If `C(s)+O_(2)(g)to CO_(2)(g),DeltaH=rand CO(g)+1/2O_(2)to CO_(2)(g),DeltaH=s` then, the heat of formation of CO is

Options

A. (a) `r + s`
B. (b) `r - s`
C. (c) `s - r`
D. (d) `rs`

Explanation

To find the heat of formation of carbon monoxide (CO) from the given reactions, we can follow these steps:

### Step 1: Write down the given reactions and their enthalpy changes.
1. The first reaction is:
\[ C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta H = R \]
2. The second reaction is:
\[ CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g), \quad \Delta H = S \]

### Step 2: Reverse the second reaction.
To find the heat of formation of CO, we need to reverse the second reaction:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) \]
When we reverse a reaction, the sign of the enthalpy change also reverses:
\[ \Delta H = -S \]

### Step 3: Add the two reactions together.
Now, we can add the first reaction and the reversed second reaction:
1. First reaction:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \]
2. Reversed second reaction:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) \]

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

### Step 4: Simplify the equation.
The \(CO_2(g)\) on both sides cancels out:
\[ C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g) \]

### Step 5: Write the enthalpy change for the new reaction.
The enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual reactions:
\[ \Delta H = R + (-S) = R - S \]

### Conclusion:
Thus, the heat of formation of CO is:
\[ \Delta H_{f} (CO) = R - S \]