To determine the standard enthalpy change, [tex]\(\Delta H^{\circ} \mathrm{rxn}\)[/tex], for the reaction:
[tex]\[
4 \mathrm{NO}(g) + 2 \mathrm{O}_2(g) \rightarrow 4 \mathrm{NO}_2(g)
\][/tex]
we can use the given enthalpies of the following reactions:
[tex]\[
\begin{array}{ll}
\mathrm{(1)\quad} \mathrm{N}_2(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) & \Delta H^{\circ} \mathrm{rxn\_1} = +183 \text{ kJ} \\
\mathrm{(2)\quad} \frac{1}{2} \mathrm{N}_2(g) + \mathrm{O}_2(g) \rightarrow \mathrm{NO}_2(g) & \Delta H^{\circ} \mathrm{rxn\_2} = +33 \text{ kJ}
\end{array}
\][/tex]
### Step-by-Step Solution:
1. Scale the first reaction to match the target reaction for NO:
[tex]\[
\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) \quad (\Delta H^{\circ} = 183 \text{ kJ})
\][/tex]
We need [tex]\(4 \mathrm{NO}\)[/tex], so we multiply the entire reaction by 2:
[tex]\[
2 \mathrm{N}_2(g) + 2 \mathrm{O}_2(g) \rightarrow 4 \mathrm{NO}(g) \quad (\Delta H^{\circ} = 2 \times 183 \text{ kJ} = 366 \text{ kJ})
\][/tex]
Thus, the scaled enthalpy change for this reaction is [tex]\( 366 \text{ kJ} \)[/tex].
2. Scale the second reaction to match the target reaction for NO₂:
[tex]\[
\frac{1}{2} \mathrm{N}_2(g) + \mathrm{O}_2(g) \rightarrow \mathrm{NO}_2(g) \quad (\Delta H^{\circ} = 33 \text{ kJ})
\][/tex]
We need [tex]\(4 \mathrm{NO}_2\)[/tex], so we multiply the entire reaction by 4:
[tex]\[
2 \mathrm{N}_2(g) + 4 \mathrm{O}_2(g) \rightarrow 4 \mathrm{NO}_2(g) \quad (\Delta H^{\circ} = 4 \times 33 \text{ kJ} = 132 \text{ kJ})
\][/tex]
Thus, the scaled enthalpy change for this reaction is [tex]\( 132 \text{ kJ} \)[/tex].
3. Combine the scaled reactions:
Subtract the enthalpy of the NO formation reaction from the NO₂ formation reaction:
[tex]\[
\Delta H^{\circ} \mathrm{rxn} = 132 \text{ kJ} - 366 \text{ kJ} = -234 \text{ kJ}
\][/tex]
Therefore, the standard enthalpy change, [tex]\(\Delta H^{\circ} \mathrm{rxn}\)[/tex], for the reaction
[tex]\[
4 \mathrm{NO}(g) + 2 \mathrm{O}_2(g) \rightarrow 4 \mathrm{NO}_2(g)
\][/tex]
is [tex]\(-234 \text{ kJ}\)[/tex].
