The enthalpy of combustion of methane, graphite and dihydrogen at `298 K` are, `-890.3 kJ mol^(-1) -393.5 kJ mol^(-1)`, and `-285.8 kJ mol^(-1)` respectively. Enthapy of formation of `CH_(4)(g)` will be

Options

A. `-74.8 kJ mol^(-1)`
B. `-52.27 kJ mol^(-1)`
C. `+74.8 kJ mol^(-1)`
D. `+52.26 kJ mol^(-1)`

Explanation

To find the enthalpy of formation of methane (CH₄), we can use the enthalpy of combustion values for methane, graphite, and dihydrogen. Here’s a step-by-step solution:

### Step 1: Write the combustion reactions
We need to write the balanced equations for the combustion of methane, graphite, and dihydrogen.

1. **Combustion of methane (CH₄)**:
\[
\text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(l) \quad \Delta H = -890.3 \, \text{kJ/mol}
...