OCR A-Level Chemistry AS Breadth in chemistry (01), June 2022: Question 25
9 marks · Medium difficulty · Structured Questions
Calculate the enthalpy change of reaction for barium hydroxide and nitric acid, explain why it is not the enthalpy change of neutralisation, and analyze a Boltzmann distribution curve for reaction rate.
Practise this questionQuestion
Question text
25 This question is about enthalpy changes and reaction rates.
(a) Aqueous barium hydroxide, Ba(OH)2(aq), reacts with dilute nitric acid, HNO3(aq), as in
Equation 25.1.
Ba(OH)2(aq) + 2HNO3(aq) → Ba(NO3)2(aq) + 2H2O(l) Equation 25.1
A student carries out an experiment to determine the enthalpy change of this reaction, ∆rH.
The student measures out:
• 25.0 cm3 of 2.00 mol dm–3 Ba(OH) (aq) and
• 50.0 cm3 of 2.00 mol dm–3 HNO (aq).
The temperature of each solution is the same.
The student mixes both solutions in a polystyrene cup, stirs the mixture and records the
maximum temperature.
Temperature readings
Initial temperature = 20.5 °C
Maximum temperature = 39.0 °C
(i) Calculate ∆ H, in kJ mol–1, for the reaction shown in Equation 25.1.
r
Give your answer to 3 significant figures.
Assume that the density and specific heat capacity, c, of the solutions are the same as
for water.
∆ H = … kJ mol–1 [4]
r
(ii) The student looked back at Equation 25.1 and noticed that the reaction was a
neutralisation.
The student concluded that ∆rH is the enthalpy change of neutralisation.
Explain why the student’s conclusion is incorrect and determine the correct value for
the enthalpy change of neutralisation.
enthalpy change of neutralisation = … kJ mol–1 [2]
(b) The Boltzmann distribution model can be used by chemists to explain how the rate of a
reaction is affected by temperature.
Fig. 25.1 shows the Boltzmann distribution for a gas at room temperature.
Fig. 25.1
Label the axes on Fig. 25.1 and add a second curve to show the Boltzmann distribution of
the gas at a higher temperature.
Explain why the Boltzmann distribution shows that the rate of a reaction is affected by
temperature.
… [3]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
25 (a) (i) FIRST, CHECK THE ANSWER ON ANSWER LINE 4 ANNOTATE ANSWER WITH TICKS AND
IF ∆ H = –116 (kJ mol–1) award 4 marks CROSSES
r
IF ∆ H = +116 (kJ mol–1) award 3 marks ----------------------------------------------------------------
r
----------------------------------------------------------------
Energy released in J OR kJ
= 75.0 × 4.18 × 18.5 = 5799.75 (J) OR 5.79975 (kJ) AO2.4 ALLOW 5799.8 OR 5800 J OR 5.7998 OR 5.8 kJ
DO NOT ALLOW < 3 SF EXCEPT 5.8
Correctly calculates n(Ba(OH)2) OR n(HNO3) (trailing zeroes)
25.0
n(Ba(OH)2) = 2 × = 0.05(00) (mol) IGNORE any sign
1000
OR
50.0 IGNORE units i.e. ALLOW correctly calculated
n(HNO3) = 2 × = 0.1(00) (mol) AO2.4 number in J OR kJ OR no units
1000
∆H per mole Ba(OH)2 in J OR kJ
Answer MUST divide energy by n(Ba(OH)2 OR 2 × n(HNO3))
5799.75 5799.75
± OR ± 2 × = ±115995 (J) ALLOW 3SF or more OR use of 5800 J OR 5.8 kJ
0.05 0.1
OR
5.79975 5.79975
± OR ± 2 × = ±115.995 (kJ) AO2.8 Sign NOT needed
0.05 0.1
∆H in kJ mol-1 to 3 SF AND – sign 3 SF needed
∆ H = –116 (kJ mol–1) -----------------------------------------------------------
r
AO2.8 Common errors
3 marks
5799.75
0.1 → –58.0 no 2 × using 0.1
5799.75
0.15 → –38.7 ÷ by 0.05 + 0.10
5799.75
2 × → –77.3
0.15
AO
element
20 2 marks for answers above with wrong sign or not
to 3 SF
Other multiples by using m as 50 or 25:
Mark using same principal
Use of 50 → –77.3 3 marks
Use of 25 → –38.7 3 marks
(ii) Reason for incorrect conclusion 2 AO3.2
neutralisation forms 1 mol H2O ×1
OR ∆rH forms 2 mol H2O H2O essential
answer to 25a(i) –1 IGNORE sign, even if wrong
Value for ∆neutH = ± (kJ mol )
2 SF or more ALLOW 2 SF, e.g. 58
AO
element
(b) 3 ANNOTATE ANSWER WITH TICKS AND
21 CROSSES
----------------------------------------------------------------
NOTE: Look for marking criteria within annotations
