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Question 7

Use the standard reaction enthalpies given below to determine [tex]$\Delta H^{\circ}_{\text{rxn}}$[/tex] for the following reaction:

[tex]\[ 4 \text{NO (g)} + 2 \text{O}_2 \text{(g)} \rightarrow 4 \text{NO}_2 \text{(g)} \quad \Delta H^{\circ}_{\text{rxn}} =\text{ ? } \][/tex]

Given:
[tex]\[
\begin{array}{ll}
\text{N}_2 \text{(g)} + \text{O}_2 \text{(g)} \rightarrow 2 \text{NO (g)} & \Delta H^{\circ}_{\text{rxn}} = +183 \text{kJ} \\
\frac{1}{2} \text{N}_2 \text{(g)} + \text{O}_2 \text{(g)} \rightarrow \text{NO}_2 \text{(g)} & \Delta H^{\circ}_{\text{rxn}} = +33 \text{kJ}
\end{array}
\][/tex]



Answer :

GinnyAnswer
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].