Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest. (a) $E_{a}=75 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-20 \mathrm{~kJ} / \mathrm{mol}$ (b) $E_{a}=100 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=+30 \mathrm{~kJ} / \mathrm{mol}$ (c) $E_{a}=85 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-50 \mathrm{~kJ} / \mathrm{mol}$

Short Answer

Expert verified
The ranking of the reactions from slowest to fastest, based on their activation energies and energy changes, is: (b), (c), (a).

Step by step solution

01

Comparing Activation Energies

To rank the reactions from slowest to fastest, we can first look at their activation energies. The reaction with the highest activation energy is expected to be the slowest because it requires more energy for the reaction to proceed. Recall that the activation energies for the three reactions are: (a) \(E_{a}=75 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=100 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=85 \mathrm{~kJ} / \mathrm{mol}\) From these values, we can see that reaction (b) has the highest activation energy, followed by reaction (c) and then reaction (a).
02

Considering Energy Changes

Now let's consider the energy changes for each reaction. The energy change indicates whether the reaction is endothermic (positive \(\Delta E\)) or exothermic (negative \(\Delta E\)). In general, exothermic reactions tend to be faster than endothermic reactions, as they release energy and promote more particle collisions. The energy changes for the three reactions are: (a) \(\Delta E=-20 \mathrm{~kJ} / \mathrm{mol}\) (b) \(\Delta E=+30 \mathrm{~kJ} / \mathrm{mol}\) (c) \(\Delta E=-50 \mathrm{~kJ} / \mathrm{mol}\) From these values, we can see that reaction (a) and (c) are exothermic while reaction (b) is endothermic. Since exothermic reactions are generally faster, we can tentatively rank the reactions as: (c) > (a) > (b).
03

Combining Activation Energies and Energy Changes

Finally, we can combine the rankings from the activation energies and energy changes to determine the overall ranking from slowest to fastest. Taking into account the activation energy ranking from Step 1 and the energy change ranking from Step 2, we have: Slowest to Fastest: (b) > (c) > (a) Therefore, the ranking of the reactions from slowest to fastest is: (b), (c), (a).

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Most popular questions from this chapter

For the elementary process $\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)$ the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are $154 \mathrm{~kJ} / \mathrm{mol}\( and \)136 \mathrm{~kJ} / \mathrm{mol}$, respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

(a) Can an intermediate appear as a reactant in the first step of a reaction mechanism? (b) On a reaction energy profile diagram, is an intermediate represented as a peak or a valley? (c) If a molecule like \(\mathrm{Cl}_{2}\) falls apart in an elementary reaction, what is the molecularity of the reaction?

From the following data for the second-order gas-phase decomposition of HI at \(430^{\circ} \mathrm{C}\), calculate the second-order rate constant and half- life for the reaction: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { [HI]/mol } \mathrm{dm}^{-3} \\ \hline 0 & 1 \\ 100 & 0.89 \\ \hline 200 & 0.8 \\ \hline 300 & 0.72 \\ \hline 400 & 0.66 \\ \hline \end{array} $$

A certain enzyme catalyzes a biochemical reaction. In water, without the enzyme, the reaction proceeds with a rate constant of $6.50 \times 10^{-4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C} .$ In the presence of the enzyme in water, the reaction proceeds with a rate constant of $1.67 \times 10^{4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C}$. Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

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