OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2022: Question 17
14 marks · Hard difficulty · Structured Questions
Calculate maximum temperature change, standard enthalpy of formation, standard entropy change, enthalpy change, and feasibility of reactions involving energy changes, enthalpy, and entropy.
Practise this questionQuestion
Question text
17 This question is about energy changes.
(a) Magnesium reacts with aqueous silver nitrate, AgNO3(aq) as shown below.
Mg(s) + 2AgNO (aq) 2Ag(s) + Mg(NO ) (aq) ∆H = –678 kJ mol–1
33 2
A student adds an excess of magnesium to 100.0 cm3 of 0.400 mol dm–3 AgNO (aq).
The initial temperature is 20.0 °C.
(i) Determine the maximum temperature reached in this reaction.
Give your answer to 3 significant figures.
Assume that the specific heat capacity and density of the solution are the same as for
water, and that there are no heat losses.
maximum temperature reached = … °C [4]
(ii) The student wants to repeat the experiment, but there is not enough AgNO3(aq) left to
use another 100.0 cm3 portion.
The student decides to modify the method by adding an excess of magnesium to
50.0 cm3 of 0.400 mol dm–3AgNO (aq).
Predict, with reasons, how this modification would affect the maximum temperature
reached. Assume that there are no heat losses.
… [1]
(b) Nitric acid is manufactured from ammonia in a multi-stage process.
The equation for the first stage in this process is shown in Reaction 17.1.
4NH (g) + 5O (g) 4NO(g) + 6H O(l) ∆H ө= –1172 kJ mol–1 Reaction 17.1
32 2
Some standard enthalpy changes of formation are shown in the table.
Compound ∆ H ө / kJ mol–1
f
NH3(g) –46
H2O(l) –286
(i) Explain the term enthalpy change of formation.
… [1]
(ii) Calculate the standard enthalpy change of formation, ∆ H ө, of NO(g).
f
∆ H ө of NO(g) = … kJ mol–1 [2]
f
(c) Carbon disulfide, CS2, reacts with dinitrogen oxide, N2O, as shown in Reaction 17.2.
4CS2(l) + 8N2O(g) S8(s) + 4CO2(g) + 8N2(g) Reaction 17.2
Standard entropies, S ө, are shown in the table.
Substance CS2(l) N2O(g) S8(s) CO2(g) N2(g)
S ө/ J K–1 mol–1 151 220 256 214 192
(i) Explain the term entropy.
… [1]
(ii) The free energy change, ∆G, of Reaction 17.2 is –2672 kJ mol–1 at 25 °C.
Calculate the enthalpy change, ∆H, of Reaction 17.2, in kJ mol–1.
∆H = … kJ mol–1 [3]
(iii) A student concludes that Reaction 17.2 is feasible at all temperatures.
Explain whether the student is correct or not.
… [2]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
17 (a) (i) FIRST, CHECK THE ANSWER ON ANSWER LINE 4 FULL ANNOTATIONS MUST BE USED
IF T = 52.4 ºC OR 52.5 ºC award 4 marks ----------------------------------------------------------------
IF T = 32.4 (ºC) award 3 marks ALLOW ECF throughout
---------------------------------------------------------------- ------------------------------------------------------------
Correctly calculates n(AgNO3)
100.0
= 0.400 × OR 0.04(00) (mol) AO1.2
1000
Energy released per mole of AgNO3 in J OR kJ
678 × 0.0400 ALLOW 13.6 kJ OR 13600 J (to 3SF)
= OR 13.56 (kJ) OR 13560 (J) AO2.4 DO NOT ALLOW < 3 SF
IGNORE any sign and units
Correctly calculates ∆T i.e. ALLOW correctly calculated value in J OR kJ
-----------------------------------------------------
13560
∆T = OR 32.4 (ºC) AO2.8
100 x 4.18
Maximum temperature reached
= 32.4… + 20.0 = 52.4 ºC AO2.8
3 SF required ALLOW ECF ONLY from calculated ∆T +20 ºC
Common errors
3 marks
o ∆H
84.9 C (not divided )
(a) (ii) Maximum temperature is the same 1 AO3.4 ALLOW response that links the same
AND proportionality/ratio of volume/mass and
Half the energy/ moles AND half the mass/volume 13 energy/moles
ALLOW if seen by a calculation
