Calculate the enthalpy of combustion of nitric oxide (NO) from the following data:
`1/2 N_(2)(g) + 1/2O_(2)(g) to NO(g) , Delta_(r)H^(@) = 90.7 kJ`
`1/2 N_(2)(g) + O_(2)(g) to NO_(2)(g) Delta_(r)H^(@) = 34.0 kJ`

Explanation

To calculate the enthalpy of combustion of nitric oxide (NO), 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. **Identify the combustion reaction of nitric oxide (NO)**:
The combustion of nitric oxide can be represented as:
\[ 2 \, \text{NO}(g) + \text{O}_2(g) \rightarrow 2 \, \text{NO}_2(g) \]
We need to find the enthalpy change (\(\Delta H\)) for this reaction.

2. **Write the given reactions**:
We have the following reactions with their enthalpy changes:
- Reaction 1:
\[ \frac{1}{2} \, \text{N}_2(g) + \frac{1}{2} \, \text{O}_2(g) \rightarrow \text{NO}(g), \quad \Delta H_1 = +90.7 \, \text{kJ} \]
- Reaction 2:
\[ \frac{1}{2} \, \text{N}_2(g) + \text{O}_2(g) \rightarrow \text{NO}_2(g), \quad \Delta H_2 = +34.0 \, \text{kJ} \]

3. **Reverse Reaction 1**:
Since we need NO to be a reactant in the combustion reaction, we reverse Reaction 1:
\[ \text{NO}(g) \rightarrow \frac{1}{2} \, \text{N}_2(g) + \frac{1}{2} \, \text{O}_2(g), \quad \Delta H = -90.7 \, \text{kJ} \]

4. **Use Reaction 2 as it is**:
We keep Reaction 2 as it is since it already has NO2 as a product:
\[ \frac{1}{2} \, \text{N}_2(g) + \text{O}_2(g) \rightarrow \text{NO}_2(g), \quad \Delta H = +34.0 \, \text{kJ} \]

5. **Combine the reactions**:
Now, we can add the modified Reaction 1 and Reaction 2:
\[ \text{NO}(g) + \left(\frac{1}{2} \, \text{N}_2(g) + \text{O}_2(g)\right) \rightarrow \left(\frac{1}{2} \, \text{N}_2(g) + \frac{1}{2} \, \text{O}_2(g)\right) + \text{NO}_2(g) \]
The \(\frac{1}{2} \, \text{N}_2(g)\) cancels out, leading to:
\[ \text{NO}(g) + \text{O}_2(g) \rightarrow \text{NO}_2(g) \]

6. **Calculate the total enthalpy change**:
Now, we can calculate the total enthalpy change for the combustion reaction:
\[ \Delta H_{\text{combustion}} = -90.7 \, \text{kJ} + 34.0 \, \text{kJ} = -56.7 \, \text{kJ} \]

### Final Result:
The enthalpy of combustion of nitric oxide (NO) is:
\[ \Delta H_{\text{combustion}} = -56.7 \, \text{kJ/mol} \]