Based on the following reaction C(graphite) +`O_2(g)rarrCO_2(g),DeltaH=-394KJ//mol...(i) `
`2CO(g)+O_2(g)rarr2CO_2(g),DeltaH=-569KJ//mol...(ii)`
The heat of formation of CO will be

Options

A. `+109.5 kJ`
B. `0219kJ`
C. `-109.5KJ`
D. `-219KJ`

Answer

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

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

1. \( C(\text{graphite}) + O_2(g) \rightarrow CO_2(g), \Delta H = -394 \, \text{kJ/mol} \) (Reaction 1)
2. \( 2CO(g) + O_2(g) \rightarrow 2CO_2(g), \Delta H = -569 \, \text{kJ/mol} \) (Reaction 2)

### Step 2: Identify the target reaction
We need to find the heat of formation of CO, which is represented by the reaction:

\[ C(\text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g) \]

### Step 3: Manipulate the given reactions
To derive the target reaction from the given reactions, we can manipulate them as follows:

1. **From Reaction 2**, we need to reverse it to get CO on the left side. When we reverse the reaction, we change the sign of ΔH:

\[ 2CO_2(g) \rightarrow 2CO(g) + O_2(g), \Delta H = +569 \, \text{kJ/mol} \]

Since we only need 1 mole of CO, we will divide the entire reaction by 2:

\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g), \Delta H = +\frac{569}{2} \, \text{kJ/mol} = +284.5 \, \text{kJ/mol} \] (Reaction 3)

### Step 4: Add Reaction 1 and Reaction 3
Now we will add Reaction 1 and Reaction 3 to obtain the target reaction:

- Reaction 1:
\[ C(\text{graphite}) + O_2(g) \rightarrow CO_2(g), \Delta H = -394 \, \text{kJ/mol} \]

- Reaction 3:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g), \Delta H = +284.5 \, \text{kJ/mol} \]

When we add these two reactions, the \( CO_2(g) \) cancels out:

\[ C(\text{graphite}) + O_2(g) + CO_2(g) \rightarrow CO_2(g) + CO(g) + \frac{1}{2}O_2(g) \]

This simplifies to:

\[ C(\text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g) \]

### Step 5: Calculate the overall ΔH
Now we can calculate the overall ΔH for the target reaction:

\[ \Delta H = \Delta H_1 + \Delta H_3 \]

Substituting the values:

\[ \Delta H = -394 \, \text{kJ/mol} + 284.5 \, \text{kJ/mol} \]

Calculating this gives:

\[ \Delta H = -394 + 284.5 = -109.5 \, \text{kJ/mol} \]

### Final Answer
The heat of formation of CO is:

\[ \Delta H_f^\circ (CO) = -109.5 \, \text{kJ/mol} \]