(1) For the given heat of reaction,
(i) `C(s)+O_(2)(g)=CO_(2)(g)+97 kcal`
(ii) `CO_(2)(g)+C(s)=2CO(g)-39 kcal`
the heat of combustion of `CO(g)` is:

Options

A. 68 kcal
B. `-68` kcal
C. `+48` kcal
D. none of these

Explanation

To find the heat of combustion of carbon monoxide (CO), we will use the given reactions and manipulate them accordingly. Let's break down the solution step by step.

### Step 1: Write down the given reactions
We have two reactions provided:

1. \( C(s) + O_2(g) \rightarrow CO_2(g) + 97 \, \text{kcal} \) (Reaction 1)
2. \( CO_2(g) + C(s) \rightarrow 2CO(g) - 39 \, \text{kcal} \) (Reaction 2)

### Step 2: Reverse Reaction 2
Since we need CO on the reactant side for the combustion reaction, we will reverse Reaction 2. When we reverse a reaction, the sign of the enthalpy change (\( \Delta H \)) also changes.

Reversed Reaction 2:
\[ 2CO(g) \rightarrow CO_2(g) + C(s) + 39 \, \text{kcal} \]

### Step 3: Add Reaction 1 and the reversed Reaction 2
Now we will add Reaction 1 and the reversed Reaction 2 together:

1. \( C(s) + O_2(g) \rightarrow CO_2(g) + 97 \, \text{kcal} \) (Reaction 1)
2. \( 2CO(g) \rightarrow CO_2(g) + C(s) + 39 \, \text{kcal} \) (Reversed Reaction 2)

Adding these two reactions:
- On the left side: \( C(s) + O_2(g) + 2CO(g) \)
- On the right side: \( CO_2(g) + 97 \, \text{kcal} + CO_2(g) + C(s) + 39 \, \text{kcal} \)

This simplifies to:
\[ 2CO(g) + O_2(g) \rightarrow 2CO_2(g) + 136 \, \text{kcal} \]

### Step 4: Divide the entire equation by 2
Since we need the combustion of one mole of CO, we will divide the entire equation by 2:

\[ CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g) + 68 \, \text{kcal} \]

### Step 5: Conclusion
The heat of combustion of carbon monoxide (CO) is therefore:
\[ \Delta H = 68 \, \text{kcal} \]

### Final Answer
The heat of combustion of CO is **68 kilocalories**.

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