`C(s)+O_(2)(g)rarr CO_(2)(g)+94.0` K cal.
`CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g), Delta H=-67.7` K cal. From the above reactions find how much heat (Kcal `"mole"^(-1)`)would be produced in the following reaction : `C(s)+(1)/(2)O_(2)(g)rarr CO(g)`

Options

A. `20.6`
B. `26.3`
C. `44.2`
D. `161.6`

Answer

To find the heat produced in the reaction \( C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \), 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.

### Step-by-Step Solution:

1. **Identify the Given Reactions:**
- Reaction 1:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_1 = -94.0 \text{ kcal} \]
- Reaction 2:
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad \Delta H_2 = -67.7 \text{ kcal} \]

2. **Reverse Reaction 2:**
To find the enthalpy change for the reaction we need, we will reverse Reaction 2:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g) \quad \Delta H_2' = +67.7 \text{ kcal} \]

3. **Combine the Reactions:**
Now, we will add Reaction 1 and the reversed Reaction 2:
- From Reaction 1:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_1 = -94.0 \text{ kcal} \]
- From the reversed Reaction 2:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g) \quad \Delta H_2' = +67.7 \text{ kcal} \]

Adding these two reactions:
\[ C(s) + O_2(g) + CO_2(g) \rightarrow CO_2(g) + CO(g) + \frac{1}{2} O_2(g) \]

The \( CO_2(g) \) cancels out:
\[ C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \]

4. **Calculate the Total Enthalpy Change:**
Now, we can calculate the total enthalpy change for the desired reaction:
\[ \Delta H = \Delta H_1 + \Delta H_2' = -94.0 \text{ kcal} + 67.7 \text{ kcal} = -26.3 \text{ kcal} \]

### Final Answer:
The heat produced in the reaction \( C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \) is:
\[ \Delta H = -26.3 \text{ kcal} \]