The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are `- 890 .3 kJmol^(-1), -393.5kJ mol^(-1)` and `- 28.5 kJ mol^(-1)` respectively. Enthalpy of formation of `CH_(4)(g)` will be `:`
(i) `-74.8 kJ mol^(-1)` (ii) `-52.27 kJ mol^(-1)` (iii) `+74.8 kJ mol^(-1)` (iv) `+ 52 .26 kJ mol^(-1)`

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 Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Write the Reaction for Enthalpy of Formation**:
The enthalpy of formation of methane can be represented by the following reaction:
\[
\text{C (s) + 2 H}_2 \text{(g) } \rightarrow \text{CH}_4 \text{(g)}
...