Topic of the Question:
The topic of this question is chemical thermodynamics, focusing on the relationship between enthalpy change and internal energy change during a chemical reaction.
Step 1 : Understanding the Question:
We are given the enthalpy of reaction ($\Delta H$) for the combustion of liquid ethanol at $27^\circ\text{C}$ as $-1366.5 \text{ kJ mol}^{-1}$. We are asked to calculate the corresponding change in internal energy ($\Delta U$) for the reaction at this same temperature.
Step 2 : Key Formulas and Approach:
The thermodynamic relation between the enthalpy change ($\Delta H$) and the internal energy change ($\Delta U$) at a constant temperature $T$ is: $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging the formula to solve for the internal energy change gives: $\Delta U = \Delta H - \Delta n_g RT$.
The change in the number of moles of gaseous species is: $\Delta n_g = \sum n_{g}(\text{gaseous products}) - \sum n_{g}(\text{gaseous reactants})$.
The universal gas constant $R$ is $8.314 \text{ J mol}^{-1} \text{ K}^{-1}$, which must be converted to $\text{kJ mol}^{-1} \text{ K}^{-1}$ by dividing by $1000$.
Temperature must be converted from Celsius to Kelvin: $T(\text{K}) = T(^\circ\text{C}) + 273$.
Step 3 : Detailed Explanation:
We begin by analyzing the balanced chemical equation to calculate the change in gaseous moles: $\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$.
The only gaseous product is $2 \text{ moles of } \text{CO}_2(g)$, and the only gaseous reactant is $3 \text{ moles of } \text{O}_2(g)$. The liquid species, $\text{C}_2\text{H}_5\text{OH}(l)$ and $\text{H}_2\text{O}(l)$, are excluded.
This gives: $\Delta n_g = 2 - 3 = -1$.
Next, we convert the temperature to Kelvin: $T = 27 + 273 = 300 \text{ K}$.
We express the gas constant in terms of kilojoules: $R = 8.314 \times 10^{-3} \text{ kJ mol}^{-1} \text{ K}^{-1}$.
Now we substitute these values into our rearranged equation for internal energy change: $\Delta U = -1366.5 - [(-1) \times (8.314 \times 10^{-3}) \times 300]$.
Simplifying the terms: $\Delta U = -1366.5 + [8.314 \times 10^{-3} \times 300] = -1366.5 + 2.4942$.
This calculation yields: $\Delta U \approx -1364.0 \text{ kJ mol}^{-1}$.
Step 4 : Final Answer:
The internal energy change for the reaction is $-1364.0 \text{ kJ mol}^{-1}$, which matches Option (C).