Calculate the heat of formation of methane in `kcalmol^(-1)` using the following thermo chemical reactions
`C(s)+O_(2)toCO_(2)(g) , DeltaH=-94.2 kcal mol^(-1)`
`H_(2)(g)+1/2O_(2)(g)toH_(2)O(l) , DeltaH=-68.3 kcal mol^(-1)`
`CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l) , DeltaH=-210.8 kcal mol^(-1)`

Explanation

To calculate the heat of formation of methane (CH₄) using the given thermochemical reactions, we will follow these steps:

### Step 1: Write down the given reactions and their enthalpy changes.

1. \( C(s) + O_2(g) \rightarrow CO_2(g) \)
\(\Delta H = -94.2 \, \text{kcal/mol}\)

2. \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \)
\(\Delta H = -68.3 \, \text{kcal/mol}\)

3. \( CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \)
\(\Delta H = -210.8 \, \text{kcal/mol}\)

### Step 2: Manipulate the reactions to form methane.

We need to form methane from its elements, which can be represented as:

\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

To achieve this, we will manipulate the given reactions:

- **Multiply the second reaction by 2** to get the formation of 2 moles of water:

\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \Delta H = 2 \times (-68.3) = -136.6 \, \text{kcal/mol} \]

- **Add the modified second reaction to the first reaction**:

\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -94.2 \, \text{kcal/mol} \]

Adding these two reactions gives:

\[ C(s) + 2H_2(g) + O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H = -94.2 + (-136.6) = -230.8 \, \text{kcal/mol} \]

### Step 3: Subtract the third reaction from the combined reactions.

Now, we will subtract the third reaction from the combined reactions:

\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H = -210.8 \, \text{kcal/mol} \]

When we subtract this reaction, we have:

\[ (C(s) + 2H_2(g) + O_2(g)) - (CH_4(g) + 2O_2(g)) \rightarrow CO_2(g) + 2H_2O(l) - (CO_2(g) + 2H_2O(l)) \]

This simplifies to:

\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

### Step 4: Calculate the heat of formation of methane.

Now, we can find the heat of formation of methane:

\[ \Delta H_{f} = -230.8 \, \text{kcal/mol} - (-210.8 \, \text{kcal/mol}) = -20.0 \, \text{kcal/mol} \]

### Final Answer:

The heat of formation of methane is:

\[ \Delta H_{f} = -20.0 \, \text{kcal/mol} \]

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