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 carbon monoxide (CO) based on the given reactions, we can follow these steps:

### Step 1: Write down the given reactions and their enthalpy changes.
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: Write the formation reaction for carbon monoxide.
The formation reaction for carbon monoxide can be expressed as:
\[ C(\text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g) \]
This is what we want to find the enthalpy change for, denoted as \( \Delta H_f \).

### Step 3: Manipulate the given reactions to obtain the desired reaction.
To derive the formation reaction for CO, we need to manipulate the second reaction. We can reverse Reaction 2 and divide it by 2 to get the formation of 1 mole of CO:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g), \Delta H = +\frac{569}{2} \, \text{kJ/mol} = +284.5 \, \text{kJ/mol} \]

### Step 4: Add the modified Reaction 2 to Reaction 1.
Now we can add the modified Reaction 2 to Reaction 1:
1. \( C(\text{graphite}) + O_2(g) \rightarrow CO_2(g), \Delta H = -394 \, \text{kJ/mol} \)
2. \( CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g), \Delta H = +284.5 \, \text{kJ/mol} \)

When we add these two reactions:
- The \( CO_2(g) \) cancels out.
- We are left with:
\[ C(\text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g) \]

### Step 5: Calculate the overall enthalpy change.
Now we can calculate the overall enthalpy change (\( \Delta H_f \)) for the formation of CO:
\[ \Delta H_f = (-394 \, \text{kJ/mol}) + (+284.5 \, \text{kJ/mol}) = -394 + 284.5 = -109.5 \, \text{kJ/mol} \]

### Conclusion
The heat of formation of carbon monoxide (CO) is:
\[ \Delta H_f = -109.5 \, \text{kJ/mol} \]