1. What is measured in a bomb calorimeter?

In a bomb calorimeter, the reaction occurs at constant volume, so the heat measured is:

$q_v = \Delta U$

Given that the combustion of ethanol produces heat of magnitude $1364.47\,\text{kJ mol}^{-1}$, the internal energy change is:

$\Delta U = -1364.47\,\text{kJ mol}^{-1}$

(The sign is negative because heat is released.)

2. Relation between enthalpy and internal energy

For reactions involving gases,

$\Delta H = \Delta U + \Delta n_g RT$

where:

• $\Delta n_g =$ moles of gaseous products $-$ moles of gaseous reactants
• $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1} = 0.008314\,\text{kJ mol}^{-1}\text{K}^{-1}$
• $T = 25^\circ C = 298\,\text{K}$

3. Calculate $\Delta n_g$

Reaction:

$C_2H_5OH(l) + 3O_2(g) \to 2CO_2(g) + 3H_2O(l)$

Only gaseous species count:

• Gaseous products = $2$ (from $2CO_2$)
• Gaseous reactants = $3$ (from $3O_2$)

So,

$\Delta n_g = 2 - 3 = -1$

4. Compute $\Delta n_g RT$

$\Delta n_g RT = (-1)(0.008314)(298)$

$= -2.4776\,\text{kJ mol}^{-1} \approx -2.48\,\text{kJ mol}^{-1}$

5. Calculate enthalpy of combustion

$\Delta H = \Delta U + \Delta n_g RT$

$\Delta H = -1364.47 + (-2.48)$

$\Delta H = -1366.95\,\text{kJ mol}^{-1}$

6. Match with options

The correct option is:

$\boxed{\text{C: } -1366.95\,\text{kJ mol}^{-1}}$