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Chapter 12: Problem 65

The activation energy for the reaction $$ \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ 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 \(-91 \mathrm{~kJ/mol}\).

Step by step solution

01

Recall the principle of microscopic reversibility

According to the principle of microscopic reversibility, the difference in activation energy for the forward and reverse reactions is equal to the change in energy, \(\Delta E\), for the reaction. Mathematically, it can be expressed as: \[E_\text{a, reverse} - E_\text{a, forward} = \Delta E\] Where: \(E_\text{a, forward}\) = Activation energy of the forward reaction \(E_\text{a, reverse}\) = Activation energy of the reverse reaction \(\Delta E\) = Change in energy of the reaction
02

Substitute known values into the equation

We are given the activation energy for the forward reaction, \(E_\text{a, forward} = 125 \mathrm{~kJ/mol}\), and the change in energy for the overall reaction, \(\Delta E = -216\mathrm{~kJ/mol}\). We want to find the activation energy of the reverse reaction, \(E_\text{a, reverse}\). Substitute these values into the equation: \[E_\text{a, reverse} - 125\mathrm{~kJ/mol} = -216\mathrm{~kJ/mol}\]
03

Solve for the activation energy of the reverse reaction

Now we can solve the equation for \(E_\text{a, reverse}\): \[E_\text{a, reverse} = 125\mathrm{~kJ/mol} - 216\mathrm{~kJ/mol}\] \[E_\text{a, reverse} = -91\mathrm{~kJ/mol}\] So, the activation energy for the reverse reaction is \(-91 \mathrm{~kJ/mol}\).

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Most popular questions from this chapter

The central idea of the collision model is that molecules must collide in order to react. Give two reasons why not all collisions of reactant molecules result in product formation.

The rate law for the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) is $$ \text { Rate }=-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right] $$ It takes 120. s for \(1.00 M \mathrm{PH}_{3}\) to decrease to \(0.250 \mathrm{M}\). How much time is required for \(2.00 M \mathrm{PH}_{3}\) to decrease to a concentration of \(0.350 M ?\)

The decomposition of hydrogen iodide on finely divided gold at \(150^{\circ} \mathrm{C}\) is zero order with respect to HI. The rate defined below is constant at \(1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). $$ \begin{gathered} 2 \mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t}=k=1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{gathered} $$ a. If the initial HI concentration was \(0.250 \mathrm{~mol} / \mathrm{L}\), calculate the concentration of HI at 25 minutes after the start of the reaction. b. How long will it take for all of the \(0.250 \mathrm{M}\) HI to decompose?

For the reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\), explain at least two ways in which the rate law could be zero order in chemical A.

How does temperature affect \(k\), the rate constant? Explain.
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