Heat of formation of `CH_(4)` are:
If given heat:
`C(s)+ O_(2)(g) rarr CO_(2) (g) " "DeltaH =-394 KJ`
`2H_(2) (g)+ O_(2)(g) rarr 2H_(2)O(l) rarr 2H_(2)O(l) " "DeltaH =-568 KJ`
`CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l) " " DeltaH =- 892 KJ`

Options

A. `-70 KJ`
B. `-16.7 Kcal`
C. `-244 KJ`
D. `-50 Kcal`

Explanation

To find the heat of formation of methane (CH₄), we will manipulate the given reactions and their corresponding enthalpy changes (ΔH). Here’s the step-by-step solution:

### Step 1: Write down the given reactions and their ΔH values.

1. \( C(s) + O_2(g) \rightarrow CO_2(g) \) \(\Delta H = -394 \, \text{kJ}\)
2. \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \) \(\Delta H = -568 \, \text{kJ}\)
3. \( CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \) \(\Delta H = -892 \, \text{kJ}\)

### Step 2: Identify the desired reaction.

The desired reaction for the formation of methane is:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

### Step 3: Manipulate the given reactions.

To derive the desired reaction, we can add the first two reactions and then subtract the third reaction.

- **Add Reaction 1 and Reaction 2:**
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \text{(1)} \]
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \text{(2)} \]
Adding these gives:
\[ C(s) + O_2(g) + 2H_2(g) \rightarrow CO_2(g) + 2H_2O(l) \]

- **Subtract Reaction 3:**
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \text{(3)} \]
When we subtract this reaction, we reverse it:
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \]

### Step 4: Combine the reactions.

Now, combining the manipulated reactions:
\[ C(s) + O_2(g) + 2H_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \text{(from 1 and 2)} \]
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \quad \text{(reversed 3)} \]

This gives us:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

### Step 5: Calculate ΔH for the desired reaction.

Now we need to calculate the ΔH for the desired reaction:
\[ \Delta H = \Delta H_1 + \Delta H_2 - \Delta H_3 \]
Substituting the values:
\[ \Delta H = (-394 \, \text{kJ}) + (-568 \, \text{kJ}) - (-892 \, \text{kJ}) \]
\[ \Delta H = -394 - 568 + 892 \]
\[ \Delta H = -70 \, \text{kJ} \]

### Step 6: Convert ΔH to kilocalories.

To convert kJ to kilocalories:
\[ \Delta H = -70 \, \text{kJ} \times 0.24 \, \text{kcal/kJ} = -16.8 \, \text{kcal} \]

### Final Answer:

The heat of formation of CH₄ is:
\[ \Delta H = -70 \, \text{kJ} \quad \text{or} \quad -16.8 \, \text{kcal} \]