The heat of formations for `CO_(2)(g),H_(2)O(l)` and `CH_(4)(g)` are `-400kJ"mol"^(-1), -280kJ"mol"^(-1)` and `-70kJ"mol"^(-1)` respectively. The heat of combustion of `CH_(4)` in `kJ"mol"^(-1)` is

Options

A. 890
B. `-160`
C. `-890`
D. `-90`

Explanation

To find the heat of combustion of methane (CH₄), we can use the heat of formation values provided for carbon dioxide (CO₂), water (H₂O), and methane (CH₄). The heat of combustion can be calculated using the following balanced chemical equation for the combustion of methane:

\[ \text{CH}_4(g) + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]

### Step-by-Step Solution:

1. **Write the balanced equation for the combustion of CH₄:**
\[ \text{CH}_4(g) + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]

2. **Identify the heat of formation values:**
- ΔH_f (CO₂) = -400 kJ/mol
- ΔH_f (H₂O) = -280 kJ/mol
- ΔH_f (CH₄) = -70 kJ/mol
- ΔH_f (O₂) = 0 kJ/mol (since it is in its elemental form)

3. **Apply Hess's Law to find the heat of combustion (ΔH_combustion):**
The heat of combustion can be calculated using the formula:
\[ \Delta H_{\text{combustion}} = \left( \Delta H_f \text{(products)} \right) - \left( \Delta H_f \text{(reactants)} \right) \]
For our reaction:
\[ \Delta H_{\text{combustion}} = \left( \Delta H_f \text{(CO₂)} + 2 \times \Delta H_f \text{(H₂O)} \right) - \left( \Delta H_f \text{(CH₄)} + 3 \times \Delta H_f \text{(O₂)} \right) \]

4. **Substitute the values into the equation:**
\[ \Delta H_{\text{combustion}} = \left( -400 + 2 \times (-280) \right) - \left( -70 + 3 \times 0 \right) \]
\[ = \left( -400 - 560 \right) - (-70) \]
\[ = -960 + 70 \]
\[ = -890 \text{ kJ/mol} \]

5. **Final Result:**
The heat of combustion of CH₄ is:
\[ \Delta H_{\text{combustion}} = -890 \text{ kJ/mol} \]