Answered

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Consider the chemical equations shown here:

[tex]\[
\begin{array}{l}
NO (g) + O_3 (g) \rightarrow NO_2 (g) + O_2 (g) \quad \Delta H_1 = -198.9 \, \text{kJ} \\
\frac{3}{2} O_2 (g) \rightarrow O_3 (g) \quad \Delta H_2 = 142.3 \, \text{kJ} \\
O (g) \rightarrow \frac{1}{2} O_2 (g) \quad \Delta H_3 = -247.5 \, \text{kJ}
\end{array}
\][/tex]

What is [tex]\(\Delta H_{\text{rxn}}\)[/tex] for the reaction shown below?

[tex]\[
NO (g) + O (g) \rightarrow NO_2 (g)
\][/tex]

[tex]\[\boxed{\Delta H_{\text{rxn}} = ?}\][/tex]

Sagot :

GinnyAnswer
To determine the enthalpy change ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) for the reaction

[tex]\[ \text{NO(g) + O(g) → NO}_2\text{(g)}, \][/tex]

we will use the given reactions and their associated enthalpy changes:

1. [tex]\[\text{NO(g) + O}_3\text{(g) → NO}_2\text{(g) + O}_2\text{(g)} \quad \Delta H_1 = -198.9 \, \text{kJ} \][/tex]
2. [tex]\[\frac{3}{2} \text{O}_2\text{(g) → O}_3\text{(g)} \quad \Delta H_2 = 142.3 \, \text{kJ} \][/tex]
3. [tex]\[\text{O(g) → } \frac{1}{2} \text{O}_2\text{(g)} \quad \Delta H_3 = -247.5 \, \text{kJ} \][/tex]

Our goal is to derive the reaction [tex]\(\text{NO(g) + O(g) → NO}_2\text{(g)}\)[/tex]. Let's proceed step-by-step:

1. Start with Reaction 1:
[tex]\[ \text{NO(g) + O}_3\text{(g) → NO}_2\text{(g) + O}_2\text{(g)} \][/tex]
[tex]\[ \Delta H_1 = -198.9 \, \text{kJ} \][/tex]

2. To eliminate [tex]\(\text{O}_3\)[/tex] and introduce [tex]\(\text{O(g)}\)[/tex], we use Reaction 2 and Reaction 3. First, reverse Reaction 2:

[tex]\[ \text{O}_3\text{(g) → } \frac{3}{2} \text{O}_2\text{(g)} \quad \Delta H_2 \text{ (reversed)} = -142.3 \, \text{kJ} \][/tex]

3. Next, take half of Reaction 3 to match the amount of oxygen atoms:

[tex]\[ \frac{1}{2} \text{O(g) → }\frac{1}{4} \text{O}_2\text{(g)} \right) \][/tex]

When halving [tex]\(\Delta H_3\)[/tex]:

[tex]\[ \Delta H_3 \text{ (halved)} = \frac{-247.5}{2} \, \text{kJ} = -123.75 \, \text{kJ} \][/tex]

4. Now, we combine all these enthalpy changes ([tex]\(\Delta H_1\)[/tex], [tex]\(\Delta H_2\)[/tex] reversed, and [tex]\(\Delta H_3\)[/tex] halved):

[tex]\[ \Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 \text{ (reversed)} + \Delta H_3 \text{ (halved)} \][/tex]

[tex]\[ \Delta H_{\text{rxn}} = -198.9 \, \text{kJ} + (-142.3 \, \text{kJ}) + (-123.75 \, \text{kJ}) \][/tex]

5. Adding these values together:

[tex]\[ \Delta H_{\text{rxn}} = -198.9 + -142.3 + -123.75 \][/tex]

[tex]\[ \Delta H_{\text{rxn}} = -464.95 \, \text{kJ} \][/tex]

Therefore, the enthalpy change [tex]\(\Delta H_{\text{rxn}}\)[/tex] for the reaction [tex]\(\text{NO(g) + O(g) → NO}_2\text{(g)}\)[/tex] is

[tex]\[ \boxed{-464.95 \, \text{kJ}} \][/tex]
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