OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2018: Question 18

10 marks · Medium difficulty · Calculations

Write an expression for Kc, calculate the amount of NO2 at equilibrium given Kc and moles, predict enthalpy change from Kp at different temperatures, and explain the effect of pressure change on equilibrium position in terms of Kp.

Practise this question

Question

Chemistry exam question with three parts based on the equilibrium reaction 2NO(g) + O2(g) = 2NO2(g). Part (a) asks to write an expression for Kc and state its units. Part (b) is a calculation of the amount of NO2 in mol given initial moles and volume at equilibrium with a given Kc value. Part (c)(i) asks to predict whether the forward reaction is exothermic or endothermic using a table of Kp values at 298K and 1000K, and part (c)(ii) asks to explain how equilibrium position changes when pressure is increased in terms of Kp.
Question text

18 Nitrogen monoxide, NO, and oxygen, O2, react to form nitrogen dioxide, NO2, in the reversible

reaction shown in equilibrium 18.1.

2NO(g) + O2(g) 2NO2(g) Equilibrium 18.1

(a) Write an expression for Kc for this equilibrium and state the units.

Kc =

Units = … [2]

(b) A chemist mixes together nitrogen and oxygen and pressurises the gases so that their total

gas volume is 4.0 dm3.

• The mixture is allowed to reach equilibrium at constant temperature and volume.

• The equilibrium mixture contains 0.40 mol NO and 0.80 mol O2.

• Under these conditions, the numerical value of Kc is 45.

Calculate the amount, in mol, of NO2 in the equilibrium mixture.

amount of NO2 = … mol [4]

(c) The values of Kp for equilibrium 18.1 at 298 K and 1000 K are shown below.

2NO(g) + O2(g) 2NO2(g) Equilibrium 18.1

Temperature / K K / atm–1

p

298 K = 2.19 × 1012

p

1000 K = 2.03 × 10–1

p

(i) Predict, with a reason, whether the forward reaction is exothermic or endothermic.

… [1]

(ii) The chemist increases the pressure of the equilibrium mixture at the same temperature.

State, and explain in terms of Kp, how you would expect the equilibrium position to

change.

… [3]

Mark scheme

Show the mark scheme Mark scheme showing answers for question 18. Part (a) gives the Kc expression and dm3 mol-1 units for 2 marks. Part (b) outlines the calculation steps leading to 1.2 mol for 4 marks. Part (c)(i) requires Exothermic and noting Kp decreases as temperature increases for 1 mark. Part (c)(ii) requires stating equilibrium shifts to the right, explaining the effect on the Kp expression ratio, and explaining how the ratio is restored for 3 marks.

Question Answer Marks Guidance

18 (a) [NO ]2 2 Must be square brackets

Kc = [NO]2 [O ] IGNORE state symbols

3 –1 ALLOW mol–1 dm3

Units = dm mol

ALLOW mol dm–3 as ECF from inverted K expression

c

(b) FIRST CHECK THE ANSWER ON THE ANSWER 4 ANNOTATIONS MUST BE USED

LINE IF answer = 1.2 (mol) award 4 marks For all parts, ALLOW numerical answers from 2 significant

figures up to the calculator value

Unless otherwise stated, marks are for correctly

calculated values. Working shows how values Ignore rounding errors after second significant figure

have been derived.

1st mark is for realising that concentrations need to be

0.40 –3 calculated.

[NO] = 4.0 = 0.1(0) (mol dm )

AND ALLOW ECF

0.80 –3

[O2] = 4.0 = 0.2(0) (mol dm ) Correct numerical answer with no working would score

all previous calculation marks

[NO ]2 = 45 0.102 0.20 OR = 0.09(0)

[NO ] = (45 0.102 0.20) OR = 0.3(0) (mol dm–3) Making point 2 subsumes point 1

amount NO = 0.30 4 = 1.2 (mol) Making point 3 subsumes points 2 and 1

Common errors

9.6 = 3 marks mol of NO and O2 used

0.36 = 3 marks mol of NO calculated from [NO ]2

2.4 = 2 marks mol of NO and O2 used and no mol of NO2

calculated

(c) (i) Exothermic 1 ALLOW Kc for Kp

AND

Kp decreases as temperature increases ALLOW Equilibrium shifts to left hand side as temperature

increases

(c) (ii) 3 FULL ANNOTATIONS NEEDED

ALLOW Kc for Kp throughout the response.

