0.16 g of methane was subjected to combustion at `27^(@)` C in a bomb calorimeter. The temperature of calorimeter system (including water) was found to rise by `0.5^(@)` C. Calculate the heat of combustion of methane at (i) constant volume and (ii) constant pressure. The thermal capacity of the calorimeter system is `17.7 kJ K^(-1)`.

Answer

To solve the problem of calculating the heat of combustion of methane at constant volume and constant pressure, we will follow these steps:

### Step-by-Step Solution

1. **Write the Combustion Reaction**:
The balanced chemical equation for the combustion of methane (CH₄) is:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]

2. **Calculate the Heat Released at Constant Volume (ΔE)**:
The heat released during the combustion can be calculated using the formula:
\[ Q = C \times \Delta T \]
where:
- \( Q \) is the heat absorbed by the calorimeter,
- \( C \) is the heat capacity of the calorimeter system (given as 17.7 kJ/K),
- \( \Delta T \) is the change in temperature (given as 0.5 °C).

Substituting the values:
\[ Q = 17.7 \, \text{kJ/K} \times 0.5 \, \text{K} = 8.85 \, \text{kJ} \]

3. **Convert Heat to Per Mole Basis**:
Since we need the heat of combustion per mole of methane, we first need to calculate the number of moles of methane in 0.16 g. The molar mass of methane (CH₄) is approximately 16 g/mol.
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.16 \, \text{g}}{16 \, \text{g/mol}} = 0.01 \, \text{mol} \]

Now, we can find the heat of combustion per mole:
\[ \Delta E = \frac{Q}{\text{Number of moles}} = \frac{8.85 \, \text{kJ}}{0.01 \, \text{mol}} = 885 \, \text{kJ/mol} \]
Since this is an exothermic reaction, we take it as negative:
\[ \Delta E = -885 \, \text{kJ/mol} \]

4. **Calculate the Heat of Combustion at Constant Pressure (ΔH)**:
We can relate the change in enthalpy (ΔH) to the change in internal energy (ΔE) using the formula:
\[ \Delta H = \Delta E + \Delta n \cdot R \cdot T \]
where:
- \( \Delta n \) is the change in the number of moles of gas,
- \( R \) is the universal gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K)),
- \( T \) is the temperature in Kelvin (27 °C = 300 K).

For the reaction:
- Moles of gaseous products = 1 (CO₂) + 2 (H₂O) = 3
- Moles of gaseous reactants = 1 (CH₄) + 2 (O₂) = 3
- Thus, \( \Delta n = 1 - 3 = -2 \).

Now substituting the values:
\[ \Delta H = -885 \, \text{kJ/mol} + (-2) \cdot (0.008314 \, \text{kJ/(mol·K)}) \cdot (300 \, \text{K}) \]
\[ \Delta H = -885 \, \text{kJ/mol} - 4.99 \, \text{kJ/mol} = -889.99 \, \text{kJ/mol} \]

### Final Answers
- **Heat of combustion at constant volume (ΔE)**: \(-885 \, \text{kJ/mol}\)
- **Heat of combustion at constant pressure (ΔH)**: \(-889.99 \, \text{kJ/mol}\)