Kinetics and Equilibrium Problem Solving

This lesson brings together kinetics and equilibrium in the kind of multi-step problems that appear in Edexcel A-level exams. We will work through extended calculations combining rate equations, the Arrhenius equation, ICE tables, and Kp calculations, building the fluency you need for the exam.

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Multi-Step Rate Equation Problems

Problem 1: Determining the Rate Equation and Calculating k

The reaction X + 2Y → Z was studied using the initial rate method:

Experiment	[X] / mol dm⁻³	[Y] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	3.2 × 10⁻³
2	0.10	0.20	1.28 × 10⁻²
3	0.20	0.10	6.4 × 10⁻³

Step 1: Determine order with respect to Y.

Compare Experiments 1 and 2 ([X] constant):

[Y] doubles (0.10 → 0.20). Rate changes: 1.28 × 10⁻² / 3.2 × 10⁻³ = 4.0 = 2²

Order with respect to Y = 2

Step 2: Determine order with respect to X.

Compare Experiments 1 and 3 ([Y] constant):

[X] doubles (0.10 → 0.20). Rate changes: 6.4 × 10⁻³ / 3.2 × 10⁻³ = 2.0 = 2¹

Order with respect to X = 1

Step 3: Write the rate equation.

Rate = k[X][Y]² (overall order = 3)

Step 4: Calculate k using data from any experiment (use Experiment 1).

k = Rate / ([X][Y]²) = 3.2 × 10⁻³ / (0.10 × (0.10)²) = 3.2 × 10⁻³ / 1.0 × 10⁻³ = 3.2 mol⁻² dm⁶ s⁻¹

Step 5: Verify k with another experiment (Experiment 2).

k = 1.28 × 10⁻² / (0.10 × (0.20)²) = 1.28 × 10⁻² / (0.10 × 0.04) = 1.28 × 10⁻² / 4.0 × 10⁻³ = 3.2 mol⁻² dm⁶ s⁻¹ ✓

Step 6: Predict the rate for [X] = 0.30, [Y] = 0.15.

Rate = 3.2 × (0.30) × (0.15)² = 3.2 × 0.30 × 0.0225 = 0.0216 mol dm⁻³ s⁻¹

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Problem 2: Rate Equation with Mechanism Deduction

The reaction 2A + B → C + D has the rate equation Rate = k[A][B].

The overall order is 2, and both A and B appear in the rate equation with order 1 each.

What does this tell us about the mechanism?

Since the rate equation is first order in A (not second order as the stoichiometric coefficient might suggest), the rate-determining step involves one molecule of A and one molecule of B:

Slow step (RDS): A + B → intermediate

Fast step: intermediate + A → C + D

This mechanism is consistent with the rate equation because only species in or before the RDS appear in the rate equation.

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Arrhenius Equation Calculations

Problem 3: Finding Ea from Two Temperatures

For a first-order reaction, k₁ = 2.5 × 10⁻³ s⁻¹ at 290 K and k₂ = 1.0 × 10⁻² s⁻¹ at 310 K.

Step 1: Use the two-temperature form.

ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)

ln(1.0 × 10⁻² / 2.5 × 10⁻³) = (Ea/8.314)(1/290 − 1/310)

Step 2: Calculate each side.

ln(4.0) = 1.386

1/290 − 1/310 = (310 − 290)/(290 × 310) = 20/89900 = 2.225 × 10⁻⁴ K⁻¹

Step 3: Solve for Ea.

Ea = 1.386 × 8.314 / 2.225 × 10⁻⁴ = 11.52 / 2.225 × 10⁻⁴ = 51 800 J mol⁻¹ = 51.8 kJ mol⁻¹

Problem 4: Predicting k at a New Temperature

Using the Ea calculated above (51.8 kJ mol⁻¹), predict k at 330 K.

ln(k₃/k₁) = (Ea/R)(1/T₁ − 1/T₃)

ln(k₃/2.5 × 10⁻³) = (51800/8.314)(1/290 − 1/330)

1/290 − 1/330 = (330 − 290)/(290 × 330) = 40/95700 = 4.179 × 10⁻⁴ K⁻¹

ln(k₃/2.5 × 10⁻³) = 6230 × 4.179 × 10⁻⁴ = 2.604

k₃/2.5 × 10⁻³ = e^2.604 = 13.52

k₃ = 13.52 × 2.5 × 10⁻³ = 0.0338 s⁻¹

Problem 5: Arrhenius with Half-Life Data

A first-order reaction has t½ = 100 s at 300 K and t½ = 25 s at 320 K.

