Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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]
[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]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
