Question 18M.2.hl.TZ2.5
| Date | May 2018 | Marks available | [Maximum mark: 8] | Reference code | 18M.2.hl.TZ2.5 |
| Level | hl | Paper | 2 | Time zone | TZ2 |
| Command term | Calculate, Determine, Outline | Question number | 5 | Adapted from | N/A |
Enthalpy changes depend on the number and type of bonds broken and formed.
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]
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]
The table lists the standard enthalpies of formation, , for some of the species in the reaction above.
Outline why no value is listed for H2(g).
[1]
for any element = 0 «by definition»
OR
no energy required to form an element «in its stable form» from itself
[1 mark]
Determine the value of ΔHΘ, in kJ, for the reaction using the values in the table.
[1]
ΔHΘ « = (products) – (reactants) = –111 + 0 – [–74.0 + (–242)]»
= «+» 205 «kJ»
[1 mark]
Enthalpy changes depend on the number and type of bonds broken and formed.
The table lists standard entropy, SΘ, values.
Calculate the standard entropy change for the reaction, ΔSΘ, in J K−1.
CH4(g) + H2O(g) → 3H2(g) + CO(g)
[1]
«ΔSΘ = ΣSΘproducts – ΣSΘreactants = 198 + 3 × 131 – (186 + 189) =» «+» 216 «J K–1»
[1 mark]
Calculate the standard free energy change, ΔGΘ, in kJ, for the reaction at 298 K using your answer to (b)(ii).
[1]
«ΔGΘ = ΔHΘ – TΔSΘ = 205 kJ – 298 K × kJ K–1 =» «+» 141 «kJ»
[1 mark]
Determine the temperature, in K, above which the reaction becomes spontaneous.
[1]
«ΔHΘ = TΔSΘ»
«»
«T =» 949 «K»
Do not award a mark for negative value of T.
[1 mark]