Edexcel IAL Chemistry 6.10 Deriving mean bond enthalpy

List bonds broken and formed with stoichiometric counts, substitute them into ΔH=broken−formed and isolate the unknown before dividing by its bond multiplicity.

[Maximum number: 3]

Fuels

Fuels burn in oxygen to release a lot of energy.
Many hydrocarbons and alcohols are used as fuels. During complete combustion, they produce carbon dioxide and water.

Petrol contains 2,2,4-trimethylpentane, an isomer of octane, that promotes smooth combustion.

Figure for Question 6.10—Mean bond enthalpies from enthalpy changes of reaction question 1 — Edexcel A-Level Chemistry AS

2,2,4-trimethylpentane

Alcohols, such as methanol and ethanol, can be used as fuels either on their own or as additives in petrol.

The enthalpy changes for the conversion of four compounds in the gas phase into their constituent atoms are shown.

\begin{aligned} \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) & \rightarrow 2 \mathrm{H}(\mathrm{~g})+\mathrm{O}(\mathrm{~g}) & \Delta_{\mathrm{r}} H & =+928 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{CH}_{4}(\mathrm{~g}) & \rightarrow \mathrm{C}(\mathrm{~g})+4 \mathrm{H}(\mathrm{~g}) & \Delta_{\mathrm{r}} H & =+1740 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{CH}_{3} \mathrm{OH}(\mathrm{~g}) & \rightarrow \mathrm{C}(\mathrm{~g})+4 \mathrm{H}(\mathrm{~g})+\mathrm{O}(\mathrm{~g}) & \Delta_{\mathrm{r}} H & =+2105 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{~g}) & \rightarrow 2 \mathrm{C}(\mathrm{~g})+6 \mathrm{H}(\mathrm{~g})+\mathrm{O}(\mathrm{~g}) & \Delta_{\mathrm{r}} H & =+3322 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

Calculate the bond enthalpy of the C-C bond, in $\mathrm{kJ} \mathrm{mol}^{-1}$ .
You must show your working.

calculation of bond energies of O-H and C-H

- calculation of bond energy of C-O

- calculation of bond energy of C-C

Example of calculation:
Method 1 bond energy $\mathrm{O}-\mathrm{H}=928 / 2=(+) 464\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
and
bond energy C-H $=1740 / 4=(+) 435\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
bond energy $\mathrm{C}-\mathrm{O}=2105-(3 \times 435)-464$ $=(+) 336\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
TE on M1
bond energy $\mathrm{C}-\mathrm{C}=3322-(5 \times 435)-464-336$ $=(+) 347\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
Method 2
$3322-2105=1217=\mathrm{C}-\mathrm{C}+2 \times \mathrm{C}-\mathrm{H}(1)$
bond energy $2 \times \mathrm{C}-\mathrm{H}=1740 / 2=(+) 870\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)(1)$ $\mathrm{C}-\mathrm{C}=1217-870=(+) 347\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)(1)$
M3 TE on M1 and M2 in both methods
Correct answer with some working scores

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6.10—Mean bond enthalpies from enthalpy changes of reaction question 1