To find the enthalpy of formation of carbon monoxide (CO), we can use the enthalpy of combustion values given in the question. The enthalpy of formation can be derived from the enthalpy of combustion using Hess's law.
### Step-by-Step Solution:
1. **Write the combustion reactions:**
- The combustion of carbon (C) can be represented as:
\[
C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -393.5 \, \text{kJ/mol}
\]
- The combustion of carbon monoxide (CO) can be represented as:
\[
2CO(g) + O_2(g) \rightarrow 2CO_2(g) \quad \Delta H = -283 \, \text{kJ/mol}
\]
2. **Determine the enthalpy change for the formation of CO:**
- The formation of carbon monoxide from its elements can be represented as:
\[
C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)
\]
- We need to find the enthalpy change for this reaction, which we can denote as \( \Delta H_f \) for CO.
3. **Use Hess's law:**
- According to Hess's law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We can manipulate the combustion reactions to find the enthalpy of formation of CO.
- From the combustion of carbon, we can see that:
\[
C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -393.5 \, \text{kJ/mol}
\]
- For the combustion of CO, we can write:
\[
2CO(g) + O_2(g) \rightarrow 2CO_2(g) \quad \Delta H = -283 \, \text{kJ/mol}
\]
- If we reverse the combustion of CO, we get:
\[
2CO_2(g) \rightarrow 2CO(g) + O_2(g) \quad \Delta H = +283 \, \text{kJ/mol}
\]
- Now, we can combine these reactions:
- From the combustion of carbon, we have \( C(s) + O_2(g) \rightarrow CO_2(g) \) with \( \Delta H = -393.5 \, \text{kJ/mol} \).
- From the reversed combustion of CO, we have \( 2CO_2(g) \rightarrow 2CO(g) + O_2(g) \) with \( \Delta H = +283 \, \text{kJ/mol} \).
4. **Combine the reactions:**
- Combining these gives:
\[
C(s) + O_2(g) \rightarrow CO_2(g) \quad (-393.5 \, \text{kJ/mol})
\]
\[
2CO_2(g) \rightarrow 2CO(g) + O_2(g) \quad (+283 \, \text{kJ/mol})
\]
- The net reaction is:
\[
C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)
\]
- The total enthalpy change is:
\[
\Delta H_f = -393.5 + \frac{1}{2}(283) = -393.5 + 141.5 = -252 \, \text{kJ/mol}
\]
### Final Answer:
The enthalpy of formation of carbon monoxide (CO) is \(-252 \, \text{kJ/mol}\).
