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To determine the enthalpy of the overall chemical equation [tex]\( \text{NO (g) + O (g) → NO}_2 \text{(g)} \)[/tex], we need to combine the given intermediate reactions and their respective enthalpies.
We are given the following intermediate chemical equations with their enthalpies:
1. [tex]\(\text{NO (g) + O}_3 \text{(g)} \rightarrow \text{NO}_2 \text{(g) + O}_2 \text{(g)}, \Delta H_1 = -198.9 \, \text{kJ}\)[/tex]
2. [tex]\(\frac{3}{2} \text{O}_2 \text{(g)} \rightarrow \text{O}_3 \text{(g)}, \Delta H_2 = 142.3 \, \text{kJ}\)[/tex]
3. [tex]\(\text{O (g)} \rightarrow \frac{1}{2} \text{O}_2 \text{(g)}, \Delta H_3 = -247.5 \, \text{kJ}\)[/tex]
We need to manipulate these equations to obtain the overall reaction:
[tex]\[ \text{NO (g) + O (g) → NO}_2 \text{(g)} \][/tex]
### Step-by-Step Solution:
1. Reverse the second equation to match the required product [tex]\(\text{O}_3\)[/tex]:
[tex]\[ \text{O}_3 \text{(g)} \rightarrow \frac{3}{2} \text{O}_2 \text{(g)}, \Delta H = -142.3 \, \text{kJ} \][/tex]
2. Sum the enthalpies of the first and third equations as they directly contribute to forming [tex]\(\text{NO}_2 \text{(g)}\)[/tex] and [tex]\(\text{O}_2 \text{(g)}\)[/tex]:
1. [tex]\(\text{NO (g) + O}_3 \text{(g)} \rightarrow \text{NO}_2 \text{(g) + O}_2 \text{(g)}, \Delta H_1 = -198.9 \, \text{kJ}\)[/tex]
2. [tex]\(\text{O} \text{(g)} \rightarrow \frac{1}{2} \text{O}_2 \text{(g)}, \Delta H_3 = -247.5 \, \text{kJ}\)[/tex]
3. By combining the [tex]\(\Delta H\)[/tex] values of equation 1 and equation 3:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_3 \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -198.9 \, \text{kJ} + (-247.5 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -446.4 \, \text{kJ} \][/tex]
Therefore, the enthalpy of the overall chemical equation [tex]\( \text{NO (g) + O (g) → NO}_2 \text{(g)} \)[/tex] is:
[tex]\[ \boxed{-446.4 \, \text{kJ}} \][/tex]
However, checking the given options, our calculated answer (-446.4 kJ) does not match any of the options provided. Depending on a reassessment or potential options mismatch, it's crucial to note this detail if further context or data is given subsequently.
We are given the following intermediate chemical equations with their enthalpies:
1. [tex]\(\text{NO (g) + O}_3 \text{(g)} \rightarrow \text{NO}_2 \text{(g) + O}_2 \text{(g)}, \Delta H_1 = -198.9 \, \text{kJ}\)[/tex]
2. [tex]\(\frac{3}{2} \text{O}_2 \text{(g)} \rightarrow \text{O}_3 \text{(g)}, \Delta H_2 = 142.3 \, \text{kJ}\)[/tex]
3. [tex]\(\text{O (g)} \rightarrow \frac{1}{2} \text{O}_2 \text{(g)}, \Delta H_3 = -247.5 \, \text{kJ}\)[/tex]
We need to manipulate these equations to obtain the overall reaction:
[tex]\[ \text{NO (g) + O (g) → NO}_2 \text{(g)} \][/tex]
### Step-by-Step Solution:
1. Reverse the second equation to match the required product [tex]\(\text{O}_3\)[/tex]:
[tex]\[ \text{O}_3 \text{(g)} \rightarrow \frac{3}{2} \text{O}_2 \text{(g)}, \Delta H = -142.3 \, \text{kJ} \][/tex]
2. Sum the enthalpies of the first and third equations as they directly contribute to forming [tex]\(\text{NO}_2 \text{(g)}\)[/tex] and [tex]\(\text{O}_2 \text{(g)}\)[/tex]:
1. [tex]\(\text{NO (g) + O}_3 \text{(g)} \rightarrow \text{NO}_2 \text{(g) + O}_2 \text{(g)}, \Delta H_1 = -198.9 \, \text{kJ}\)[/tex]
2. [tex]\(\text{O} \text{(g)} \rightarrow \frac{1}{2} \text{O}_2 \text{(g)}, \Delta H_3 = -247.5 \, \text{kJ}\)[/tex]
3. By combining the [tex]\(\Delta H\)[/tex] values of equation 1 and equation 3:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_3 \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -198.9 \, \text{kJ} + (-247.5 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -446.4 \, \text{kJ} \][/tex]
Therefore, the enthalpy of the overall chemical equation [tex]\( \text{NO (g) + O (g) → NO}_2 \text{(g)} \)[/tex] is:
[tex]\[ \boxed{-446.4 \, \text{kJ}} \][/tex]
However, checking the given options, our calculated answer (-446.4 kJ) does not match any of the options provided. Depending on a reassessment or potential options mismatch, it's crucial to note this detail if further context or data is given subsequently.
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