Edexcel A-Level Chemistry Paper 3, November 2021: Question 7
19 marks · Hard difficulty · Calculations
Determine the enthalpy change for the thermal decomposition of sodium hydrogencarbonate using experimental calorimetry data and Hess's law.
Practise this questionQuestion
Question text
7 The enthalpy change for the decomposition of sodium hydrogencarbonate can be
determined indirectly using Hess’s Law.
∆rH
2NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g)
A student carried out two experiments.
(a) Experiment 1 involved the reaction between sodium hydrogencarbonate and
hydrochloric acid.
The student used the following procedure:
• use a measuring cylinder to measure 50 cm3 of 2.00 mol dm−3
hydrochloric acid and pour it into a polystyrene cup
• measure the initial temperature of the acid
• weigh the test tube containing sodium hydrogencarbonate
• tip the sodium hydrogencarbonate into the hydrochloric acid in the
polystyrene cup, stir the mixture and record the lowest temperature reached
• weigh the empty test tube.
Results
Measurement Value
Mass of test tube + NaHCO3 / g 21.23
Mass of empty test tube / g 15.61
Mass of NaHCO3 used / g
Initial temperature / °C 21.0
Final temperature / °C 14.4
Temperature fall / °C
(i) Complete the table.
(1)
(ii) Show, by calculation, that the hydrochloric acid is in excess.
You must show your working.
NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
(2)
(iii) Calculate the enthalpy change for the reaction between
sodium hydrogencarbonate and hydrochloric acid, using the results
of the experiment.
Include a sign and units in your answer.
Assume: mass of reaction mixture = 50.0g
f −1 −1 p
specific heat capacity of the reaction mixture = 4.18J g °C
(3)
(b) Experiment 2 involved the reaction between sodium carbonate and
20 hydrochloric acid.
The student repeated the procedure for*P67806A02036*Experiment 1but used
sodium carbonate instead of sodium hydrogencarbonate and measured the
maximum temperature rise.
Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
The student calculated the enthalpy change for this reaction as −29.4 kJ mol−1.
(i) Complete the Hess cycle with appropriate formulae and labelled arrows.
(2)
∆rH
2NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g)
(ii) Calculate the enthalpy change for the decomposition of
sodium hydrogencarbonate.
Include a sign and units in your answer.
∆rH
2NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g)
(3)
(c) Another student carried out the same two experiments and obtained
a value for the enthalpy change of decomposition of
sodium hydrogencarbonate of +74 kJ mol−1.
The data book value for this enthalpy change is +90 kJ mol−1.
(i) Calculate the percentage error in this student’s value.
(1)
*P67806A02136*
22(ii) Calculate the percentage uncertainties in measuring 50 cm3 of
hydrochloric acid using a burette and using a measuring cylinder.*P67806A02236*
(1)
Apparatus Uncertainty Percentage uncertainty
Measuring cylinder ±0.5 cm3 for each volume measured
Burette ±0.05 cm3 for each reading
(iii) Give a reason why using a burette rather than a measuring cylinder will not
improve the accuracy of the experiment.
(1)
(iv) Describe changes to the method and how the data is used that would improve 23
the accuracy of the determination of the temperature change in*P67806A02336*Experiment 2.
Your description should involve the use of a clock and plotting a graph.
