Question

Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.274 g of naphthalene burns in a bomb calorimeter, the temperature rises from 26.214 ∘C to 30.284 ∘C.

Find ΔrH for the combustion of naphthalene at 298 K. When considering phase, assume all reactants and products are at 298 K.

Answer 1

Answer on Question #81387, Chemistry/ General Chemistry

Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.274 g of naphthalene burns in a bomb calorimeter, the temperature rises from 26.214 °C to 30.284 °C.

Find $\Delta \mathrm{rH}$ for the combustion of naphthalene at 298 K. When considering phase, assume all reactants and products are at 298 K.

Solution

It seems like some data is missing in the task like heat capacity of calorimeter, but we can take some value as an example to show how to calculate $\Delta_{\mathrm{rH}}$, for example, let heat capacity of calorimeter to be $5.11\,\mathrm{kJ}/^{\circ}\mathrm{C}$.

Then, the amount of heat released upon combustion to calorimeter:

$$q_{\mathrm{cal}} = (\text{heat capacity of calorimeter}) \times \Delta T = 5.11\,\mathrm{kJ}/^{\circ}\mathrm{C} \times (30.284 - 26.214)^{\circ}\mathrm{C} = 20.78\,\mathrm{kJ}$$ $$q_{\mathrm{rxn}} = -q_{\mathrm{cal}}$$ $$q_{\mathrm{rxn}} = -20.78\,\mathrm{kJ}$$ $$n(C_{10}H_8) = \frac{m}{M} = 1.274\,\mathrm{g}/128\,\mathrm{g/mol} = 0.00995\,\mathrm{mol}$$ $$q_{\mathrm{rxn}}(\text{per mole}) = \frac{q_{\mathrm{rxn}}}{n(C_{10}H_8)} = -20.78\,\mathrm{kJ}/0.00995\,\mathrm{mol} = -2088.44\,\mathrm{kJ/mol}$$ $$\begin{array}{l} C_{10}H_8(s) + 12\,O_2(g) \rightarrow 10\,CO_2(g) + 4H_2O(l) = -2088.44\,\mathrm{kJ} \\ \Delta n = 10 - 12 = -2\,\mathrm{mol} \\ \Delta H = \Delta E + \Delta nRT = -2088440\,\mathrm{J} + (-2\,\mathrm{mol}) \times 8.314\,\mathrm{J/mol} \times K \times 298\,\mathrm{K} = -2093395\,\mathrm{J} = -2093\,\mathrm{kJ} \\ \end{array}$$

Answer: -2093 kJ

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