Four grams of graphite is burnt in a bomb calorimeter of heat capacity `30kJK^(-1)` is excess of oxygen at 1 atmospheric pressure. The temperature rises from 300 to 304 K. What is the enthalpy of combustion of graphite (in kJ `"mol"^(-1)`)?

Options

A. 360
B. 1440
C. `-360`
D. `-1440`

Explanation

To find the enthalpy of combustion of graphite, we can follow these steps:

### Step 1: Calculate the change in temperature (ΔT)
The temperature rises from 300 K to 304 K.
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 304 \, K - 300 \, K = 4 \, K \]

### Step 2: Use the heat capacity of the calorimeter (C)
The heat capacity of the bomb calorimeter is given as \(C = 30 \, \text{kJ/K}\).

### Step 3: Calculate the heat absorbed by the calorimeter (q)
Using the formula:
\[ q = C \cdot \Delta T \]
Substituting the values:
\[ q = 30 \, \text{kJ/K} \cdot 4 \, K = 120 \, \text{kJ} \]

### Step 4: Calculate the number of moles of graphite burnt
The molar mass of graphite (carbon) is approximately \(12 \, \text{g/mol}\). Given that 4 g of graphite is burnt:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \, \text{g}}{12 \, \text{g/mol}} = \frac{1}{3} \, \text{mol} \]

### Step 5: Calculate the enthalpy of combustion (ΔH)
The enthalpy of combustion can be calculated using the formula:
\[ \Delta H = \frac{q}{\text{number of moles}} \]
Substituting the values:
\[ \Delta H = \frac{120 \, \text{kJ}}{\frac{1}{3} \, \text{mol}} = 120 \, \text{kJ} \cdot 3 = 360 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of combustion of graphite is \(-360 \, \text{kJ/mol}\) (the negative sign indicates that it is an exothermic reaction).

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