on Boltzmann distribution diagram
IGNORE slight inflexion on the curve
Curve at higher temperature 1 mark
Curve starts close to zero
AND
does not touch x axis at high energy
AND
maximum to right AND lower than provided curve For labels,
AND ALLOW number of particles
finishing higher than provided curve AO1.2 ALLOW amount of molecules/particles
IGNORE number of atoms
Labels 1 mark ALLOW kinetic energy
Axes labels correct: IGNORE enthalpy for energy
• Number of molecules AND Energy AO1.1
ORA at lower temperature
Explanation 1 mark ALLOW more molecules have the energy to react
More molecules have energy greater than Ea more molecules can overcome/reach Ea
OR IGNORE atoms
Greater area under curve above Ea
Could be in diagram AO1.1 IGNORE more successful collisions
OR collide more frequently
If not stated, assume higher temperature
DO NOT ALLOW explanation is in terms of two
activation energies (i.e. ‘catalyst explanation)
How to answer it
Enthalpy Changes & Reaction Rates Study Guide
What this question tests
This question assesses core physical chemistry competencies: calculating enthalpy changes from calorimetry temperature data, understanding the precise definition of enthalpy change of neutralisation, and interpreting the Boltzmann distribution model to explain the effect of temperature on reaction rates.
Calculating Enthalpy Change (ΔH)
✅ Correct Answer
ΔH = -116 kJ mol⁻¹
💡 Key Knowledge
- Use q = m × c × ΔT to find energy transferred.
- Total mass ( m ) = sum of volumes mixed ( 25.0 + 50.0 = 75.0 cm³ ), assuming density of 1.00 g cm⁻³ .
- ΔT = Maximum temp - Initial temp ( 39.0 - 20.5 = 18.5 °C ).
- Specific heat capacity ( c ) = 4.18 J g⁻¹ K⁻¹ .
📐 Step-by-Step Calculation
- Calculate heat energy released (q):
q = 75.0 × 4.18 × 18.5 = 5799.75 J ( 5.79975 kJ ) - Calculate moles of reactants:
n(Ba(OH)₂) = 25.0 × 2.00 / 1000 = 0.050 mol
n(HNO₃) = 50.0 × 2.00 / 1000 = 0.100 mol - Calculate ΔH per mole of reaction:
Divide energy by moles of limiting reagent/stoichiometric ratio: 5799.75 / 0.05 = 115995 J mol⁻¹ ( 116 kJ mol⁻¹ ) - Apply correct sign and significant figures:
Temperature increases, so the reaction is exothermic ( - ). Express to 3 SF: -116 kJ mol⁻¹ .
❌ Common Errors & Exam Technique
- Incorrect Moles: Forgetting to divide by the stoichiometric coefficient (using 0.1 instead of 0.05 for Ba(OH)₂) leads to half the value ( -58.0 kJ mol⁻¹ ).
- Sign Omission: Forgetting the negative sign loses 1 mark.
- Rounding too early: Always keep full numbers in your calculator until the final rounding step to avoid rounding errors.
Evaluating Enthalpy Change of Neutralisation
✅ Correct Answer
Reason: Neutralisation forms 1 mol of H₂O (whereas Equation 25.1 forms 2 mol of H₂O).
Value: -58 kJ mol⁻¹ (or -58.0 kJ mol⁻¹ )
🧠 Exam Technique & Insight
- The enthalpy change of neutralisation is strictly defined as the enthalpy change when solutions of an acid and an alkali react under standard conditions to form 1 mole of water.
- Because Equation 25.1 produces 2 H₂O , the enthalpy change calculated in (a)(i) represents the reaction for 2 moles of water formed. You must divide your previous answer by 2.
Interpreting the Boltzmann Distribution
💡 Key Knowledge & Diagram Requirements
- Axes Labels: y-axis = Number of molecules (or particles), x-axis = Energy .
- Higher Temperature Curve: Must start at the origin (0,0), peak must be lower and shifted to the right, and the tail must cross above the original curve and stay above it without touching the x-axis.
✅ Explanation of Rate Increase
- At a higher temperature, more molecules have energy greater than or equal to the activation energy ( E ≥ E_a ).
- Therefore, there is a greater proportion of successful collisions per unit time, increasing the rate of reaction.
Topics
Module 3: Periodic table and energy · Practical Activity Groups · 3.2 Physical chemistry · PAG 3: Enthalpy determination
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.