Calculating heat of combustion in a bomb calorimeter
In a constant volume bomb-calorimeter, we can make chemical reactions take place. If we let chemical reactions freely take place in it, then the total internal energy change is zero. This is due to the fact there is no expansion work since the system is a constant volume by definition and the heat transferred is zero since the bomb calorimeter is well insulated. The heat released in a reaction is automatically absorbed by the bomb calorimeter device. Hence the total internal energy change is zero.
Now, consider another path of the reaction. We start with reactants and turn them into products under constant volume and constant temperature conditions (*) and then these products we raise the temperature as to reach the final temperature we got in the previous path.
The energy associated with moving from reactants to products isothermally is $\Delta U_{rcxtn}$ and the energy change for raising up the products to the temperature of first path is $ \Delta U_{heat}$
Now, we can write:
$$ \Delta U_{heat} = C_{v}^{cal} \Delta T$$
And,
$$ \Delta U_{rcxtn} + \Delta  U_{heat} = 0$$
Or,
$$ \Delta U_{rcxtn} = - C_{v}^{cal} \Delta T$$
Now, consider the enthalpy change of reaction:
$$ \Delta H_{rcxtn} = \Delta U_{rcxtn} + \Delta (PV)$$
Now, if we were to assume our reactants as ideal gases a, then $PV=nRT$ and also considering constant temperature, we get:
$$ \Delta (PV) = \Delta (nRT) = RT \Delta n$$
Using the above result and expression for internal energy, we finally arrive at the following equation:
$$ \Delta H_{rcxtn} =  - C_{v}^{cal} \Delta T + RT \Delta n$$

Refernces:
Around 42:49 of this video
Page-2 of these notes