To find the enthalpy change [tex]\(\Delta H_{\text{rxn}}\)[/tex] for the reaction [tex]\(NO (g) + O (g) \rightarrow NO_2 (g)\)[/tex], we utilize Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the steps that lead to the final reaction.
Given:
1. [tex]\(NO (g) + O_3 (g) \rightarrow NO_2 (g) + O_2 (g)\)[/tex] with [tex]\(\Delta H_1 = -198.9 \, \text{kJ}\)[/tex]
2. [tex]\(\frac{3}{2} O_2 (g) \rightarrow O_3 (g)\)[/tex] with [tex]\(\Delta H_2 = 142.3 \, \text{kJ}\)[/tex]
3. [tex]\(O (g) \rightarrow \frac{1}{2} O_2 (g)\)[/tex] with [tex]\(\Delta H_3 = -247.5 \, \text{kJ}\)[/tex]
We need to rearrange and combine these equations to derive the target reaction: [tex]\(NO (g) + O (g) \rightarrow NO_2 (g)\)[/tex].
### Step-by-Step Solution
1. Original Reaction (1):
[tex]\[
NO (g) + O_3 (g) \rightarrow NO_2 (g) + O_2 (g) \quad \Delta H_1 = -198.9 \, \text{kJ}
\][/tex]
2. Reverse Reaction (2):
By reversing the second reaction, we obtain:
[tex]\[
O_3 (g) \rightarrow \frac{3}{2} O_2 (g)
\][/tex]
Since we reversed the reaction, we also reverse the sign of [tex]\(\Delta H_2\)[/tex]:
[tex]\[
\Delta H_2 (\text{reversed}) = -142.3 \, \text{kJ}
\][/tex]
3. Original Reaction (3):
[tex]\[
O (g) \rightarrow \frac{1}{2} O_2 (g) \quad \Delta H_3 = -247.5 \, \text{kJ}
\][/tex]
4. Combining the Equations:
[tex]\[
\begin{aligned}
NO (g) + O_3 (g) &\rightarrow NO_2 (g) + O_2 (g) \quad \Delta H_1 = -198.9 \, \text{kJ} \\
O_3 (g) &\rightarrow \frac{3}{2} O_2 (g) \quad \Delta H_2 (\text{reversed}) = -142.3 \, \text{kJ} \\
O (g) &\rightarrow \frac{1}{2} O_2 (g) \quad \Delta H_3 = -247.5 \, \text{kJ}
\end{aligned}
\][/tex]
5. Combining and Canceling Intermediates:
[tex]\[
NO (g) + O_3 (g) + O_3 (g) \rightarrow NO_2 (g) + O_2 (g) + \frac{3}{2} O_2 (g) + \frac{1}{2} O_2 (g)
\][/tex]
After rearranging and canceling out the intermediate substances ([tex]\(O_3\)[/tex] and [tex]\(O_2\)[/tex] that appear on both sides), the remaining reaction is:
[tex]\[
NO (g) + O (g) \rightarrow NO_2 (g)
\][/tex]
6. Summing Up the Enthalpy Changes:
[tex]\[
\Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 (\text{reversed}) + \Delta H_3
\][/tex]
[tex]\[
\Delta H_{\text{rxn}} = -198.9 \, \text{kJ} + (-142.3 \, \text{kJ}) + (-247.5 \, \text{kJ})
\][/tex]
[tex]\[
\Delta H_{\text{rxn}} = -588.7 \, \text{kJ}
\][/tex]
Therefore, the enthalpy change [tex]\(\Delta H_{\text{rxn}}\)[/tex] for the reaction [tex]\(NO (g) + O (g) \rightarrow NO_2 (g)\)[/tex] is [tex]\(-588.7 \, \text{kJ}\)[/tex].
