To find the enthalpy of combustion of graphite, we need to follow these steps:
1. **Calculate the heat absorbed by the calorimeter:**
The heat absorbed by the calorimeter (\(q\)) can be calculated using the heat capacity (\(C\)) and the change in temperature (\(\Delta T\)):
\[
q = C \times \Delta T
\]
Here:
- \(C = 30 \text{ kJ/K}\)
- \(\Delta T = 304 \text{ K} - 300 \text{ K} = 4 \text{ K}\)
So:
\[
q = 30 \text{ kJ/K} \times 4 \text{ K} = 120 \text{ kJ}
\]
2. **Determine the moles of graphite burned:**
The molar mass of graphite (carbon) is approximately \(12 \text{ g/mol}\).
The number of moles of graphite burned (\(n\)) is:
\[
n = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{12 \text{ g/mol}} = \frac{1}{3} \text{ mol}
\]
3. **Calculate the enthalpy of combustion per mole:**
The enthalpy of combustion (\(\Delta H_{\text{comb}}\)) is the heat absorbed per mole of substance.
Since 120 kJ of heat is absorbed by burning \(\frac{1}{3}\) mol of graphite, the enthalpy of combustion per mole is:
\[
\Delta H_{\text{comb}} = \frac{q}{n}
\]
So:
\[
\Delta H_{\text{comb}} = \frac{120 \text{ kJ}}{\frac{1}{3} \text{ mol}} = 120 \text{ kJ} \times 3 = 360 \text{ kJ/mol}
\]
The enthalpy of combustion is typically reported as a negative value because combustion is an exothermic process. Therefore:
\[
\Delta H_{\text{comb}} = -360 \text{ kJ/mol}
\]
So, the correct answer is **(C) −360 kJ/mol**.