(b) (i) (Enthalpy change) when 1 mole of a compound 1 AO1.1 ALLOW energy required OR energy released
is formed from its elements ×1
ALLOW one mole of product/substance
DO NOT ALLOW 1 mole of element
DO NOT ALLOW
is formed from its gaseous elements
when 1 mole of a solid compound
when 1 mole of a gaseous compound
(ii) FIRST, CHECK THE ANSWER ON ANSWER LINE 2
If answer = (+)90 award 2 marks
--------------------------------------------------------------------
4(∆ Ho.NO) = –1172 – 6(–286) + 4(–46) AO2.2
f
= –1172 + 1716 – 184 ×2
= (+)360 (kJ mol–1)
o 360 –1 ALLOW ECF providing all values are used
∆fH .NO = = (+)90 (kJ mol ) ALLOW one transcription error in the values
used for M2
Common error
1 mark
-90 (wrong sign)
(c) (i) a measure of the dispersal of energy (in a system) 1 AO1.1 ALLOW a measure/degree of the disorder (of a
system) ORA
(ii) FIRST CHECK THE ANSWER ON ANSWER LINE 3 AO2.2
If answer = –2587 (kJ mol–1) award 3 marks ×3 ALLOW ECF throughout
---------------------------------------------------------------------
∆So
∆So = 256 + 4(214) + 8(192) – 4(151) – 8(220)
= (+)284 (J K–1 mol–1)
OR (+)0.284 (kJ K–1 mol–1)
Use of
T = 298 (K) M2 is for unit conversions seen anywhere.
AND
∆S = 0.284 (kJ K–1 mol–1 )
∆H = (∆G + T∆S)
= –2587 (kJ mol–1) ALLOW 3SF up to the calculator value
-2587.368 (kJ mol–1)
ALLOW ECF from incorrect unit conversions or
incorrect ∆S.
15 Common errors
2 marks
-2664.9 (kJ mol–1) (Use of 25oC)
81960 (kJ mol–1) (Use of ∆S 284)
4428 (kJ mol–1) (Use of 25oC and ∆S 284)
-2756.632 (kJ mol–1) (Use of ΔS = -0.284)
(iii) ∆S is positive/ + AND ∆H is negative/ – AO3.1 ALLOW ∆H is exothermic
ALLOW ‘-T∆S’ is negative’
∆G is negative (– at all temperatures) AO3.2
OR ∆G is (always) negative/ – 2 ∆G comment is dependent on on the signs
assigned to ∆S AND ∆H (either in answer or from
17 cii).
ALLOW ECF from incorrect signs for ∆S and/or
∆H from c(ii)
i.e.
∆S is positive/ + AND ∆H is positive/ +
Reaction is feasible only at high temperatures
∆S is negative/ - AND ∆H is negative/ -
Reaction is feasible only at low temperatures
IGNORE ∆S is negative/ - AND ∆H is positive/ +
(-∆G given in 17 cii)
----------------------------------------------------------------
Alternative Approach
ALLOW use of ∆G=0 for 2 marks
i.e. calculates T = - 9109K
It is always feasible above - 9109K / calculated
-ve value and all temperatures are above this
Total 14
How to answer it
Energy Changes & Thermodynamics Study Guide
What this question tests
This comprehensive multi-part question assesses core thermodynamic and energetic skills: calculating enthalpy changes from calorimetry data, applying Hess's Law / enthalpy cycles using standard enthalpies of formation, defining key thermodynamic terms (enthalpy of formation and entropy), calculating entropy changes (ΔS), applying the Gibbs free energy equation (ΔG = ΔH - TΔS), and predicting reaction feasibility at different temperatures.
Determining Maximum Temperature Reached
✅ Correct Answer
Maximum temperature reached = 52.4 °C
📐 Step-by-Step Calculation
- Moles of limiting reagent:
n(AgNO₃) = 0.400 × (100.0 / 1000) = 0.0400 mol - Energy released:
From equation, 2 moles of AgNO₃ release 678 kJ.
Energy = (678 × 0.0400) / 2 = 13.56 kJ = 13560 J - Temperature change (ΔT):
q = mcΔT → ΔT = 13560 / (100.0 × 4.18) = 32.4 °C - Final temperature:
Initial + ΔT = 20.0 + 32.4 = 52.4 °C (to 3 SF)
❌ Common Errors
- Forgetting to divide the enthalpy change by the stoichiometric coefficient (2) for AgNO₃, leading to 84.9 °C.
- Failing to add the initial temperature (20.0 °C) back onto ΔT.
- Rounding intermediate values too early or giving an incorrect number of significant figures (must be 3 SF).
