Energetics

WCH12Topic 2.13 questions

Enthalpy changes, calorimetry, Hess's law and mean bond enthalpies.

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Question 15 marks
A spirit burner containing propan-1-ol, $\text{C}_3\text{H}_7\text{OH}$ , was used to heat $200\,\text{g}$ of water in a copper can. Burning $0.580\,\text{g}$ of propan-1-ol raised the temperature of the water by $15.4\,^\circ\text{C}$ .

Calculate the enthalpy change of combustion of propan-1-ol, in $\text{kJ mol}^{-1}$ . Give your answer to 3 significant figures.

[Specific heat capacity of water $= 4.18\,\text{J g}^{-1}\,^\circ\text{C}^{-1}$ ]
Mark scheme
- $Q = mc\Delta T = 200 \times 4.18 \times 15.4$ (1)
- $Q = 12\,874\,\text{J} = 12.9\,\text{kJ}$ (1)
Question 24 marks
Use the mean bond enthalpies below to calculate the enthalpy change for the hydrogenation of ethene:

$\text{C}_2\text{H}_4(\text{g}) + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_6(\text{g})$

12874J=

12.9kJ

n(\text{C}_3\text{H}_7\text{OH}) = 0.580 \div 60.0 = 9.67 \times 10^{-3}\,\text{mol}

(1)

\Delta_c H = -12.874 \div (9.67 \times 10^{-3})

(1)

\Delta_c H = -1330\,\text{kJ mol}^{-1}

(negative sign and 3 s.f. required) (1)

H2(g)→

C2H6(g)

bond	mean bond enthalpy / $\text{kJ mol}^{-1}$
C=C	612
C–C	347
C–H	413
H–H	436

The data-book value is $-137\,\text{kJ mol}^{-1}$ . Suggest why your answer differs from this.

Mark scheme

Bonds broken: $\text{C{=}C} + \text{H{-}H} = 612 + 436 = 1048\,\text{kJ mol}^{-1}$ (1)
Bonds made: $\text{C{-}C} + 2(\text{C{-}H}) = 347 + 826 = 1173\,\text{kJ mol}^{-1}$ (1)
$\Delta H = 1048 - 1173 = -125\,\text{kJ mol}^{-1}$ (1)
Mean bond enthalpies are averages over many compounds, so they differ from the actual bond enthalpies in ethene and ethane (1)