Step 1: Calculate k at each temperature.

k₁ = ln 2 / t½ = 0.693 / 100 = 0.00693 s⁻¹ (at 300 K) k₂ = ln 2 / 25 = 0.0277 s⁻¹ (at 320 K)

Step 2: Use the Arrhenius two-temperature form.

ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)

ln(0.0277/0.00693) = (Ea/8.314)(1/300 − 1/320)

ln(4.0) = (Ea/8.314)(2.083 × 10⁻⁴)

1.386 = (Ea/8.314)(2.083 × 10⁻⁴)

Ea = 1.386 × 8.314 / 2.083 × 10⁻⁴ = 55 300 J mol⁻¹ = 55.3 kJ mol⁻¹

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ICE Table Calculations

Problem 6: Kc with a Perfect Square

0.200 mol of A and 0.200 mol of B are placed in a 1.00 dm³ flask. At equilibrium, Kc = 4.00 for A(g) + B(g) ⇌ C(g) + D(g). Find [C] at equilibrium.

	A	B	C	D
Initial	0.200	0.200	0	0
Change	−x	−x	+x	+x
Equilibrium	0.200−x	0.200−x	x	x

Kc = [C][D] / ([A][B]) = x² / (0.200 − x)² = 4.00

Taking the square root: x / (0.200 − x) = 2.00

x = 2.00(0.200 − x) = 0.400 − 2.00x

3.00x = 0.400

x = 0.133 mol dm⁻³

[C] = 0.133 mol dm⁻³ (and [A] = [B] = 0.200 − 0.133 = 0.067 mol dm⁻³)

Check: Kc = (0.133)² / (0.067)² = 0.01769 / 0.004489 = 3.94 ≈ 4.00 ✓ (rounding)

Problem 7: ICE Table Requiring the Quadratic Formula

0.300 mol of A and 0.100 mol of B in a 1.00 dm³ flask. Kc = 10.0 for A + B ⇌ C.

	A	B	C
Initial	0.300	0.100	0
Change	−x	−x	+x
Equilibrium	0.300 − x	0.100 − x	x

Kc = x / ((0.300 − x)(0.100 − x)) = 10.0

x = 10.0 × (0.0300 − 0.300x − 0.100x + x²)

x = 10.0 × (0.0300 − 0.400x + x²)

x = 0.300 − 4.00x + 10.0x²

10.0x² − 5.00x + 0.300 = 0

Using the quadratic formula: x = (5.00 ± √(25.00 − 12.00)) / 20.0 = (5.00 ± √13.00) / 20.0

x = (5.00 ± 3.606) / 20.0

x = 8.606/20.0 = 0.430 (rejected -- exceeds [B]₀ = 0.100)

x = 1.394/20.0 = 0.0697 mol dm⁻³

Check: Kc = 0.0697 / ((0.300 − 0.0697)(0.100 − 0.0697)) = 0.0697 / (0.2303 × 0.0303) = 0.0697 / 0.00698 = 9.99 ≈ 10.0 ✓

Key insight: Always check both roots of the quadratic. Reject any root that gives a negative concentration.

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Kp Calculations from Percentage Conversion

Problem 8: Kp for a Dissociation Equilibrium

2.00 mol of SO₃ is placed in a container and heated to 1000 K at a total pressure of 1.50 atm. At equilibrium, 40.0% of SO₃ has dissociated:

2SO₃(g) ⇌ 2SO₂(g) + O₂(g)

Step 1: ICE table in moles.

40% of 2.00 = 0.80 mol SO₃ dissociates.

	2SO₃	2SO₂	O₂
Initial	2.00	0	0
Change	−0.80	+0.80	+0.40
Equilibrium	1.20	0.80	0.40

(Ratio: 2:2:1, so 0.80 mol SO₂ and 0.40 mol O₂ formed)

Step 2: Total moles and mole fractions.

Total = 1.20 + 0.80 + 0.40 = 2.40 mol

x(SO₃) = 1.20/2.40 = 0.500 x(SO₂) = 0.80/2.40 = 0.333 x(O₂) = 0.40/2.40 = 0.167

Check: 0.500 + 0.333 + 0.167 = 1.000 ✓

Step 3: Partial pressures.

p(SO₃) = 0.500 × 1.50 = 0.750 atm p(SO₂) = 0.333 × 1.50 = 0.500 atm p(O₂) = 0.167 × 1.50 = 0.250 atm

Check: 0.750 + 0.500 + 0.250 = 1.500 atm ✓

Step 4: Calculate Kp.

