Question

A 0.502 g sample of liquid C6H12 was combusted completely using excess oxygen inside a bomb (constant volume) calorimeter, with the products being carbon dioxide and liquid water. The calorimeter's heat capacity is 4.886 kJ °C-1. If the temperature inside the calorimeter increased from 25.0 °C to 37.6 °C, determine ΔH for this reaction with respect to the system in kJ mol-1. Do not worry about how realistic the final answer is.

Answer 1

Heat absorbed by calorimeter

$$\begin{array}{l} q = C * (\Delta T) \\ q = 4.886 \, \mathrm{kJ} / ^\circ \mathrm{C} * (37.6^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C}) \\ q = +61.6 \, \mathrm{kJ} \text{ absorbed by calorimeter} \\ \end{array}$$

so therefore $q = -61.6 \, \mathrm{kJ}$ for heat released from combustion of $\mathrm{C_6H_{12}}$

and per mole

Heat evolved = $(-61.6 \, \mathrm{kJ} / 0.502 \, \mathrm{g} \, \mathrm{C_6H_{12}}) / (84 \, \mathrm{g/mol})$

$\Delta \mathrm{U} = -10.31 \times 10^3 \, \mathrm{kJ/mol}$ of $\mathrm{C_6H_{12}}$

Reaction

$\mathrm{C_6H_{12}}(l) + 9\mathrm{O_2}(g) \rightarrow 6\mathrm{CO_2}(g) + 6\mathrm{H_2O}(l)$

$\Delta \mathrm{U} = -10.31 \times 10^3 \, \mathrm{kJ/mol}$

Next section will show that if the same amount of gas is on reactant and product side then $\Delta \mathrm{U} = \Delta \mathrm{H}$ so for above $\Delta \mathrm{H} = -10.31 \times 10^3 \, \mathrm{kJ/mol}$.

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