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Consider the following intermediate chemical equations:

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

What is the enthalpy of the overall chemical equation [tex]\( NO (g) + O (g) \rightarrow NO_2 (g) \)[/tex]?

A. [tex]\(-305 \, \text{kJ}\)[/tex]

B. [tex]\(-304.1 \, \text{kJ}\)[/tex]

C. [tex]\(-93.7 \, \text{kJ}\)[/tex]

D. [tex]\(588.7 \, \text{kJ}\)[/tex]



Answer :

GinnyAnswer
To determine the enthalpy change for the overall chemical reaction [tex]\( \text{NO(g) + O(g) → NO}_2\text{(g)} \)[/tex], we need to combine the given intermediate reactions, making sure they cancel out intermediate species and sum to the desired reaction. Here is the detailed step-by-step process:

1. List the given intermediate reactions and their enthalpy changes:

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

2. Manipulate the intermediate reactions to cancel out unwanted species:

- The desired final reaction is [tex]\( \text{NO(g) + O(g) → NO}_2\text{(g)} \)[/tex].

- We need to reverse the second reaction to eliminate [tex]\( \text{O}_3\text{(g)} \)[/tex]. The reversed reaction is:
[tex]\[ \text{O}_3\text{(g) → }\frac{3}{2}\text{O}_2\text{(g)} \quad \Delta H = -142.3 \ \text{kJ} \][/tex]

3. Write the intermediate reactions with appropriate signs:

[tex]\[ \begin{array}{ll} \text{NO(g) + O}_3\text{(g) → NO}_2\text{(g) + O}_2\text{(g)} & \Delta H = -198.9 \ \text{kJ} \\ \text{O}_3\text{(g) → }\frac{3}{2}\text{O}_2\text{(g)} & \Delta H = -142.3 \ \text{kJ} \\ \text{O(g) → }\frac{1}{2}\text{O}_2\text{(g)} & \Delta H = -247.5 \ \text{kJ} \\ \end{array} \][/tex]

4. Sum the three reactions to yield the overall reaction:

- Add the species on the left side of the arrows:
[tex]\[ \text{NO(g) + O}_3\text{(g) + O}_3\text{(g) + O(g)} \][/tex]

- Add the species on the right side of the arrows:
[tex]\[ \text{NO}_2\text{(g) + O}_2\text{(g) + }\frac{3}{2}\text{O}_2\text{(g) + }\frac{1}{2}\text{O}_2\text{(g)} \][/tex]

- Simplify by canceling out common species:
[tex]\[ \text{NO(g) + O(g) → NO}_2\text{(g)} \][/tex]

5. Sum the enthalpy changes to find the overall enthalpy:

[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + \Delta H_3 \][/tex]

6. Sum the specific enthalpy changes:

[tex]\[ \Delta H_{\text{total}} = -198.9 \ \text{kJ} + (-142.3 \ \text{kJ}) + (-247.5 \ \text{kJ}) \][/tex]

[tex]\[ \Delta H_{\text{total}} = -304.1 \ \text{kJ} \][/tex]

Thus, the enthalpy change for the overall reaction [tex]\( \text{NO(g) + O(g) → NO}_2\text{(g)} \)[/tex] is [tex]\(-304.1 \ \text{kJ}\)[/tex].

The correct answer is:
[tex]\[ -304.1 \ \text{kJ} \][/tex]

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