Given that
`C(s)+O_(2)(g)rarrCO_(2)(g), DeltaH=-394 kJ`
`2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l), DeltaH=-568kJ`
`CH_(4)(g)+2O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l)DeltaH=-892 kJ`
Heat of formation of `CH_(4)` is

Options

A. `-70` kJ
B. 71.8 kJ
C. `-244` kJ
D. `+782` kJ

Explanation

To find the heat of formation of methane (CH₄), we need to manipulate the given thermochemical equations. The heat of formation of a compound is defined as the change in enthalpy when one mole of the compound is formed from its elements in their standard states.

### Step-by-Step Solution:

1. **Write the Formation Reaction for CH₄:**
The formation reaction for methane (CH₄) from its elements is:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

2. **Identify the Given Reactions:**
We have the following reactions with their respective enthalpy changes (ΔH):
- Reaction 1:
\[ C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta H = -394 \, \text{kJ} \]
- Reaction 2:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l), \quad \Delta H = -568 \, \text{kJ} \]
- Reaction 3:
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l), \quad \Delta H = -892 \, \text{kJ} \]

3. **Manipulate the Reactions:**
To find the heat of formation for CH₄, we need to manipulate the equations:
- Use Reaction 1 as it is.
- Use Reaction 2 as it is.
- Reverse Reaction 3 to express it in terms of CH₄ formation:
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g), \quad \Delta H = +892 \, \text{kJ} \]

4. **Combine the Reactions:**
Now, we can add the modified equations:
- From Reaction 1:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -394 \, \text{kJ} \]
- From Reaction 2:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \Delta H = -568 \, \text{kJ} \]
- From the reversed Reaction 3:
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \quad \Delta H = +892 \, \text{kJ} \]

Adding these equations together:
\[ C(s) + O_2(g) + 2H_2(g) + O_2(g) + CO_2(g) + 2H_2O(l) \rightarrow CO_2(g) + 2H_2O(l) + CH_4(g) + 2O_2(g) \]

The \(CO_2(g)\) and \(2H_2O(l)\) on both sides cancel out, leaving:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

5. **Calculate the Total ΔH:**
Now, we can calculate the total change in enthalpy:
\[ \Delta H = -394 \, \text{kJ} - 568 \, \text{kJ} + 892 \, \text{kJ} \]
\[ \Delta H = -394 - 568 + 892 = -70 \, \text{kJ} \]

### Final Answer:
The heat of formation of CH₄ is:
\[ \Delta H_f(CH_4) = -70 \, \text{kJ} \]