`C(s)+O_(2)(g)toCO_(2)(g),DeltaH=-94` kcal
`2CO(g)+O_(2)to2CO_(2),DeltaH=-135.2` kcal
The heat of formation of `CO(g)` is

Options

A. `-26.4` kcal
B. `41.2 kcal`
C. `26.4 kcal`
D. `229.2 kcal`

Explanation

To find the heat of formation of carbon monoxide (CO), we can use Hess's law of constant heat summation. Here’s a step-by-step solution:

### Step 1: Write the reactions and their enthalpy changes

1. The first reaction is:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -94 \text{ kcal} \]

2. The second reaction is:
\[ 2CO(g) + O_2(g) \rightarrow 2CO_2(g) \quad \Delta H = -135.2 \text{ kcal} \]

### Step 2: Reverse the second reaction

To find the heat of formation of CO, we need to reverse the second reaction (since we want CO on the product side):
\[ 2CO_2(g) \rightarrow 2CO(g) + O_2(g) \quad \Delta H = +135.2 \text{ kcal} \]

### Step 3: Divide the reversed reaction by 2

Since we want the formation of 1 mole of CO, we divide the entire reaction by 2:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) \quad \Delta H = \frac{135.2}{2} = 67.6 \text{ kcal} \]

### Step 4: Write the first reaction again

The first reaction remains the same:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -94 \text{ kcal} \]

### Step 5: Add the two reactions

Now, we can add the modified second reaction to the first reaction:
\[ \begin{align*} C(s) + O_2(g) & \rightarrow CO_2(g) \quad \Delta H = -94 \text{ kcal} \\ CO_2(g) & \rightarrow CO(g) + \frac{1}{2}O_2(g) \quad \Delta H = +67.6 \text{ kcal} \\ \hline C(s) + \frac{1}{2}O_2(g) & \rightarrow CO(g) \quad \Delta H = -94 + 67.6 \text{ kcal} \end{align*} \]

### Step 6: Calculate the total enthalpy change

Now, calculate the total enthalpy change:
\[ \Delta H = -94 + 67.6 = -26.4 \text{ kcal} \]

### Conclusion

The heat of formation of CO(g) is:
\[ \Delta H_f = -26.4 \text{ kcal} \]