Equilibrium shift

(Equilibrium position) shifts to right / forward /

towards products

Effect of increased pressure on Kp expression

Ratio (in Kp expression) decreases ALLOW Kp (initially) decreases for second marking point IF

OR Kp is seen to be restored later in the process.

Denominator/bottom of Kp expression increases

more (than numerator/top)

Equilibrium shift (Kp expression) ALLOW more NO2 / product formed to restore Kp

Ratio (in Kp expression) increases to restore Kp ALLOW ratio adjusts to restore Kp

OR

Numerator/top of Kp expression increases to

restore Kp

Total 10

How to answer it

Equilibrium Constants (Kc and Kp) & Le Chatelier's Principle

What this question tests

This question assesses your mastery of chemical equilibria: writing expression constants (Kc), calculating equilibrium amounts from moles and volume, determining units, linking temperature changes to thermodynamic favourability (exothermic vs endothermic via Kp), and applying Le Chatelier's principle through the lens of equilibrium expressions.

Part (a)

Kc Expression and Units

✅ Correct Answer

Kc expression: [NO₂]² / ([NO]² [O₂])

Units: dm³ mol⁻¹ (or L mol⁻¹ )

2 marks available: 1 for the correct algebraic expression using square brackets, 1 for correct units.

💡 Key Knowledge

  • Products go on the numerator, reactants on the denominator.
  • Stoichiometric coefficients from the balanced equation become powers.
  • State symbols (g, l, s, aq) are ignored inside the Kc expression.

❌ Common Errors

  • Using round brackets ( ) instead of mandatory square brackets [ ] .
  • Inverting the expression (putting reactants on top).
  • Cancelling unit terms incorrectly (e.g., writing mol dm⁻³ instead).
Part (b)

Equilibrium Amount Calculation

📐 Step-by-Step Calculation

  1. Convert moles to concentration:
    [NO] = 0.40 mol / 4.0 dm³ = 0.10 mol dm⁻³
    [O₂] = 0.80 mol / 4.0 dm³ = 0.20 mol dm⁻³
  2. Rearrange the Kc expression for [NO₂]²:
    [NO₂]² = Kc × [NO]² × [O₂]
    [NO₂]² = 45 × (0.10)² × 0.20 = 0.090
  3. Find [NO₂] by taking the square root:
    [NO₂] = √(0.090) = 0.30 mol dm⁻³
  4. Scale concentration back to total volume (4.0 dm³):
    Amount = 0.30 mol dm⁻³ × 4.0 dm³ = 1.2 mol
4 marks available. ECF allowed. Correct final answer of 1.2 mol with no working still scores full marks!

🧠 Exam Technique & Traps

  • The Volume Trap: Forgetting to divide the initial moles by 4.0 dm³ to find concentrations before substituting into Kc. (This generates common wrong answers like 9.6 or 2.4).
  • Significant Figures: Give your final answer to 2 significant figures, matching the precision of the input data (0.40 mol, 0.80 mol, 4.0 dm³).
Part (c)(i)

Predicting Enthalpy Change from Kp

✅ Correct Answer

Exothermic

Reason: As temperature increases (from 298 K to 1000 K), the value of Kp decreases (from 2.19 × 10¹² to 2.03 × 10⁻¹).

1 mark available. Must state exothermic AND link it to Kp decreasing as T increases.

💡 Key Knowledge

  • For an exothermic forward reaction, raising temperature shifts equilibrium to the left, decreasing the yield of products and thus decreasing the equilibrium constant (Kp or Kc).
  • Conversely, for an endothermic reaction, raising temperature increases the equilibrium constant.
Part (c)(ii)

Effect of Pressure on Equilibrium Position and Kp

✅ Correct Answer

  1. Equilibrium shift: Shifts to the right (towards products / forward).
  2. Kp expression effect: The denominator increases more than the numerator, causing the pressure ratio to momentarily decrease.
  3. Restoration: Equilibrium shifts right to increase the numerator, restoring Kp to its constant value at constant temperature.
3 marks available. Full explanations must mention the specific shift, the immediate mathematical impact on the Kp fraction, and how equilibrium responds to keep Kp constant.

🧠 Examiner Insights

  • Top-level students easily distinguish between factors that change the value of K (only temperature) and factors that cause a temporary shift without altering K (pressure and concentration).
  • Always frame your explanation using numerator/denominator terminology when discussing expression ratios!

Topics

Module 5: Physical chemistry and transition elements · 5.1 Rates, equilibrium and pH

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.