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Chapter 11: Problem 65

The activation energy for the reaction $$\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$ is \(125 \mathrm{kJ} / \mathrm{mol},\) and \(\Delta E\) for the reaction is \(-216 \mathrm{kJ} / \mathrm{mol}\). What is the activation energy for the reverse reaction \(\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?\)

Short Answer

Expert verified
The activation energy for the reverse reaction is \(-91 kJ/mol\). Since the activation energy is negative, it indicates that the reverse reaction is exothermic and has a lower activation energy compared to the endothermic forward reaction.

Step by step solution

01

Write the equation relating activation energies and change in energy

The relationship between the activation energies of the forward and reverse reactions, and the change in energy for the reaction is given by the following equation: \(E_{a(reverse)} = E_{a(forward)} + \Delta E\)
02

Input the given values

Given the values for the activation energy of the forward reaction (\(E_{a(forward)} = 125 kJ/mol\)) and the change in energy for the reaction (\(\Delta E = -216 kJ/mol\)), plug them into the equation: \(E_{a(reverse)} = 125 kJ/mol - 216 kJ/mol\)
03

Calculate the activation energy for the reverse reaction

Now, we just need to perform the subtraction: \(E_{a(reverse)} = -91 kJ/mol\)
04

Interpret the result

The activation energy for the reverse reaction is \(-91 kJ/mol\). Since the activation energy is negative, it means that the reverse reaction is exothermic, and it has a lower activation energy compared to the forward reaction, which requires an input of energy (endothermic).

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Most popular questions from this chapter

The reaction $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-}$$ in a certain solvent is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to OH \(^{-} .\) In several experiments, the rate constant \(k\) was determined at different temperatures. A plot of \(\ln (k)\) versus \(1 / T\) was constructed resulting in a straight line with a slope value of \(-1.10 \times 10^{4} \mathrm{K}\) and \(y\) -intercept of 33.5. Assume \(k\) has units of \(\mathrm{s}^{-1}\).a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor \(A\) c. Calculate the value of \(k\) at \(25^{\circ} \mathrm{C}\)

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