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

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

Determine the value of ΔH^Θ, in kJ, for the reaction using the values in the table.

The table lists standard entropy, S^Θ, values.

Calculate the standard entropy change for the reaction, ΔS^Θ, in J K⁻¹.

CH₄(g) + H₂O(g) → 3H₂(g) + CO(g)

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

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

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]

\(\Delta H_{\text{f}}^\Theta \) for any element = 0 «by definition»

OR

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

[1 mark]

ΔH^Θ « = \(\sum {\Delta H_{\text{f}}^\Theta } \)_(products) – \(\sum {\Delta H_{\text{f}}^\Theta } \)_(reactants) = –111 + 0 – [–74.0 + (–242)]»

= «+» 205 «kJ»

[1 mark]

«ΔS^Θ = ΣS^Θ_products – ΣS^Θ_reactants = 198 + 3 × 131 – (186 + 189) =» «+» 216 «J K^–1»

[1 mark]

«ΔG^Θ = ΔH^Θ – TΔS^Θ = 205 kJ – 298 K × \(\frac{{216}}{{1000}}\) kJ K^–1 =» «+» 141 «kJ»

[1 mark]

«ΔH^Θ = TΔS^Θ»

«\({\text{T}} = \frac{{\Delta {H^\Theta }}}{{\Delta {S^\Theta }}} = \frac{{205000{\text{ J}}}}{{216{\text{ J }}{{\text{K}}^{ - 1}}}}\)»

«T =» 949 «K»

Do not award a mark for negative value of T.

[1 mark]