To find the net change in enthalpy (ΔH) for the formation of one mole of nitric oxide (NO) from nitrogen, hydrogen, and oxygen through the given reactions, we will break down the process into two steps and analyze the enthalpy change for each step.
### Step 1: Formation of Ammonia
[tex]\( \text{Reaction:} \quad N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \quad \Delta H = -92 \, \text{kJ} \)[/tex]
In this reaction, 2 moles of ammonia (NH_3) are produced with an enthalpy change (ΔH) of -92 kJ. We need to express this per mole of NO formed. Given that in step 2, 1 mole of NH_3 is required to form 1 mole of NO, we calculate ΔH per mole of NH_3:
[tex]\[ \Delta H_{\text{per mole NH}_3} = \frac{-92 \, \text{kJ}}{2 \, \text{moles NH}_3} = -46 \, \text{kJ/mole NH}_3 \][/tex]
### Step 2: Formation of Nitric Oxide
[tex]\( \text{Reaction:} \quad 4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g) \quad \Delta H = -905 \, \text{kJ} \)[/tex]
In this reaction, 4 moles of nitric oxide (NO) are produced with an enthalpy change (ΔH) of -905 kJ. We need to express this per mole of NO:
[tex]\[ \Delta H_{\text{per mole NO}} = \frac{-905 \, \text{kJ}}{4 \, \text{moles NO}} = -226.25 \, \text{kJ/mole NO} \][/tex]
### Net Change in Enthalpy
The net change in enthalpy (ΔH) for the formation of one mole of NO is the sum of the enthalpy changes from both steps per mole of NO:
[tex]\[
\Delta H_{\text{net per mole NO}} = \Delta H_{\text{per mole NO (Step 2)}} + \Delta H_{\text{per mole NH}_3 \text{ (Step 1)}}
\][/tex]
Substituting the calculated values:
[tex]\[
\Delta H_{\text{net per mole NO}} = -226.25 \, \text{kJ/mole NO} + (-46 \, \text{kJ/mole NH}_3)
\][/tex]
[tex]\[
\Delta H_{\text{net per mole NO}} = -226.25 \, \text{kJ/mole NO} - 46 \, \text{kJ/mole NH}_3
\][/tex]
[tex]\[
\Delta H_{\text{net per mole NO}} = -272.25 \, \text{kJ/mole NO}
\][/tex]
### Rounded Answer
Finally, rounding the net change in enthalpy to the nearest kJ:
[tex]\[
\Delta H_{\text{net per mole NO rounded}} = -272 \, \text{kJ}
\][/tex]
Thus, the net change in enthalpy for the formation of one mole of nitric oxide (NO) from nitrogen, hydrogen, and oxygen is [tex]\(\boxed{-272\text{kJ}}\)[/tex].