Kp = p(SO₂)² × p(O₂) / p(SO₃)²

Kp = (0.500)² × 0.250 / (0.750)²

Kp = 0.250 × 0.250 / 0.5625

Kp = 0.0625 / 0.5625

Kp = 0.111 atm

Units: atm² × atm / atm² = atm ✓

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Combining Kinetics and Equilibrium

Problem 9: Linking Rate Constants to Kc

For the equilibrium: A + B ⇌ C

The forward rate equation is: Rate_forward = k_f[A][B]

The reverse rate equation is: Rate_reverse = k_r[C]

At equilibrium, Rate_forward = Rate_reverse:

k_f[A][B] = k_r[C]

Rearranging: [C] / ([A][B]) = k_f / k_r = Kc

This shows that Kc = k_f / k_r -- the equilibrium constant equals the ratio of the forward and reverse rate constants. This is a fundamental link between kinetics and equilibrium.

Worked Example: At 500 K, k_f = 0.040 mol⁻¹ dm³ s⁻¹ and k_r = 0.0080 s⁻¹.

Kc = k_f / k_r = 0.040 / 0.0080 = 5.0 mol⁻¹ dm³

Implication: If temperature increases and Ea(forward) < Ea(reverse), then k_f increases more slowly than k_r, so Kc decreases. This is consistent with the exothermic forward reaction being disfavoured by higher temperature.

Problem 10: Effect of Temperature on Both k and Kc

For A ⇌ B with ΔH = −30 kJ mol⁻¹ (exothermic forward):

At 300 K: k_f = 0.050 s⁻¹, k_r = 0.010 s⁻¹. Kc = 0.050/0.010 = 5.0

The forward reaction has Ea(f) = 60 kJ mol⁻¹. Since ΔH = Ea(f) − Ea(r), we get Ea(r) = 60 + 30 = 90 kJ mol⁻¹.

At 350 K, both k values increase, but k_r increases more (it has higher Ea):

Using Arrhenius for k_f: ln(k_f'/0.050) = (60000/8.314)(1/300 − 1/350) = 7216 × 4.762 × 10⁻⁴ = 3.437

k_f' = 0.050 × e^3.437 = 0.050 × 31.1 = 1.55 s⁻¹

Using Arrhenius for k_r: ln(k_r'/0.010) = (90000/8.314)(1/300 − 1/350) = 10824 × 4.762 × 10⁻⁴ = 5.154

k_r' = 0.010 × e^5.154 = 0.010 × 173.2 = 1.73 s⁻¹

New Kc = 1.55/1.73 = 0.90

Kc has decreased from 5.0 to 0.90 -- consistent with Le Chatelier's prediction that increasing temperature shifts an exothermic equilibrium to the left.

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Exam Strategy and Common Mistakes

Systematic Approach Checklist

Problem type	Key steps	Common error to avoid
Rate equation from data	Compare pairs with one variable constant	Comparing experiments where both concentrations change
Calculate k	k = Rate / [A]^m[B]^n; verify with all experiments	Wrong units for k
Arrhenius (two temps)	Use ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)	Ea in kJ instead of J; T in °C instead of K
ICE table for Kc	Moles → concentrations; check stoichiometry	Forgetting to divide by volume
Kp calculation	Moles → mole fractions → partial pressures → Kp	Mole fractions not summing to 1
Quadratic ICE	Use quadratic formula; reject unphysical root	Accepting a root that gives negative concentration
Kc from rate constants	Kc = k_f / k_r	Inverting the ratio

Golden Rules

1. Read the question carefully -- identify what type of calculation is required.
2. Write out the relevant equation before substituting numbers.
3. Check units throughout -- convert kJ to J, °C to K.
4. For ICE tables, always check stoichiometric ratios carefully.
5. For Kp, always verify that mole fractions sum to 1 and partial pressures sum to P_total.
6. Show your working -- marks are awarded for method even if the final answer is wrong.
7. Verify your answer -- substitute back to check Kc or Kp matches the given value.

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Summary

This lesson has demonstrated how to combine rate equations, Arrhenius calculations, ICE tables, and Kp calculations in extended problems. The link Kc = k_f/k_r connects kinetics and equilibrium quantitatively. Practice these problem types until the procedures become fluent -- they are heavily examined.

