Enthalpy of Combustion from Bomb Calorimeter Experiment

To find the numerical value of the enthalpy of combustion of the gas, we proceed with the following steps:

1. Calculate the number of moles of the gas:
   The mass of the gas is $m = 3.5\text{ g}$ and its molar mass is $M = 28\text{ g mol}^{-1}$.
   $n = \frac{m}{M} = \frac{3.5\text{ g}}{28\text{ g mol}^{-1}} = 0.125\text{ mol}$

2. Calculate the heat absorbed by the calorimeter:
   The heat capacity of the calorimeter is $C = 2.5\text{ kJ K}^{-1}$, and the rise in temperature is:
   $\Delta T = 298.45\text{ K} - 298.0\text{ K} = 0.45\text{ K}$

   The total heat absorbed by the calorimeter is:
   $q_v = C \Delta T = 2.5\text{ kJ K}^{-1} \times 0.45\text{ K} = 1.125\text{ kJ}$

3. Determine the internal energy change of combustion ($\Delta U$):
   Since the reaction is carried out in a constant volume calorimeter (bomb calorimeter), the heat released during the reaction is equal to the change in internal energy:
   $\Delta U = -\frac{q_v}{n} = -\frac{1.125\text{ kJ}}{0.125\text{ mol}} = -9\text{ kJ mol}^{-1}$

4. Enthalpy of combustion ($\Delta H$):
   The relationship between enthalpy of combustion and internal energy change is:
   $\Delta H = \Delta U + \Delta n_g RT$
   Since the specific molecular formula of the gas is not provided, the $\Delta n_g RT$ contribution is taken to be negligible (or $\Delta n_g \approx 0$), so:
   $\Delta H \approx \Delta U = -9\text{ kJ mol}^{-1}$

Taking the numerical value (magnitude) of the enthalpy of combustion:
$\text{Numerical value} = 9$