If `C(s)+O_(2)(g)to CO_(2)(g)+94.2` Kcal
`H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l)+68.3` Kcal
`CH_(4)(g)+2O_(2)(g)to CO_(2)(g)+2H_(2)O+210.8` Kcal
then the heat of formation of methane will be

Options

A. 47.3 Kcal
B. `-20.0` Kcal
C. 45.9 Kcal
D. `-47.3` Kcal

Answer

To find the heat 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 reactions. We will use the given reactions and their corresponding heat changes to derive the heat of formation for methane.

### Step-by-Step Solution:

1. **Write down the given reactions and their enthalpy changes:**
- Reaction 1: \( C(s) + O_2(g) \rightarrow CO_2(g) \) \(\Delta H_1 = -94.2 \, \text{kcal}\)
- Reaction 2: \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \) \(\Delta H_2 = -68.3 \, \text{kcal}\)
- Reaction 3: \( CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \) \(\Delta H_3 = -210.8 \, \text{kcal}\)

2. **Identify the heat of formation of the products:**
- The heat of formation of \( CO_2(g) \) is given by Reaction 1:
\[ \Delta H_f (CO_2) = -94.2 \, \text{kcal} \]
- The heat of formation of \( H_2O(l) \) is given by Reaction 2:
\[ \Delta H_f (H_2O) = -68.3 \, \text{kcal} \]

3. **Set up the equation using Hess's law for Reaction 3:**
- The enthalpy change for Reaction 3 can be expressed in terms of the heats of formation:
\[ \Delta H_3 = \Delta H_f (CO_2) + 2 \Delta H_f (H_2O) - \Delta H_f (CH_4) \]
- Plugging in the known values:
\[ -210.8 \, \text{kcal} = -94.2 \, \text{kcal} + 2(-68.3 \, \text{kcal}) - \Delta H_f (CH_4) \]

4. **Calculate the total heat of formation for the products:**
- Calculate \( 2 \times -68.3 \):
\[ 2 \times -68.3 = -136.6 \, \text{kcal} \]
- Substitute this back into the equation:
\[ -210.8 = -94.2 - 136.6 - \Delta H_f (CH_4) \]

5. **Combine and solve for \( \Delta H_f (CH_4) \):**
- Combine the terms on the right:
\[ -210.8 = -230.8 - \Delta H_f (CH_4) \]
- Rearranging gives:
\[ \Delta H_f (CH_4) = -230.8 + 210.8 \]
- Calculate:
\[ \Delta H_f (CH_4) = -20.0 \, \text{kcal} \]

### Final Answer:
The heat of formation of methane (CH₄) is \(-20.0 \, \text{kcal}\).