Edexcel A-Level Chemistry Paper 1, June 2018: Question 3
5 marks · Medium difficulty · Calculations
Calculate the enthalpy change of neutralisation for the reaction between nitric acid and sodium hydroxide using experimental temperature and volume data.
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Question text
3 Nitric acid reacts with sodium hydroxide solution in a neutralisation reaction.
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
In an experiment to determine the enthalpy change of neutralisation, the following
results were obtained.
Volume of 1.00 mol dm−3 HNO = 25.0 cm3
Volume of 1.05 mol dm−3 NaOH = 25.0 cm3
Temperature rise = 6.8°C
(a) Give a reason why excess sodium hydroxide was used.
(1)
(b) Calculate the enthalpy change of neutralisation for the reaction between nitric acid
and sodium hydroxide solution, using the results of the experiment.
Give your answer to an appropriate number of significant figures.
Assume: density of the reaction mixture = 1.0 g cm−3
specific heat capacity of the reaction mixture = 4.18 J g−1 °C−1
(4)
*P52302RA0428*
(Total for Question 3 = 5 marks)
Mark scheme
Show the mark scheme
Question
Acceptable Answer Additional Guidance Mark
Number
3(a) An answer that makes reference to the following point: (1)
+ Allow (nitric acid) / HNO3 is the limiting
to make sure that (all) the (nitric) acid / HNO3 / H has reagent
reacted / been neutralised / is used up
Allow so that 0.025 mol of water / H2O
forms
Ignore to make sure that 1 mol of
water / H2O forms
Ignore just ‘to ensure that reaction is
complete’
Question
Acceptable Answer Additional Guidance Mark
Number
3(b) Example of calculation (4)
calculation of heat produced (1) heat produced = 50.0 x 4.18 x 6.8 = 1421.2( J) /
1.4212 (kJ)
calculation of amount (mol) of HNO3(1) amount HNO3 used = 25.0 x 1.00/1000
= 0.025 / 2.5 x 10−2 (mol)
Ignore moles NaOH and total moles calculated
calculation of enthalpy change (1) enthalpy change = 1421.2 = 56848 (J mol−1)
0.025
or = 1.4212 = 56.848 (kJ mol−1)
0.025
TE on heat produced and amount HNO3
negative sign final answer −57 / −60 kJ mol−1
and or −57 000 / −60 000 J mol−1
units TE on enthalpy change
and
answer to 2 / 1 SF (1) Do not award 3 SF
Correct final answer with sign, units and 2 or 1 SF but no
working scores (4)
Ignore units and sign of enthalpy change in M1 and M3
(Total for Question 3 = 5 marks)
How to answer it
Enthalpy Change of Neutralisation Study Guide
This question assesses practical chemistry, specifically calorimetric data processing for neutralisation reactions. You must understand why excess reagents are used in calorimetry, calculate heat transferred using q = mcΔT , determine moles of a limiting reactant, and correctly apply thermodynamic sign conventions, appropriate units, and significant figure rules.
Reason for using excess sodium hydroxide
✅ Correct Answer
To make sure that all of the nitric acid ( HNO₃ / H⁺ ions) has reacted, been neutralised, or been used up.
💡 Key Knowledge
In calorimetry experiments measuring enthalpy change, the reactant whose enthalpy change is being determined (the limiting reagent) must react completely. Adding an excess of the other reactant ensures 100% reaction of the target species.
❌ Common Errors
Generic statements like "to ensure the reaction is complete" are too vague and do not score. You must specify that the acid is the reactant that is fully used up.
Calculation of Enthalpy Change of Neutralisation
📐 Step-by-Step Calculation
- Calculate total mass ( m ) of the reaction mixture:
Volume = 25.0 cm³ + 25.0 cm³ = 50.0 cm³
Density = 1.0 g cm⁻³ , so mass = 50.0 g - Calculate heat produced ( q ):
q = m × c × ΔT
q = 50.0 × 4.18 × 6.8 = 1421.2 J (or 1.4212 kJ ) - Calculate moles of limiting reactant ( HNO₃ ):
Moles = (25.0 / 1000) × 1.00 = 0.0250 mol
Note: Ignore the concentration of NaOH ; it is in excess! - Calculate enthalpy change per mole ( ΔH ):
ΔH = -q / moles = -1421.2 / 0.025 = -56848 J mol⁻¹ = -56.848 kJ mol⁻¹
✅ Final Answer Requirements
- Value: -57 or -60 kJ mol⁻¹ (or -57000 / -60000 J mol⁻¹ )
- Significant Figures: 2 or 1 SF only (due to ΔT = 6.8 °C having 2 significant figures). Do NOT give 3 SF!
- Sign & Units: Must include the negative sign ( - ) and correct units ( kJ mol⁻¹ or J mol⁻¹ ).
🧠 Exam Technique & Mark Breakdown
- M1: Calculation of heat produced ( 1421.2 J ).
- M2: Calculation of moles of HNO₃ ( 0.025 mol ).
- M3: Calculation of enthalpy change value.
- M4: Correct negative sign, correct units, and strict adherence to 2 or 1 SF.
❌ Common Calculation Traps
- Using the total moles of both solutions or picking the wrong concentration for moles.
- Forgetting to apply the negative sign (exothermic reactions require a minus sign for enthalpy changes).
- Rounding to 3 SF because of habit, thereby losing the final mark.
- Using 25.0 cm³ instead of total volume ( 50.0 cm³ ) for mass in q = mcΔT .
Topics
Physical Chemistry · Core Practicals · Topic 8: Energetics I · Topic 5: Formulae, Equations and Amounts of Substance · Core Practical 8: Determine the enthalpy change of a reaction using Hess’s law
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.