The heat of combustion of `CH_(4(g)), C_((g))` and `H_(2(g))` at `25^(@)C` are `-212.4` K cal, -94.0 K cal and -68.4 K cal respectively, the heat of formation of `CH_(4)` will be -

Options

A. `+54.4` K cal
B. `-18.4` K cal
C. `-375.2` K cal
D. `+212.8` K cal

Answer

To find the heat of formation of methane (CH₄), we can use the heat of combustion values provided for CH₄, carbon (C), and hydrogen (H₂). The heat of formation can be calculated using the following equation:

\[ \Delta H_f^\circ (\text{CH}_4) = \Delta H_c^\circ (\text{C}) + 2 \Delta H_c^\circ (\text{H}_2) - \Delta H_c^\circ (\text{CH}_4) \]

Where:
- \(\Delta H_f^\circ\) is the heat of formation.
- \(\Delta H_c^\circ\) is the heat of combustion.

### Step-by-Step Solution:

1. **Identify the given values:**
- Heat of combustion of CH₄: \(\Delta H_c^\circ (\text{CH}_4) = -212.4 \, \text{kcal}\)
- Heat of combustion of C: \(\Delta H_c^\circ (\text{C}) = -94.0 \, \text{kcal}\)
- Heat of combustion of H₂: \(\Delta H_c^\circ (\text{H}_2) = -68.4 \, \text{kcal}\)

2. **Substitute the values into the equation:**
\[ \Delta H_f^\circ (\text{CH}_4) = (-94.0) + 2 \times (-68.4) - (-212.4) \]

3. **Calculate \(2 \times (-68.4)\):**
\[ 2 \times (-68.4) = -136.8 \]

4. **Now substitute this back into the equation:**
\[ \Delta H_f^\circ (\text{CH}_4) = -94.0 - 136.8 + 212.4 \]

5. **Combine the values:**
\[ \Delta H_f^\circ (\text{CH}_4) = -94.0 - 136.8 + 212.4 = -18.4 \, \text{kcal} \]

### Final Answer:
The heat of formation of CH₄ is \(-18.4 \, \text{kcal}\).

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