Find the enthalpy of combustion of carbon ( graphite) to produce carbon monoxide( g)on the basis of data given below`:` C ( graphite ) `+ O_(2)(g) rarr CO_(2)(g) + 393. 4 kJ mol^(-1)`
`CO(g) + (1)/(2) O_(2)(g) rarr CO_(2)(g) + 283.0 kJ mol^(-1)`

Explanation

To find the enthalpy of combustion of carbon (graphite) to produce carbon monoxide (CO), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Write the Given Reactions**:
- Reaction 1:
\[ C (graphite) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_1 = -393.4 \, \text{kJ/mol} \]
- Reaction 2:
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad \Delta H_2 = -283.0 \, \text{kJ/mol} \]

2. **Rearranging the Reactions**:
- We need to find the enthalpy change for the reaction:
\[ C (graphite) + \frac{1}{2} O_2(g) \rightarrow CO(g) \]
- To do this, we can manipulate the two given reactions. We will reverse Reaction 2 to express it in terms of CO formation:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g) \quad \Delta H = +283.0 \, \text{kJ/mol} \]

3. **Combining the Reactions**:
- Now we can add Reaction 1 and the reversed Reaction 2:
\[ C (graphite) + O_2(g) \rightarrow CO_2(g) \quad (-393.4 \, \text{kJ/mol}) \]
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g) \quad (+283.0 \, \text{kJ/mol}) \]
- When we add these reactions, the \(CO_2(g)\) cancels out:
\[ C (graphite) + O_2(g) + CO_2(g) \rightarrow CO_2(g) + CO(g) + \frac{1}{2} O_2(g) \]
- This simplifies to:
\[ C (graphite) + \frac{1}{2} O_2(g) \rightarrow CO(g) \]

4. **Calculating the Enthalpy Change**:
- The total enthalpy change for the reaction is:
\[ \Delta H = \Delta H_1 + \Delta H_2 \]
\[ \Delta H = -393.4 \, \text{kJ/mol} + 283.0 \, \text{kJ/mol} \]
\[ \Delta H = -110.4 \, \text{kJ/mol} \]

5. **Final Result**:
- The enthalpy of combustion of carbon (graphite) to produce carbon monoxide (g) is:
\[ \Delta H = -110.4 \, \text{kJ/mol} \]