For complete combustion of methanol
$$CH_3OH(l) + \frac{3}{2}O_2(g) \to CO_2(g) + 2H_2O(l)$$
the amount of heat produced as measured by bomb calorimeter is $$726$$ kJ mol$$^{-1}$$ at $$27°$$C. The enthalpy of combustion for the reaction is $$-x$$ kJ mol$$^{-1}$$, where $$x$$ is ______
(Given : $$R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$)

We need to find the enthalpy of combustion of methanol given that the heat produced in a bomb calorimeter is 726 kJ/mol at 27°C.

Understand the relationship between $$\Delta U$$ and $$\Delta H$$

A bomb calorimeter measures the heat at constant volume, which gives the internal energy change ($$\Delta U$$).

$$\Delta U = -726$$ kJ/mol (negative because combustion is exothermic)

The enthalpy of combustion ($$\Delta H$$) is measured at constant pressure. The relationship is:

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

where $$\Delta n_g$$ = change in the number of moles of gaseous species.

Calculate $$\Delta n_g$$

The reaction is: $$CH_3OH(l) + \frac{3}{2}O_2(g) \to CO_2(g) + 2H_2O(l)$$

Moles of gaseous products = 1 (only $$CO_2$$ is gas; $$H_2O$$ is liquid)

Moles of gaseous reactants = $$\frac{3}{2}$$ (only $$O_2$$ is gas; $$CH_3OH$$ is liquid)

$$\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}$$

Calculate $$\Delta H$$

Temperature T = 27°C = 300 K

R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$ = 0.0083 kJ K$$^{-1}$$ mol$$^{-1}$$

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

$$= -726 + \left(-\frac{1}{2}\right)(0.0083)(300)$$

$$= -726 + (-0.5)(2.49)$$

$$= -726 - 1.245$$

$$= -727.245$$ kJ/mol

Find x

Since $$\Delta H = -x$$ kJ/mol:

$$x = 727.245 \approx 727$$

The correct answer is 727.