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A bomb calorimeter (a sealed steel vessel of fixed volume in a water bath) is first calibrated…

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A bomb calorimeter (a sealed steel vessel of fixed volume in a water bath) is first calibrated: burning $0.500\ \text{g}$ of benzoic acid ( $M = 122.12\ \text{g mol}^{-1}$ , $\Delta_c U = -3226\ \text{kJ mol}^{-1}$ ) raises its temperature by $1.321\ \text{K}$ . Then $0.300\ \text{g}$ of naphthalene, $\text{C}_{10}\text{H}_8(s)$ ( $M = 128.17\ \text{g mol}^{-1}$ ), is burnt in it, and the temperature rises by $1.205\ \text{K}$ . The water in the products is liquid, and the temperature is close to $298\ \text{K}$ . Take $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$ , $\Delta_f H^\circ(\text{CO}_2, g) = -393.5\ \text{kJ mol}^{-1}$ and $\Delta_f H^\circ(\text{H}_2\text{O}, l) = -285.8\ \text{kJ mol}^{-1}$ .

What is the heat capacity of the calorimeter, with its water?
What is $\Delta_c U$ of naphthalene?
What is $\Delta_c H$ of naphthalene?
What is the standard enthalpy of formation of naphthalene?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Calibrate, find $\Delta_c U$ from the second burning, turn it into $\Delta_c H$ , then use Hess's law.

(a) Calibration. $0.500\ \text{g}$ is $4.0943 \times 10^{-3}\ \text{mol}$ of benzoic acid, which gives out $4.0943 \times 10^{-3} \times 3226 = 13.208\ \text{kJ}$ . So

\displaystyle C = \frac{13.208\ \text{kJ}}{1.321\ \text{K}} = 10.00\ \text{kJ K}^{-1}

(b) Naphthalene. Heat given out: $9.999 \times 1.205 = 12.048\ \text{kJ}$ , from $2.3406 \times 10^{-3}\ \text{mol}$ . So

\displaystyle \Delta_c U = -\frac{12.048}{2.3406 \times 10^{-3}}\ \text{kJ mol}^{-1} = -5148\ \text{kJ mol}^{-1}

(negative: heat is given out).

(c) Enthalpy. $\text{C}_{10}\text{H}_8(s) + 12\,\text{O}_2(g) \rightarrow 10\,\text{CO}_2(g) + 4\,\text{H}_2\text{O}(l)$ , so $\Delta n_g = 10 - 12 = -2$ and

\displaystyle \Delta_c H = \Delta_c U + \Delta n_g RT = -5148 + (-2)(8.314 \times 10^{-3})(298) = -5152\ \text{kJ mol}^{-1}

(d) Formation. $\Delta_c H = 10(-393.5) + 4(-285.8) - \Delta_f H(\text{C}_{10}\text{H}_8)$ , so

\displaystyle \Delta_f H = -5078.2 - (-5152) = 74.3\ \text{kJ mol}^{-1}

Check. The same cycle the other way round: forming naphthalene from 10 C and 4 $\text{H}_2$ , burning it, and comparing with burning the elements directly, gives the same $74.3\ \text{kJ mol}^{-1}$ . The positive value means naphthalene is less stable than its elements. It is a small difference of large numbers: rounding the heat capacity to $10.00\ \text{kJ K}^{-1}$ first moves it by about $0.7\ \text{kJ mol}^{-1}$ , so the answer box accepts $\pm 1\ \text{kJ mol}^{-1}$ .

Watch out. A bomb calorimeter measures $\Delta U$ . Forgetting the $\Delta n_g RT$ term here changes the answer by about $5\ \text{kJ mol}^{-1}$ .

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