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A-Level Deep Dive: Kinetics and Equilibrium Problem Solving

Spec mapping

Edexcel 9CH0 specification, Topics 9, 10, 11 and 16 combined — synoptic kinetics and equilibrium problem solving requires candidates to integrate rate-determination methods (Topic 9), order analysis and Arrhenius (Topic 16), Le Chatelier (Topic 10) and equilibrium constants $K_c$Kc​/$K_p$Kp​ (Topics 10 and 11) into multi-part questions that may also draw on Topic 8 (energetics), Topic 17 (organic mechanisms — SN1/SN2), Topic 19 (transition metal catalysis) and CP5/CP8/CP11 (refer to the official specification document for exact wording). Synoptic kinetics-and-equilibrium questions are examined principally on Paper 3 (synoptic, ~30% of the A-Level), where candidates are expected to draw connections across the whole specification, often within a single 12+ mark question. The data booklet provides Arrhenius, $R$R and selected thermodynamic data; rate equations, $K$K expressions, Le Chatelier shifts must be derived/applied from first principles.

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Worked example with full mark scheme

Question (12 marks — synoptic):

The reaction $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$H2​(g)+I2​(g)⇌2HI(g) is studied. Initial-rate data at 700 K:

Exp	[H₂]₀ / mol dm⁻³	[I₂]₀ / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.010	0.010	1.20 × 10⁻⁵
2	0.020	0.010	2.40 × 10⁻⁵
3	0.020	0.030	7.20 × 10⁻⁵

At 700 K, equilibrium analysis gives [H₂] = 0.020, [I₂] = 0.020, [HI] = 0.140 mol dm⁻³. $K_c$Kc​ at 800 K is 50.7. $\Delta H = -10$ΔH=−10 kJ mol⁻¹.

(a) Determine orders, rate equation and $k$k at 700 K. (4)

(b) Calculate $K_c$Kc​ at 700 K. (2)

(c) Compare $K_c$Kc​ at 700 K and 800 K and explain qualitatively. (2)

(d) Calculate $E_a$Ea​ given $k(700) = 0.120$k(700)=0.120 mol⁻¹ dm³ s⁻¹ (from your part (a)) and $k(800) = 0.580$k(800)=0.580 mol⁻¹ dm³ s⁻¹. (3)

(e) Predict the effect of doubling total pressure on (i) rate; (ii) equilibrium position. (1)

Solution with mark scheme:

(a) M1 — Exp 1 vs 2: [H₂] doubles, rate doubles → 1st order in H₂. Exp 2 vs 3: [I₂] triples, rate triples → 1st order in I₂. A1 — overall 2nd order; rate = $k[\text{H}_2][\text{I}_2]$k[H2​][I2​]. M1 — substitute Exp 1: $1.20 × 10^{-5} = k × 0.010 × 0.010$1.20×10−5=k×0.010×0.010, $k = 0.120$k=0.120 mol⁻¹ dm³ s⁻¹. A1 — units derived from algebra.

(b) M1 — $K_c = [\text{HI}]^2 / ([\text{H}_2][\text{I}_2]) = (0.140)^2 / (0.020 × 0.020) = 0.0196 / 0.000400$Kc​=[HI]2/([H2​][I2​])=(0.140)2/(0.020×0.020)=0.0196/0.000400. A1 — $K_c(700) = 49.0$Kc​(700)=49.0, dimensionless ($\Delta n = 0$Δn=0).

(c) M1 — $K_c$Kc​(800) = 50.7 > $K_c$Kc​(700) = 49.0; $K_c$Kc​ increased slightly with T. A1 — but the reaction is exothermic ($\Delta H = -10$ΔH=−10 kJ mol⁻¹), so Le Chatelier predicts $K_c$Kc​ should decrease with increased T. The fact that $K_c$Kc​ barely changes (and even rises slightly) reflects the small $|\Delta H|$∣ΔH∣: the van't Hoff equation gives $\Delta \ln K = -\Delta H / R \cdot (1/T_2 - 1/T_1)$ΔlnK=−ΔH/R⋅(1/T2​−1/T1​) — for small $\Delta H$ΔH, the change is small. The slight rise may indicate experimental scatter; sign-prediction from Le Chatelier is rigorously correct only if $\Delta H$ΔH is significantly negative.

(d) M1 — Arrhenius two-point: $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$ln(k2​/k1​)=(Ea​/R)(1/T1​−1/T2​). M1 — $\ln(0.580/0.120) = \ln(4.83) = 1.575$ln(0.580/0.120)=ln(4.83)=1.575. $1/T_1 - 1/T_2 = 1/700 - 1/800 = 1.429 × 10^{-3} - 1.250 × 10^{-3} = 1.79 × 10^{-4}$1/T1​−1/T2​=1/700−1/800=1.429×10−3−1.250×10−3=1.79×10−4 K⁻¹. A1 — $E_a = R \times 1.575 / 1.79 × 10^{-4} = 8.31 \times 8800 = 73 × 10^3$Ea​=R×1.575/1.79×10−4=8.31×8800=73×103 J mol⁻¹ = 73 kJ mol⁻¹.