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Chapter 13: Problem 134

The activation energy \(\left(E_{\mathrm{a}}\right)\) for the reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=-164 \mathrm{~kJ} / \mathrm{mol} $$ is \(240 \mathrm{~kJ} / \mathrm{mol} .\) What is \(E_{\mathrm{a}}\) for the reverse reaction?

Short Answer

Expert verified
The activation energy for the reverse reaction is \(E_{a(rev)} = 240_{kJ/mol} -164_{kJ/mol} = 76_{kJ/mol}\).

Step by step solution

01

Understand the problem

In this exercise, the given values are: the activation energy, \(E_{a} = 240 \mathrm{~kJ} / \mathrm{mol}\) and change in enthalpy of the forward reaction, \(\Delta H^{\circ}=-164 \mathrm{~kJ} / \mathrm{mol}\). The task is to determine activation energy of the reverse reaction, \(E_{a(rev)}\). The relationship between these variables is given as \(E_{a(rev)} = E_{a} +\Delta H^{\circ}\).
02

Plug the given values into the formula

Now, substitute the given values into the equation like this: \(E_{a(rev)} = 240_{kJ/mol} -164_{kJ/mol}\)
03

Compute

Perform the arithmetic operation as indicated by the formula, and compute the desired value. Here, this implies subtracting the enthalpy change from activation energy of the forward reaction to get the activation energy of reverse reaction.

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Most popular questions from this chapter

When a mixture of methane and bromine is exposed to visible light, the following reaction occurs slowly: $$ \mathrm{CH}_{4}(g)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Br}(g)+\mathrm{HBr}(g) $$ Suggest a reasonable mechanism for this reaction. (Hint: Bromine vapor is deep red; methane is colorless.)

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