Heat of combustion of C(s), `H_2(g)` and `CH_4 (g)` respectively are -94, -68 and -213 kcal/mol, then`deltaH` for the reacton `[C(s) +2H_2(g) → CH_4(g)]` is

Options

A. -17 kcal
B. -34 kcal
C. -886 kcal
D. -51 kcal

Answer

To find the enthalpy change (ΔH) for the reaction \( C(s) + 2H_2(g) \rightarrow CH_4(g) \), we can utilize the heat of combustion values provided for carbon, hydrogen, and methane. The heat of combustion values are:

- Heat of combustion of \( C(s) \): \( -94 \, \text{kcal/mol} \)
- Heat of combustion of \( H_2(g) \): \( -68 \, \text{kcal/mol} \)
- Heat of combustion of \( CH_4(g) \): \( -213 \, \text{kcal/mol} \)

### Step-by-Step Solution:

1. **Write the combustion reactions for each substance:**
- For carbon:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -94 \, \text{kcal} \]
- For hydrogen:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \Delta H = -68 \, \text{kcal} \]
- For methane:
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H = -213 \, \text{kcal} \]

2. **Reverse the combustion reaction of methane:**
- When we reverse the reaction for methane, we get:
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \quad \Delta H = +213 \, \text{kcal} \]

3. **Add the combustion reactions of carbon and hydrogen to the reversed reaction of methane:**
- Now we combine the reactions:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -94 \, \text{kcal} \]
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \Delta H = -68 \, \text{kcal} \]
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \quad \Delta H = +213 \, \text{kcal} \]

4. **Combine the ΔH values:**
- The overall reaction is:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]
- The total ΔH for the reaction is:
\[ \Delta H = (-94) + (-68) + (+213) \]

5. **Calculate ΔH:**
- Combine the values:
\[ \Delta H = -94 - 68 + 213 = -17 \, \text{kcal} \]

### Final Answer:
The ΔH for the reaction \( C(s) + 2H_2(g) \rightarrow CH_4(g) \) is \( -17 \, \text{kcal} \).

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