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 provided for methane, graphite (carbon), and dihydrogen (H₂). The enthalpy of formation can be derived from the enthalpy of combustion reactions by applying Hess's law.

### Step-by-Step Solution:

1. **Write the Reaction for the Formation of Methane:**
The formation of methane from its elements can be represented as:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

2. **Write the Combustion Reactions:**
We have the following combustion reactions:
- For methane:
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad (\Delta H = -890.3 \, \text{kJ/mol}) \]
- For graphite (carbon):
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad (\Delta H = -393.5 \, \text{kJ/mol}) \]
- For dihydrogen:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad (\Delta H = -285.8 \, \text{kJ/mol}) \]

3. **Manipulate the Reactions to Find the Enthalpy of Formation:**
We need to arrange these reactions to derive the formation reaction of methane. We will reverse the combustion reactions of carbon and hydrogen since we want to form CH₄ from its elements.

- Reverse the combustion of carbon:
\[ CO_2(g) \rightarrow C(s) + O_2(g) \quad (\Delta H = +393.5 \, \text{kJ/mol}) \]
- Reverse the combustion of dihydrogen:
\[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \quad (\Delta H = +285.8 \, \text{kJ/mol}) \]

4. **Add the Reactions:**
Now, we can add the reactions:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad (+393.5 \, \text{kJ/mol}) \]
\[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \quad (+285.8 \, \text{kJ/mol}) \]
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad (-890.3 \, \text{kJ/mol}) \]

When we add these reactions, we get:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

5. **Calculate the Enthalpy of Formation:**
The enthalpy change for the formation of methane can be calculated as:
\[ \Delta H_f = \Delta H_{combustion \, of \, CH_4} + \Delta H_{combustion \, of \, C} + \Delta H_{combustion \, of \, H_2} \]
\[ \Delta H_f = (-890.3) + (393.5 + 285.8) \]
\[ \Delta H_f = -890.3 + 393.5 + 285.8 \]
\[ \Delta H_f = -890.3 + 679.3 \]
\[ \Delta H_f = -211.0 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of formation of methane (CH₄) at 298 K is \(-211.0 \, \text{kJ/mol}\).