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Chapter 6: Problem 73

Given the following data $$ \begin{aligned} 2 \mathrm{O}_{3}(g) & \longrightarrow 3 \mathrm{O}_{2}(g) & \Delta H &=-427 \mathrm{~kJ} \\ \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{O}(g) & \Delta H &=495 \mathrm{~kJ} \\ \mathrm{NO}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-199 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{NO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}_{2}(g) $$

Short Answer

Expert verified
The enthalpy change for the reaction \(NO(g) + O(g) \rightarrow NO_{2}(g)\) is \(-907.5\:kJ\).

Step by step solution

01

Analyze the given reactions

List the given reactions and their enthalpy changes: 1. \(2\: O_{3}(g) \rightarrow 3\:O_{2}(g)\) and \(\Delta H_{1} = -427\:kJ\) 2. \(O_{2}(g) \rightarrow 2\:O(g)\) and \(\Delta H_{2} = 495\:kJ\) 3. \(NO(g) + O_{3}(g) \rightarrow NO_{2}(g) + O_{2}(g)\) and \(\Delta H_{3} = -199\:kJ\) Our target reaction is: \(NO(g) + O(g) \rightarrow NO_{2}(g)\)
02

Manipulate the given reactions

We'll now create a series of manipulations to the given reactions to obtain the desired reaction: 1. Divide reaction 1 by 2: \(\frac{1}{2}(2 O_{3}(g) \rightarrow 3 O_{2}(g))\); and likewise, divide \(\Delta H_{1}\) by 2: \(\frac{1}{2}(-427\:kJ) = -213.5\:kJ\). This results in the new reaction 1: \(O_{3}(g) \rightarrow \frac{3}{2} O_{2}(g)\) and \(\Delta H_{1'} = -213.5\:kJ\) 2. Reverse reaction 2: \(2 O(g) \rightarrow O_{2}(g)\) and, consequently, reverse the sign of \(\Delta H_{2}\): \(-495\:kJ\)
03

Combine the manipulated reactions

Now, add the manipulated reactions to obtain the target reaction: \(O_{3}(g) \rightarrow \frac{3}{2} O_{2}(g)\) (\(\Delta H_{1'} = -213.5\:kJ\)) + \(2 O(g) \rightarrow O_{2}(g)\) (\(\Delta H_{2'} = -495\:kJ\)) + \(NO(g) + O_{3}(g) \rightarrow NO_{2}(g) + O_{2}(g)\) (\(\Delta H_{3} = -199\:kJ\)) \(\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\) \(NO(g) + O(g) \rightarrow NO_{2}(g)\) Next, sum up the respective \(\Delta H\) values: \(\Delta H = \Delta H_{1'} + \Delta H_{2'} + \Delta H_{3}\)
04

Calculate the enthalpy change for the target reaction

Utilizing the formula above, calculate the enthalpy change for the target reaction: \(\Delta H = (-213.5\:kJ) + (-495\:kJ) + (-199\:kJ)\) \(\Delta H = -907.5\:kJ\) So, the enthalpy change for the reaction \(NO(g) + O(g) \rightarrow NO_{2}(g)\) is \(-907.5\:kJ\).

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