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Question 20N.2.sl.TZ0.3

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Date November 2020 Marks available [Maximum mark: 5] Reference code 20N.2.sl.TZ0.3
Level sl Paper 2 Time zone TZ0
Command term Calculate, Determine Question number 3 Adapted from N/A
3.
[Maximum mark: 5]
20N.2.sl.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

This question was generally well answered. The mean mark on the question was 1.8 out of 3 marks. Mistakes included three C–C (instead of two), missing C–C bonds completely, subtracting bonds formed bonds broken, 3 and 4 instead of 6 and 8 for bonds formed coefficients, and bond enthalpies of C=O and O=O double bonds treated as single bonds.

(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

About half of the candidates obtained both marks. In the incorrect answers coefficients were sometimes ignored, some candidates used the wrong state for water, and some candidates did not realize the heat of formation of O2 was zero and inserted the bond enthalpy for O2 in the calculation.

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