0.3 g of ethane undergoes combustion at $$27°C$$ in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by $$0.5°C$$. The heat evolved during combustion of ethane at constant pressure is ______ $$\text{kJmol}^{-1}$$.
(Nearest integer)
[Given : The heat capacity of the calorimeter system is $$20 \text{ kJ K}^{-1}$$, $$R = 8.3 \text{ JK}^{-1} \text{ mol}^{-1}$$. Atomic mass of C and H are 12 and 1 $$\text{g mol}^{-1}$$ respectively]

Molar mass of ethane $$C_2H_6$$ is $$2\times12 + 6\times1 = 30\;\text{g mol}^{-1}$$.

Given mass burnt = $$0.3\;\text{g}$$,
moles burnt $$n = \frac{0.3}{30} = 0.01\;\text{mol}$$.

The bomb calorimeter works at constant volume, so the heat measured equals $$\Delta U$$ for the combustion.
Heat absorbed by calorimeter system:
$$q_V = C_{\text{cal}} \,\Delta T = 20\;\text{kJ K}^{-1}\times0.5\;\text{K} = 10\;\text{kJ}$$.

This 10 kJ is released by 0.01 mol, therefore
$$\Delta U_{\text{comb}} = \frac{-10\;\text{kJ}}{0.01\;\text{mol}} = -1000\;\text{kJ mol}^{-1}$$ $$-(1)$$

For the constant-pressure enthalpy change we use $$\Delta H = \Delta U + \Delta n_{\text{g}}RT$$.

Balanced combustion (water liquid):
$$C_2H_6(g) + \frac72 O_2(g) \rightarrow 2 CO_2(g) + 3 H_2O(l)$$.

Gas moles: reactants $$4.5$$, products $$2$$.
Hence $$\Delta n_{\text{g}} = 2 - 4.5 = -2.5$$.

Temperature $$T = 27^\circ\text{C} = 300\;\text{K}$$,
gas constant $$R = 8.3\;\text{J K}^{-1}\text{ mol}^{-1}=0.0083\;\text{kJ K}^{-1}\text{ mol}^{-1}$$.

Compute $$\Delta n_{\text{g}}RT$$:
$$\Delta n_{\text{g}}RT = (-2.5)(0.0083)(300) = -6.225\;\text{kJ mol}^{-1}$$ $$-(2)$$

Now
$$\Delta H_{\text{comb}} = -1000\;\text{kJ mol}^{-1} + (-6.225\;\text{kJ mol}^{-1}) \approx -1006.2\;\text{kJ mol}^{-1}$$.

Magnitude of heat evolved at constant pressure = $$1006\;\text{kJ mol}^{-1}$$ (nearest integer).