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Question 20N.2.hl.TZ0.3

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Date November 2020 Marks available [Maximum mark: 10] Reference code 20N.2.hl.TZ0.3
Level hl Paper 2 Time zone TZ0
Command term Calculate, Determine, Justify, Predict Question number 3 Adapted from N/A
3.
[Maximum mark: 10]
20N.2.hl.TZ0.3

An equation for the combustion of propane is given below.

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

(a)

Determine the standard enthalpy change, ∆H⦵, for this reaction, using section 11 of the data booklet.

 

[3]

Markscheme

Bonds broken: 8(C–H) + 2(C–C) + 5(O=O) / 8 × 414 «kJ mol−1» + 2 × 346 «kJ mol−1» + 5 × 498 «kJ mol−1» / 6494 «kJ» ✔

Bonds formed: 6(C=O) + 8(O–H) / 6 × 804 «kJ mol−1» + 8 × 463 «kJ mol−1» / 8528 «kJ» ✔

«Enthalpy change=bonds broken-bonds formed =6494 kJ−8528 kJ=»−2034 «kJ»  ✔

Award [3] for correct final answer.

 

Examiners report

Many candidates had difficulty determining the number and type of bonds broken or formed and consequently this was the part of question 3 that was most poorly attempted. Those that could identify these bonds performed the calculations correctly.

(b)

Calculate the standard enthalpy change, ∆H⦵, for this reaction using section 12 of the data booklet.

[2]

Markscheme

4(−241.8 «kJ») AND 3(−393.5 «kJ») AND «1»(−105 «kJ») ✔
«ΔH⦵=4(−241.8 «kJ»)+3(−393.5 «kJ»)–«1»(−105 «kJ»)=»−2043 «kJ» ✔

Award [2] for correct final answer.

Award [1 max] for −2219 «kJ».

Examiners report

Enthalpy calculations using enthalpy of formation data were generally well done.

(c)

Predict, giving a reason, whether the entropy change, ∆S⦵, for this reaction is negative or positive.

 

[1]

Markscheme

positive AND more moles «of gas» in products ✔

Examiners report

Most knew that entropy increased however some lost the mark by not including an explanation based on increase number of mol of gaseous products.

(d)

Calculate ∆S⦵ for the reaction in J K-1, using section 12 of the data booklet.

The standard molar entropy for oxygen gas is 205 J K-1 mol-1.

[2]

Markscheme

4×188.8 «J K−1» AND 3×213.8 «J K−1» AND «1×»270 «J K−1» AND 5×205 «J K−1» ✔

«∆S⦵=4(188.8 J K−1)+3(213.8 J K−1)–[1(270 J K−1)+5(205 J K−1)]=»102«J K−1»  ✔

Award [2] for correct final answer.

Examiners report

Calculating ΔSө, like most other calculations, was well done.

(e)

Calculate the standard Gibbs free energy change, ∆G⦵, in kJ, for the reaction at 5 °C, using your answers to (b) and (d). Use section 1 of the data booklet.

(If you did not obtain an answer to (b) or (d) use values of -1952 kJ and +113 J K-1 respectively, although these are not the correct answers.)

[2]

Markscheme

«T=5+273=»278 K ✔

«ΔG⦵=−2043 kJ−(278 K×0.102 kJ K−1)=»−2071 «kJ» ✔

Award [2] for correct final answer.

Examiners report

ΔGө calculations were also well done, with some not seeing that specific units were to be used.

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