Calculate the enthalpy of formation of methane from the following data : `C(s) + O_(2)(g) to CO_(2)(g) Delta_(r)H^(@) = -393.5 kJ`
`2H_(2)(g) + O_(2)(g) to 2H_(2)O(l) Delta_(r)H^(@) = -571.8 kJ`
`CH_(4)(g) + 2O_(2)(g) to CO_(2)(g) + 2H_(2)O(l) Delta_(r)H^(@) = -890.3 kJ`

Explanation

To calculate the enthalpy of formation of methane (CH₄) from the given data, we will 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.

### Given Reactions:
1. \( C(s) + O_2(g) \rightarrow CO_2(g) \)
\(\Delta_rH^(@) = -393.5 \text{ kJ}\)

2. \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \)
\(\Delta_rH^(@) = -571.8 \text{ kJ}\)

3. \( CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \)
\(\Delta_rH^(@) = -890.3 \text{ kJ}\)

### Required Reaction:
To find the enthalpy of formation of methane, we need the reaction:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

### Steps to Solve:

1. **Reverse Reaction 3**:
We need to reverse the third reaction because we want to form CH₄ from its elements.
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \]
\(\Delta_rH^(@) = +890.3 \text{ kJ}\)

2. **Add Reactions 1 and 2**:
We will use the first two reactions as they are.
\[ C(s) + O_2(g) \rightarrow CO_2(g) \]
\(\Delta_rH^(@) = -393.5 \text{ kJ}\)
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]
\(\Delta_rH^(@) = -571.8 \text{ kJ}\)

3. **Combine the Reactions**:
Now we will add the modified Reaction 3 to the first two reactions:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \]
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \]

When we add these reactions, we can cancel out the common terms:
- The \( CO_2(g) \) cancels out.
- The \( 2H_2O(l) \) cancels out with the products of the modified reaction.

4. **Final Reaction**:
After cancellation, we are left with:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

5. **Calculate the Enthalpy Change**:
Now we can calculate the enthalpy change for the formation of methane:
\[ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
\[ \Delta H = (-393.5 \text{ kJ}) + (-571.8 \text{ kJ}) + (+890.3 \text{ kJ}) \]
\[ \Delta H = -393.5 - 571.8 + 890.3 = -75.0 \text{ kJ} \]

### Conclusion:
The enthalpy of formation of methane (CH₄) is \(-75.0 \text{ kJ/mol}\).

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