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Sagot :
To solve this problem, we use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the steps that lead to the final reaction. We are given three intermediate equations with their respective enthalpy changes:
[tex]\[ \begin{array}{ll} 1) & NO ( g ) + O_3 ( g ) \rightarrow NO_2 ( g ) + O_2 ( g ) \quad \Delta H_1 = -198.9 \, \text{kJ} \\ 2) & \frac{3}{2} O_2 ( g ) \rightarrow O_3 ( g ) \quad \Delta H_2 = 142.3 \, \text{kJ} \\ 3) & O ( g ) \rightarrow \frac{1}{2} O_2 ( g ) \quad \Delta H_3 = -247.5 \, \text{kJ} \end{array} \][/tex]
We seek the enthalpy change for the overall reaction:
[tex]\[ NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \][/tex]
To find this, we need to combine the given reactions such that the resultant equation matches the desired overall reaction [tex]\( NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \)[/tex].
First, note the target reaction:
[tex]\[ NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \][/tex]
Let's examine how to combine the given equations:
1. Start with the first equation:
[tex]\[ NO ( g ) + O_3 ( g ) \rightarrow NO_2 ( g ) + O_2 ( g ) \][/tex]
2. Then, consider the reverse of the second equation (because we need [tex]\( O_3 \)[/tex] to cancel out and match oxygen atoms):
[tex]\[ O_3 ( g ) \rightarrow \frac{3}{2} O_2 ( g ) \quad \Delta H = -142.3 \, \text{kJ} \][/tex]
3. Lastly, take the third equation:
[tex]\[ O ( g ) \rightarrow \frac{1}{2} O_2 ( g ) \][/tex]
Now, add up these reactions:
[tex]\[ \begin{align*} NO ( g ) + O_3 ( g ) & \rightarrow NO_2 ( g ) + O_2 ( g ) & (\Delta H_1 = -198.9 \, \text{kJ}) \\ O_3 ( g ) & \rightarrow \frac{3}{2} O_2 ( g ) & (\Delta H = -142.3 \, \text{kJ}) \\ O ( g ) & \rightarrow \frac{1}{2} O_2 ( g ) & (\Delta H_3 = -247.5 \, \text{kJ}) \end{align*} \][/tex]
Summing these reactions:
[tex]\[ \begin{array}{ll} NO ( g ) + O_3 ( g ) & \rightarrow NO_2 ( g ) + O_2 ( g ) \\ O_3 ( g ) & \rightarrow \frac{3}{2} O_2 ( g ) \\ O ( g ) & \rightarrow \frac{1}{2} O_2 ( g ) \end{array} \][/tex]
Simplifying by canceling the common terms on both sides, we get:
[tex]\[ NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \][/tex]
The sum of the enthalpy changes is:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 + \Delta H_3 \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{\text{overall}} = -198.9 \, \text{kJ} + 142.3 \, \text{kJ} + (-247.5 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -198.9 + 142.3 - 247.5 = -304.1 \, \text{kJ} \][/tex]
Therefore, the enthalpy change for the overall reaction [tex]\( NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \)[/tex] is:
[tex]\[ \boxed{-304.1 \, \text{kJ}} \][/tex]
[tex]\[ \begin{array}{ll} 1) & NO ( g ) + O_3 ( g ) \rightarrow NO_2 ( g ) + O_2 ( g ) \quad \Delta H_1 = -198.9 \, \text{kJ} \\ 2) & \frac{3}{2} O_2 ( g ) \rightarrow O_3 ( g ) \quad \Delta H_2 = 142.3 \, \text{kJ} \\ 3) & O ( g ) \rightarrow \frac{1}{2} O_2 ( g ) \quad \Delta H_3 = -247.5 \, \text{kJ} \end{array} \][/tex]
We seek the enthalpy change for the overall reaction:
[tex]\[ NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \][/tex]
To find this, we need to combine the given reactions such that the resultant equation matches the desired overall reaction [tex]\( NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \)[/tex].
First, note the target reaction:
[tex]\[ NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \][/tex]
Let's examine how to combine the given equations:
1. Start with the first equation:
[tex]\[ NO ( g ) + O_3 ( g ) \rightarrow NO_2 ( g ) + O_2 ( g ) \][/tex]
2. Then, consider the reverse of the second equation (because we need [tex]\( O_3 \)[/tex] to cancel out and match oxygen atoms):
[tex]\[ O_3 ( g ) \rightarrow \frac{3}{2} O_2 ( g ) \quad \Delta H = -142.3 \, \text{kJ} \][/tex]
3. Lastly, take the third equation:
[tex]\[ O ( g ) \rightarrow \frac{1}{2} O_2 ( g ) \][/tex]
Now, add up these reactions:
[tex]\[ \begin{align*} NO ( g ) + O_3 ( g ) & \rightarrow NO_2 ( g ) + O_2 ( g ) & (\Delta H_1 = -198.9 \, \text{kJ}) \\ O_3 ( g ) & \rightarrow \frac{3}{2} O_2 ( g ) & (\Delta H = -142.3 \, \text{kJ}) \\ O ( g ) & \rightarrow \frac{1}{2} O_2 ( g ) & (\Delta H_3 = -247.5 \, \text{kJ}) \end{align*} \][/tex]
Summing these reactions:
[tex]\[ \begin{array}{ll} NO ( g ) + O_3 ( g ) & \rightarrow NO_2 ( g ) + O_2 ( g ) \\ O_3 ( g ) & \rightarrow \frac{3}{2} O_2 ( g ) \\ O ( g ) & \rightarrow \frac{1}{2} O_2 ( g ) \end{array} \][/tex]
Simplifying by canceling the common terms on both sides, we get:
[tex]\[ NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \][/tex]
The sum of the enthalpy changes is:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 + \Delta H_3 \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{\text{overall}} = -198.9 \, \text{kJ} + 142.3 \, \text{kJ} + (-247.5 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -198.9 + 142.3 - 247.5 = -304.1 \, \text{kJ} \][/tex]
Therefore, the enthalpy change for the overall reaction [tex]\( NO ( g ) + O ( g ) \rightarrow NO_2 ( g ) \)[/tex] is:
[tex]\[ \boxed{-304.1 \, \text{kJ}} \][/tex]
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