(5)
(Total for Question 7 = 19 marks)
Mark scheme
Show the mark scheme
Question Answer Additional Guidance Mark
Number
7(a)(i) (1)
• mass and temperature fall correct Mass of NaHCO3 used / g 5.62
Temperature fall / °C (−)6.6
Number
7(a)(ii) Example of calculation (2)
• calculation of amount of NaHCO3 amount NaHCO3 = 5.62
and 23 + 1 + 12 + (3 x 16)
calculation of amount of hydrochloric acid (1)
= 0.0669 (mol)
TE on mass of NaHCO3 in (a)(i)
and
amount HCl = 50 x 2.00 = 0.10 (mol)
1000
Ignore SF including 1SF
• 0.0669 mol NaHCO3 needs 0.0699 mol HCl for (1) Allow mol ratio = 1 : 1 so HCl is in excess
reaction so HCl is in excess Allow just more moles of HCl used
Allow 0.10 > 0.0669 (mol)
Allow HCl is in excess by 0.033 (mol)
Number
7(a)(iii) Example of calculation (3)
• calculation of heat absorbed (1) heat absorbed = 50.0 x 4.18 x 6.6
= 1379.4 (J) / 1.3794 (kJ)
Ignore sign
• calculation of enthalpy change (1) enthalpy change = 1379.4
0.0669
= 20619 (J mol−1)
or = 1.3794
0.0669
= 20.619 (kJ mol−1)
TE on heat absorbed and amount NaHCO3 in (a)(ii)
(1) Final answer +20.6(19) kJ mol−1
• positive sign and units
or +20619 J mol−1
TE on enthalpy change
Allow +19.7(06) kJ mol−1 from 0.07 mol in (a)(ii)
Allow kJ mol− / J mol−
Ignore SF except 1 SF
Ignore incorrect / missing units in M1 and M2
Correct answer with sign and units scores (3)
Number
7(b)(i) Example of Hess cycle (2)
∆rH
2NaHCO3(s) Na2CO3(s) +
H2O(l) + CO2(g)
2NaCl(aq) + 2H2O(l) + 2CO2(g)
• correct species and balancing numbers in lower box (1) Ignore missing state symbols
• both arrows pointing in correct directions (1) Stand alone mark
Ignore labels on arrows and inclusion of HCl
Number
7(b)(ii) Example of calculation (3)
• expression for ∆rH (1) ∆rH = 2x∆H1 − ∆H2
• substitution of values into expression with both values (1) ∆rH = 2 x 20.619 – (−29.4)
in same units or
∆rH = 2 x 20619 – (−29 400)
M1 can be scored from values substituted into correct
expression in M2
TE on ∆H1 in (a)(iii) and expression in M1
No TE on incorrect arrows in cycle
(1) ∆ H = +70.638 kJ mol−1
• calculation of ∆rH r
and or
∆ H = +70638 J mol−1
sign r
and TE on ∆H1 in (a)(iii) and expression in M1 provided it
units is a +ve answer
Ignore SF except 1 SF
Correct answer with sign and units scores (3)
Number
7(c)(i) Example of calculation (1)
• calculation of percentage error percentage error
= (90 – 74) x 100 = 17.778 / 17.8 / 18(%)
Allow 17.7 recurring
Ignore SF except 1SF
Do not award 17.7
Number
7(c)(ii) Example of calculation (1)
• calculation of percentage uncertainty using measuring percentage uncertainty using measuring cylinder
cylinder and burette = 0.5 x 100 = 1(%)
and
percentage uncertainty using burette
= 2 x 0.05 x 100 = 0.2(%)
Ignore SF / ±
Number
7(c)(iii) An answer that makes reference to the following point (1)
Allow the uncertainty using the burette is not
• the difference in the uncertainty in using the burette significantly less than using the measuring cylinder
compared with the measuring cylinder is very much smaller
than the % error in the value obtained (so other factors are Allow uncertainty represents a spread of values
more significant) whereas the error is the difference of the true
value and value obtained
Allow just ‘hydrochloric acid / HCl is in excess‘
Ignore heat loss
Number
7(c)(iv) A description that makes reference to the following Allow different times in M1, M2 and M3 or (5)
points: measure the temperature at regular time intervals
• measure the temperature of the (hydrochloric) acid every (1) Allow use of a lid / additional insulation
30 s for 2½ minutes
• add the sodium carbonate / solid (at exactly 3 minutes) (1)
• (stir and) measure the temperature (of the mixture) every
30 s for another 5 minutes (1)
• plot a graph of temperature against time (1) M4 & M5 can be awarded from a suitably labelled
sketch graph
• (join the two sets of points with 2 best fit straight lines and) (1)
extrapolate the lines to the time of mixing
and
determine the maximum temperature change / rise at that
time
(Total for Question 7 = 19 marks)
How to answer it
Hess's Law and Enthalpy Changes Study Guide
What this question tests
This comprehensive question assesses core practical and thermochemical skills: processing experimental calorimetry data, proving reactant excess via mole calculations, applying Hess's Law cycles to determine indirect enthalpy changes, calculating percentage errors and uncertainties, and designing robust experimental methods using temperature-time cooling/heating graphs and extrapolation.
