In a constant volume calorimeter, `3.5 g` of a gas with molecular weight `28` was burnt in excess oxygen at `298.0 K`. The temperature of the calorimeter was found to increase from `298.0 K to 298.45 K` due to the combustion process. Given that the heat capacity of the calorimeter is `2.5 kJ K^(-1)`, find the numerical value for the enthalpy of combustion of the gas in `kJ mol^(-1)`

Answer

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

### Step 1: Calculate the number of moles of the gas
The number of moles (n) can be calculated using the formula:
\[ n = \frac{\text{mass of gas}}{\text{molecular weight of gas}} \]
Given:
- Mass of gas = 3.5 g
- Molecular weight of gas = 28 g/mol

Substituting the values:
\[ n = \frac{3.5 \, \text{g}}{28 \, \text{g/mol}} = 0.125 \, \text{mol} \]

### Step 2: Calculate the temperature change (ΔT)
The temperature change (ΔT) is given by:
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \]
Given:
- \(T_{\text{initial}} = 298.0 \, \text{K}\)
- \(T_{\text{final}} = 298.45 \, \text{K}\)

Calculating ΔT:
\[ \Delta T = 298.45 \, \text{K} - 298.0 \, \text{K} = 0.45 \, \text{K} \]

### Step 3: Calculate the heat absorbed by the calorimeter (q)
The heat absorbed by the calorimeter can be calculated using the formula:
\[ q = C \cdot \Delta T \]
Where:
- \(C\) is the heat capacity of the calorimeter (2.5 kJ/K)

Substituting the values:
\[ q = 2.5 \, \text{kJ/K} \cdot 0.45 \, \text{K} = 1.125 \, \text{kJ} \]

### Step 4: Calculate the enthalpy of combustion per mole
The enthalpy of combustion (ΔH) can be calculated using the formula:
\[ \Delta H = \frac{q}{n} \]
Substituting the values:
\[ \Delta H = \frac{1.125 \, \text{kJ}}{0.125 \, \text{mol}} = 9 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of combustion of the gas is:
\[ \Delta H = 9 \, \text{kJ/mol} \]

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