Topic of the Question:
This question deals with thermochemistry, focusing on the interpretation of calorimetric measurements and the calculation of reaction enthalpy from constant-volume heat data.
Step 1 : Understanding the Question:
The combustion of liquid ethanol is studied in a bomb calorimeter at $25^\circ\text{C}$, yielding a heat output of $1364.47 \text{ kJ mol}^{-1}$. We need to determine the standard enthalpy of combustion ($\Delta H_C$) for this reaction.
Step 2 : Key Formulas and Approach:
A bomb calorimeter operates under rigid, constant-volume conditions. Therefore, the heat released in this device is equal to the internal energy change of the reaction: $q_v = \Delta U$.
Since the reaction is exothermic (heat is produced), the internal energy change is negative: $\Delta U = -1364.47 \text{ kJ mol}^{-1}$.
We relate the enthalpy change to the internal energy change using the formula: $\Delta H_C = \Delta U + \Delta n_g RT$.
$\Delta n_g$ is the difference between the stoichiometric coefficients of gaseous products and gaseous reactants.
The temperature must be converted to Kelvin: $T = 25^\circ\text{C} + 273.15 = 298.15\text{ K}$.
Step 3 : Detailed Explanation:
We analyze the balanced equation for the complete combustion of ethanol: $\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$.
The products include $2 \text{ moles of } \text{CO}_2(g)$ and $3 \text{ moles of } \text{H}_2\text{O}(l)$. The reactants include $1 \text{ mole of } \text{C}_2\text{H}_5\text{OH}(l)$ and $3 \text{ moles of } \text{O}_2(g)$.
Only the gaseous species are considered for calculating $\Delta n_g$: $\Delta n_g = n_{g}(\text{products}) - n_{g}(\text{reactants}) = 2 - 3 = -1$.
We write the gas constant in terms of kilojoules: $R = 8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}$.
We substitute these parameters into the main relationship: $\Delta H_C = -1364.47 + [(-1) \times (8.314 \times 10^{-3}) \times 298.15]$.
Evaluating the second term: $\Delta n_g RT = -1 \times 0.008314 \times 298.15 \approx -2.4788 \text{ kJ mol}^{-1}$.
Adding this to the internal energy change gives: $\Delta H_C = -1364.47 - 2.4788 \approx -1366.95 \text{ kJ mol}^{-1}$.
Step 4 : Final Answer:
The standard enthalpy of combustion for ethanol under these conditions is $-1366.95 \text{ kJ mol}^{-1}$, which matches Option (A).