For complete combustion of ethanol, $$C_2H_5OH(l) + 3O_2(g) \to 2CO_2(g) + 3H_2O(l)$$, the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol$$^{-1}$$ at 25°C. Assuming ideality the Enthalpy of combustion, $$\Delta_cH$$, for the reaction will be: ($$R = 8.314$$ kJ mol$$^{-1}$$)

We are told that in a bomb calorimeter the heat released per mole of ethanol is $$1364.47\ \text{kJ}$$ at $$25^{\circ}\text{C}$$. Because the bomb calorimeter works at constant volume, the measured heat is the change in internal energy of reaction. So

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

(The negative sign is inserted because heat is produced; energy leaves the system.)

To convert this internal-energy change into the enthalpy change we recall the thermodynamic relation that links enthalpy and internal energy at the same temperature:

$$\Delta H = \Delta U + \Delta n_g\,R\,T.$$

Here $$\Delta n_g$$ is the difference between the number of moles of gaseous products and gaseous reactants, $$R$$ is the gas constant and $$T$$ is the absolute temperature corresponding to $$25^{\circ}\text{C}$$, i.e. $$T = 25 + 273 = 298\ \text{K}.$$

Let us evaluate $$\Delta n_g$$ for the balanced combustion equation

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

Gaseous reactants: $$3O_2$$ ⇒ $$3$$ mol
Gaseous products: $$2CO_2$$ ⇒ $$2$$ mol

Therefore

$$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - 3 = -1.$$

Next we calculate $$\Delta n_g\,R\,T$$. The usual value of the gas constant is $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} = 0.008314\ \text{kJ mol}^{-1}\text{K}^{-1}.$$ Substituting:

$$\Delta n_g R T = (-1)(0.008314\ \text{kJ mol}^{-1}\text{K}^{-1})(298\ \text{K})$$

$$\phantom{\Delta n_g R T} = -2.4777\ \text{kJ mol}^{-1} \approx -2.48\ \text{kJ mol}^{-1}.$$

Now we can obtain the enthalpy change:

$$\Delta H = \Delta U + \Delta n_gRT$$

$$\phantom{\Delta H} = (-1364.47\ \text{kJ mol}^{-1}) + (-2.48\ \text{kJ mol}^{-1})$$

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

Thus, the enthalpy of combustion of ethanol at $$25^{\circ}\text{C}$$ is

$$\boxed{\Delta_c H = -1366.95\ \text{kJ mol}^{-1}}.$$

Hence, the correct answer is Option A.