Heat of combustion `DeltaH^(@)` for `C(s),H_(2)(g)` and `CH_(4)(g)` are `-94, -68` and `-213Kcal//mol` . Then `DeltaH^(@)` for `C(s)+2H_(2)(g)rarrCH_(4)(g)` is

Options

A. `-17` kcal/mol
B. `-111` kcal/mol
C. `-170 ` kcal/mol
D. `-85 ` kcal/mol

Explanation

To find the heat of reaction (ΔH°) for the reaction \( C(s) + 2H_2(g) \rightarrow CH_4(g) \), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Write the combustion reactions for each substance:**
- **For Carbon (C):**
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -94 \text{ kcal/mol} \]
- **For Hydrogen (H₂):**
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \quad \Delta H = 2 \times (-68) = -136 \text{ kcal/mol} \]
- **For Methane (CH₄):**
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) \quad \Delta H = -213 \text{ kcal/mol} \]

2. **Identify the reactions needed to derive the target reaction:**
- We want to combine the reactions to get:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

3. **Use Hess's Law to combine the reactions:**
- From the combustion of carbon, we have:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_1 = -94 \text{ kcal/mol} \]
- From the combustion of hydrogen, we have:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \quad \Delta H_2 = -136 \text{ kcal/mol} \]
- For methane, we reverse the reaction:
\[ CO_2(g) + 2H_2O(g) \rightarrow CH_4(g) + 2O_2(g) \quad \Delta H_3 = +213 \text{ kcal/mol} \]

4. **Add the enthalpy changes:**
- According to Hess's law:
\[ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
- Substitute the values:
\[ \Delta H = (-94) + (-136) + (+213) \]
- Calculate:
\[ \Delta H = -94 - 136 + 213 = -17 \text{ kcal/mol} \]

### Final Answer:
The heat of reaction \( \Delta H° \) for \( C(s) + 2H_2(g) \rightarrow CH_4(g) \) is \(-17 \text{ kcal/mol}\).

---