🧠 Exam Technique
Always check the answer line first! The mark scheme explicitly rewards correct final answers ( 52.4 °C or 52.5 °C due to rounding variants). Show clear unit conversions for volume and energy (Joules vs kilojoules).
Effect of Changing Reagent Volumes
✅ Correct Answer
The maximum temperature reached is the same.
💡 Key Knowledge
Halving the volume of AgNO₃ (limiting reagent) halves the number of moles reacting. Consequently, this releases half the total heat energy. However, that halved energy is absorbed by half the mass of solution. Because the ratio of energy released to mass of solution remains constant, the temperature change (ΔT) and final maximum temperature remain identical.
Enthalpy Change of Formation
✅ Correct Answer
The enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions.
❌ Common Errors
Students often lose this mark by stating "formed from its gaseous elements" or referring to "1 mole of elements/products" instead of strictly 1 mole of the compound.
Calculating Enthalpy of Formation of NO
✅ Correct Answer
ΔfH°(NO) = +90 kJ mol⁻¹
📐 Step-by-Step Calculation
- Use Enthalpy Cycle / Formula:
ΔH = Σ ΔfH°(products) - Σ ΔfH°(reactants) - Substitute values for Reaction 17.1:
-1172 = [4(ΔfH° NO) + 6(ΔfH° H₂O)] - [4(ΔfH° NH₃) + 5(ΔfH° O₂)] - Note standard states: ΔfH° for O₂(g) = 0 kJ mol⁻¹.
- Plug in given data:
-1172 = [4(ΔfH° NO) + 6(-286)] - [4(-46) + 0]
-1172 = [4(ΔfH° NO) - 1716] - [-184] - Rearrange and solve:
-1172 = 4(ΔfH° NO) - 1532
4(ΔfH° NO) = +360
ΔfH°(NO) = 360 / 4 = +90 kJ mol⁻¹
❌ Common Errors
- Forgetting to multiply the tabular values by the stoichiometric balancing numbers (e.g., multiplying by 6 for water and 4 for ammonia/NO).
- Sign errors when subtracting negative values or failing to divide the final sum by 4.
Defining Entropy
✅ Correct Answer
A measure of the dispersal of energy in a system (or degree of disorder).
💡 Key Knowledge
OCR heavily prefers "dispersal of energy among particles" over older definitions like "measure of disorder", though modern mark schemes typically accept both.
Calculating Enthalpy Change (ΔH) from Free Energy
✅ Correct Answer
ΔH = -2587 kJ mol⁻¹ (Accept -2587 to -2587.368)
📐 Step-by-Step Calculation
- Calculate ΔS° for the reaction:
ΔS° = ΣS°(products) - ΣS°(reactants)
ΔS° = [S°(S₈) + 4S°(CO₂) + 8S°(N₂)] - [4S°(CS₂) + 8S°(N₂O)]
ΔS° = [256 + 4(214) + 8(192)] - [4(151) + 8(220)]
ΔS° = [256 + 856 + 1536] - [604 + 1760] = 2648 - 2364 = +284 J K⁻¹ mol⁻¹ - Convert units for ΔS° into kJ K⁻¹ mol⁻¹:
ΔS° = 284 / 1000 = +0.284 kJ K⁻¹ mol⁻¹ - Rearrange Gibbs Free Energy equation:
ΔG = ΔH - TΔS ⇒ ΔH = ΔG + TΔS - Substitute values (T = 298 K):
ΔH = -2672 + [298 × 0.284] = -2672 + 84.632 = -2587.368 kJ mol⁻¹
❌ Common Errors
- Unit Trap: Forgetting to divide ΔS by 1000 to convert Joules to kilojoules before combining with ΔG (which is in kJ mol⁻¹).
- Incorrect stoichiometry multipliers from Reaction 17.2 table values.
Feasibility at All Temperatures
✅ Correct Answer
The student is correct.
💡 Key Knowledge & Justification
- Sign identification: ΔS is positive (+284) AND ΔH is negative (-2587).
- Gibbs explanation: In the equation ΔG = ΔH - TΔS, if ΔH is negative and TΔS is positive (subtracted), then ΔG will always be negative at any temperature T. Therefore, the reaction is feasible at all temperatures.
🧠 Exam Technique
Your explanation must explicitly reference the signs of both ΔH and ΔS calculated in previous parts, and link them directly to how they affect ΔG in the equation. Mark schemes enforce that comments on ΔG being negative are dependent on correct sign tracking.
Topics
Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · 3.2 Physical chemistry · 5.2 Energy
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.