
    After: 2 × 0.400 + 0.200 = 1.000
    Why: B conserves atoms but has Q = 0.100 rather than K.

    Step 12

    Before: N₂O₄(g) ⇌ 2NO₂(g)
    Action: Write the current concentration quotient.
    After: Q = [NO₂]²/[N₂O₄]
    Why: The coefficient two becomes an exponent.

    Step 13

    Before: [NO₂] = 0.270156
    Action: Square the product concentration first.
    After: [NO₂]² = 0.270156² = 0.0729844
    Why: Square the whole concentration; do not multiply it by two.

    Step 14

    Before: Product square 0.0729844; reactant 0.364922
    Action: Divide the product square by reactant concentration.
    After: Q = 0.0729844/0.364922 = 0.2
    Why: Use the current values to calculate Q.

    Step 15

    Before: Q = 0.2; K = 0.2
    Action: Compare Q with the reference at this temperature.
    After: Q is equal to K: dynamic equilibrium.
    Why: At equilibrium the two reaction rates are equal; otherwise reaction changes the ratio towards K.

    Step 16

    Before: Report C retains the valid ICE concentrations.
    Action: Check the inventory with guard digits.
    After: 2 × 0.364922 + 0.270156 = 1.000000
    Why: Only C matches both the starting atom inventory and the supplied equilibrium condition.

    Initial nitrogen inventory: 2[N₂O₄] + [NO₂] = 1.000. A: Q = 0.200 but inventory 0.600, so impossible from this start. B: inventory 1.000 but Q = 0.100, so not at the supplied equilibrium. C: Q ≈ 0.200 and inventory 1.000000; both conditions agree. Only C passes both checks.

    Initial nitrogen inventory: 2[N₂O₄] + [NO₂] = 1.000.

    A: Q = 0.200 but inventory 0.600, so impossible from this start.

    B: inventory 1.000 but Q = 0.100, so not at the supplied equilibrium.

    C: Q ≈ 0.200 and inventory 1.000000; both conditions agree.

    Only C passes both checks.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.

  20. With original K = 0.200, derive constants for 2NO₂ ⇌ N₂O₄ and 2N₂O₄ ⇌ 4NO₂. Explain why doubled coefficients do not give 2K.

    Hint 1

    Identify whether the task concerns atoms, the quotient, an extent or changed conditions.

    Hint 2

    Write N₂O₄(g) ⇌ 2NO₂(g), then the relevant ratio or signed concentration changes.

    Hint 3

    Keep the coefficient two in powers and extents. Check bounds, conservation and Q against the appropriate temperature-dependent K.

    Show the complete working

    Step 1

    Before: Original K = [NO₂]²/[N₂O₄]
    Action: Write the expression for the reversed equation.
    After: Reverse constant = [N₂O₄]/[NO₂]²
    Why: The new reactant and product roles swap; their balanced exponents stay attached to each formula.

    Step 2

    Before: [N₂O₄]/[NO₂]²
    Action: Recognise the reciprocal of the original ratio.
    After: Reverse constant = 1/K
    Why: Multiplying the two ratios cancels matching concentration factors and gives one.

    Step 3

    Before: Reverse constant = 1/K
    Action: Substitute and calculate.
    After: 1/0.200 = 5.00
    Why: Check: 5.00 × 0.200 = 1.00.

    Step 4

    Before: Balanced coefficients are now two and four.
    Action: Apply each coefficient as an exponent.
    After: New constant = [NO₂]⁴/[N₂O₄]²
    Why: A coefficient changes an exponent, not a multiplier outside the expression.

    Step 5

    Before: [NO₂]⁴/[N₂O₄]²
    Action: Rewrite as the square of the original ratio.
    After: ([NO₂]²/[N₂O₄])² = K²
    Why: Squaring a quotient squares its numerator and denominator.

    Step 6

    Before: New constant = K²
    Action: Substitute and square the full numerical constant.
    After: 0.200² = 0.0400
    Why: Using the equilibrium pair 0.200 and 0.200: 0.200⁴/0.200² = 0.0400, confirming the new expression.

    Reverse expression [N₂O₄]/[NO₂]² = 1/K, so constant 5.00. Doubled expression [NO₂]⁴/[N₂O₄]² = ([NO₂]²/[N₂O₄])² = K² = 0.0400. Doubling coefficients doubles exponents and squares the original constant.

    Reverse expression [N₂O₄]/[NO₂]² = 1/K, so constant 5.00.

    Doubled expression [NO₂]⁴/[N₂O₄]² = ([NO₂]²/[N₂O₄])² = K² = 0.0400.

    Doubling coefficients doubles exponents and squares the original constant.

    Review your response against these criteria

    • Uses the stated physical or chemical relationship.
    • Shows the relevant working and reasoning.
    • Checks the result against the original conditions.
Extension

At fixed temperature, multiply every equilibrium concentration by f through a rapid volume change. Derive the new quotient and use f to predict direction.

Scale both concentrations, then the quotient

Optional extension task. It does not change lesson access.

For N₂O₄ ⇌ 2NO₂, write the quotient and explain its exponent. With concentrations both 0.300, predict direction at K = 0.200.

Compare your explanation

Q = [NO₂]²/[N₂O₄] = 0.300; coefficient two gives the exponent. Q > K, so net reverse change.

Initial N₂O₄ 0.100 and NO₂ 0.400: construct reverse ICE expressions and state root bounds.

Compare your explanation

Let x be N₂O₄ formed: 0.100 + x and 0.400 − 2x, with 0 ≤ x ≤ 0.200. Substitute into K and reject roots outside bounds.

Explain why an estimated extent of 0.0200 from initial reactant 0.100 fails a stated 5% small-x screen.

Compare your explanation

Depletion = 0.0200/0.100 × 100% = 20.0%; it exceeds the stated screen, so use the full equation.

Distinguish fixed-volume inert-gas addition from heating. Explain why checking K alone is insufficient. If original K is 0.200, find K for the doubled balanced equation.

Compare your explanation

Fixed-volume inert gas does not change reacting concentrations or Q in the ideal model; heating can change K. A final composition must match the starting atom inventory and imposed constraints. Doubling every balanced coefficient squares the expression, so new K = 0.200² = 0.0400.

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