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Chapter 14: Problem 61

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? Explain your answer. (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)

Short Answer

Expert verified
In conclusion, based on their activation energies, Reaction (b) is the fastest, and Reaction (c) is the slowest. This is because Reaction (b) has the lowest activation energy of 35 kJ/mol, and Reaction (c) has the highest activation energy of 55 kJ/mol.

Step by step solution

01

Understand the relationship between activation energy and reaction rate

To analyze the given reactions, we must first understand how activation energy affects the rate of a reaction. Activation energy is the minimum amount of energy required for a reaction to occur. Reactions with a lower activation energy require less energy to proceed and therefore are more likely to occur, meaning that they are faster. In contrast, reactions with a higher activation energy require more energy, making them slower.
02

Compare activation energies

To determine which reaction is the fastest and which is the slowest, we need to compare the activation energies of the three reactions: (a) Activation Energy (Ea) = 45 kJ/mol (b) Activation Energy (Ea) = 35 kJ/mol (c) Activation Energy (Ea) = 55 kJ/mol
03

Determine the fastest and slowest reactions

Based on the comparisons in Step 2, we can determine the following: - Fastest reaction: Since the reaction with the lowest activation energy is the fastest one, Reaction (b) has the lowest activation energy of 35 kJ/mol, making it the fastest reaction. - Slowest reaction: Since the reaction with the highest activation energy is the slowest, Reaction (c) has the highest activation energy of 55 kJ/mol, making it the slowest reaction. In conclusion, Reaction (b) is the fastest, and Reaction (c) is the slowest. This is because Reaction (b) has the lowest activation energy, and Reaction (c) has the highest activation energy.

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Most popular questions from this chapter

Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carboncontaining fragment such as \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2},\) and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a \(\mathrm{C}\) atom for \({ }^{\text {"R" }}\) ). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\) \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) .\) In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

Indicate whether each statement is true or false. If it is false, rewrite it so that it is true. (a) If you measure the rate constant for a reaction at different temperatures, you can calculate the overall enthalpy change for the reaction. (b) Exothermic reactions are faster than endothermic reactions. (c) If you double the temperature for a reaction, you cut the activation energy in half.

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C},\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g) .\) If the reaction is first order with a half-life of 56.3 min at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).
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