$$0.3$$ 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 ______ kJ mol$$^{-1}$$. (Nearest integer)
[Given: The heat capacity of the calorimeter system is $$20$$ kJ K$$^{-1}$$, R $$= 8.3$$ JK$$^{-1}$$ mol$$^{-1}$$. Assume ideal gas behaviour. Atomic mass of C and H are 12 and 1 g mol$$^{-1}$$ respectively]

We need to find the heat evolved during combustion of ethane at constant pressure.

Mass of ethane (C$$_2$$H$$_6$$) = 0.3 g, Temperature = 27°C = 300 K

Temperature rise = 0.5°C, Heat capacity of calorimeter = 20 kJ/K

Calculate heat at constant volume ($$q_V$$).

Molar mass of C$$_2$$H$$_6$$ = $$2(12) + 6(1) = 30$$ g/mol

Moles of ethane = $$\frac{0.3}{30} = 0.01$$ mol

$$q_V = C \times \Delta T = 20 \times 0.5 = 10$$ kJ (for 0.01 mol)

$$q_V$$ per mole = $$\frac{10}{0.01} = 1000$$ kJ/mol

Since combustion is exothermic: $$\Delta U = -1000$$ kJ/mol

Write the combustion reaction and find $$\Delta n_g$$.

$$ \text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g) \to 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) $$

$$\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}$$

$$\Delta n_g = 2 - \left(1 + \frac{7}{2}\right) = 2 - 4.5 = -2.5$$

Convert to constant pressure using $$\Delta H = \Delta U + \Delta n_g RT$$.

$$ \Delta H = -1000 + (-2.5)(8.3 \times 10^{-3})(300) $$ $$ \Delta H = -1000 + (-2.5)(2.49) $$ $$ \Delta H = -1000 - 6.225 = -1006.225 \text{ kJ/mol} $$

Find the heat evolved.

Heat evolved = $$|\Delta H| \approx 1006$$ kJ/mol (nearest integer).

The answer is $$\boxed{1006}$$.