1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation
C (graphite)`+O_(2)(g)rarrCO_(2)(g)`
During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?

Explanation

To solve the problem, we need to determine the enthalpy change (ΔH) for the combustion of 1 gram of graphite in a bomb calorimeter. Here are the steps to arrive at the solution:

### Step 1: Calculate the number of moles of graphite
The molar mass of carbon (graphite) is approximately 12 g/mol. Therefore, the number of moles (n) of 1 gram of graphite is calculated as follows:

\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{12 \text{ g/mol}} = \frac{1}{12} \text{ mol} \]
...