Calorimetry & Reactant Excess in Solution
✅ Correct Answers: (a)(i)
Mass of NaHCO₃ used: 5.62 g
Temperature fall: 6.6 °C (or -6.6 °C)
📐 Calculations: (a)(ii) Proving Excess
- Moles of NaHCO₃: 5.62 / 84.0 = 0.0669 mol
- Moles of HCl: (50 × 2.00) / 1000 = 0.100 mol
- Conclusion: Since the stoichiometric ratio is 1:1, 0.0669 mol of NaHCO₃ requires 0.0669 mol of HCl. As 0.100 mol of HCl is available, HCl is in excess.
📐 Calculations: (a)(iii) Enthalpy Change
- Heat absorbed (q): m × c × ΔT = 50.0 × 4.18 × 6.6 = 1379.4 J (1.3794 kJ)
- Enthalpy change (ΔH): -q / moles = -1379.4 J / 0.0669 mol = -20619 J mol⁻¹ = +20.6 kJ mol⁻¹
- Sign & Units: Endothermic reaction requires a positive sign (+) and units of kJ mol⁻¹ or J mol⁻¹ .
❌ Common Errors
- Forgetting the positive sign for an endothermic reaction temperature fall.
- Dividing heat energy by the moles of the substance in excess instead of the limiting reactant.
- Omitting or providing incorrect units.
Hess's Law Cycle
💡 Key Knowledge: (b)(i) Hess Cycle Construction
Complete the cycle by placing the shared aqueous products at the bottom:
2NaCl(aq) + H₂O(l) + CO₂ (g)
Ensure reaction arrows point downwards from both 2NaHCO₃(s) and Na₂CO₃(s) reacting with 2HCl(aq) .
📐 Calculations: (b)(ii) Indirect Enthalpy Change
- Hess Expression: ΔH (decomposition) = 2(ΔH₁) - ΔH₂
- Substitute values: 2(20.619) - (-29.4)
- Final Answer: +70.6 kJ mol⁻¹ (allow +70.638 kJ mol⁻¹)
🧠 Exam Technique
When applying Hess cycles, carefully track arrow directions. If travelling against an arrow, reverse its sign. Always carry unrounded intermediate values through multi-step calculations.
Percentage Error, Uncertainties, and Practical Improvements
📐 Calculations: (c)(i) & (c)(ii) Errors & Uncertainties
(c)(i) Percentage Error:
((90 - 74) / 90) × 100 = 17.8% (or 17.7%)
(c)(ii) Percentage Uncertainty:
Measuring cylinder: (0.5 × 100) / 50 = 1%
Burette (2 readings): (2 × 0.05 × 100) / 50 = 0.2%
💡 Key Knowledge: (c)(iii) Burette Evaluation
Using a burette does not improve accuracy because the percentage uncertainty from using the measuring cylinder is already significantly smaller than the error in the experimental value itself (heat loss, incomplete reaction).
🧠 Exam Technique: (c)(iv) Temperature Correction Graphs
To score all 5 marks for improving temperature change determination:
- Measure temperature of acid every 30 s for 2.5 minutes.
- Add solid sodium carbonate at exactly 3 minutes.
- Continue stirring and measuring temperature every 30 s for a further 5 minutes.
- Plot a graph of temperature against time.
- Extrapolate the cooling/heating lines back to the 3-minute mark to find the maximum temperature change, accounting for heat loss.
Topics
Physical Chemistry · Core Practicals · Core Practical 8: Determine the enthalpy change of a reaction using Hess’s law · Topic 8: Energetics I
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.