Heat of formation of `CO_(2)(g), H_(2)O(l)` and `CH_(4)(g)` are `-94.0, -68.4` and `-17.9` kcal respectively. The heat of combustion of methane is

Options

A. `-212.9` Kcal
B. `-136.8` Kcal
C. `-304.3` Kcal
D. `-105.2` Kcal

Answer

To find the heat of combustion of methane (\(CH_4\)), we can use the standard enthalpy of formation values for the products and reactants involved in the combustion reaction. The combustion of methane can be represented by the following balanced chemical equation:

\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \]

### Step-by-Step Solution:

1. **Identify the Heat of Formation Values**:
- Heat of formation of \(CO_2(g) = -94.0 \, \text{kcal}\)
- Heat of formation of \(H_2O(l) = -68.4 \, \text{kcal}\)
- Heat of formation of \(CH_4(g) = -17.9 \, \text{kcal}\)
- Heat of formation of \(O_2(g) = 0 \, \text{kcal}\) (elements in their standard state have a heat of formation of zero)

2. **Write the Formula for Heat of Combustion**:
The heat of combustion (\(\Delta H\)) can be calculated using the formula:
\[ \Delta H = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)} \]

3. **Substitute the Values into the Formula**:
For the combustion of methane:
\[ \Delta H = \left[\Delta H_f (CO_2) + 2 \times \Delta H_f (H_2O)\right] - \left[\Delta H_f (CH_4) + 2 \times \Delta H_f (O_2)\right] \]

Substituting the values:
\[ \Delta H = \left[-94.0 + 2 \times (-68.4)\right] - \left[-17.9 + 2 \times 0\right] \]

4. **Calculate the Values**:
- Calculate the products:
\[ \Delta H_f (products) = -94.0 + 2 \times (-68.4) = -94.0 - 136.8 = -230.8 \, \text{kcal} \]
- Calculate the reactants:
\[ \Delta H_f (reactants) = -17.9 + 0 = -17.9 \, \text{kcal} \]

5. **Final Calculation**:
Now, substituting back into the equation:
\[ \Delta H = -230.8 - (-17.9) = -230.8 + 17.9 = -212.9 \, \text{kcal} \]

### Conclusion:
The heat of combustion of methane is \(-212.9 \, \text{kcal}\).