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Date May 2018 Marks available 1 Reference code 18M.2.hl.TZ2.5
Level HL Paper 2 Time zone TZ2
Command term Calculate Question number 5 Adapted from N/A

Question

Enthalpy changes depend on the number and type of bonds broken and formed.

Enthalpy changes depend on the number and type of bonds broken and formed.

The table lists the standard enthalpies of formation, Δ H f Θ , for some of the species in the reaction above.

M18/4/CHEMI/SP2/ENG/TZ2/04.b

Hydrogen gas can be formed industrially by the reaction of natural gas with steam.

                                          CH4(g) + H2O(g) → 3H2(g) + CO(g)

Determine the enthalpy change, ΔH, for the reaction, in kJ, using section 11 of the data booklet.

Bond enthalpy for C≡O: 1077 kJ mol−1

[3]
a.

Outline why no value is listed for H2(g).

[1]
b.i.

Determine the value of ΔHΘ, in kJ, for the reaction using the values in the table.

[1]
b.ii.

The table lists standard entropy, SΘ, values.

M18/4/CHEMI/HP2/ENG/TZ2/05.c

Calculate the standard entropy change for the reaction, ΔSΘ, in J K−1.

CH4(g) + H2O(g) → 3H2(g) + CO(g)

[1]
c.

Calculate the standard free energy change, ΔGΘ, in kJ, for the reaction at 298 K using your answer to (b)(ii).

[1]
d.

Determine the temperature, in K, above which the reaction becomes spontaneous.

[1]
e.

Markscheme

bonds broken: 4(C–H) + 2(H–O)/4(414) + 2(463)/2582 «kJ»

bonds made: 3(H–H) + C≡O/3(436) + 1077/2385 «kJ»

ΔH «= ΣBE(bonds broken) – ΣBE(bonds made) = 2582 – 2385» = «+» 197 «kJ»

 

Award [3] for correct final answer.

Award [2 max] for –197 «kJ».

[3 marks]

a.

Δ H f Θ for any element = 0 «by definition»

OR

no energy required to form an element «in its stable form» from itself

[1 mark]

b.i.

ΔHΘ « = Δ H f Θ (products) – Δ H f Θ (reactants) = –111 + 0 – [–74.0 + (–242)]»

«+» 205 «kJ»

[1 mark]

b.ii.

«ΔSΘ = ΣSΘproducts – ΣSΘreactants = 198 + 3 × 131 – (186 + 189) =» «+» 216 «J K–1»

[1 mark]

c.

«ΔGΘ = ΔHΘ – TΔSΘ = 205 kJ – 298 K × 216 1000  kJ K–1 =» «+» 141 «kJ»

[1 mark]

d.

«ΔHΘ = TΔSΘ»

« T = Δ H Θ Δ S Θ = 205000  J 216  J  K 1 »

«T =» 949 «K»

 

Do not award a mark for negative value of T.

[1 mark]

e.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.
[N/A]
e.

Syllabus sections

Additional higher level (AHL) » Topic 15: Energetics/thermochemistry » 15.2 Entropy and spontaneity
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Additional higher level (AHL) » Topic 15: Energetics/thermochemistry
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