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Chapter 14: Problem 86

What does a catalyst do to the time it takes for a reaction to reach equilibrium? Explain how it does this.

Short Answer

Expert verified
A catalyst accelerates a chemical reaction and shortens the time it takes to reach equilibrium by lowering the activation energy barrier and providing an alternative reaction pathway. This increases the probability of successful reactions during molecular collisions, helping the system reach a state of equilibrium faster. However, the catalyst does not affect the final concentrations or position of the equilibrium.

Step by step solution

01

Definition of a Catalyst

A catalyst is a substance that increases the rate of a chemical reaction without being consumed or permanently changed. Catalysts work by lowering the activation energy barrier for a reaction, which increases the frequency of collisions between reacting molecules and therefore accelerates the reaction rate.
02

Activation Energy and Reaction Rate

Activation energy is the minimum energy required for two molecules to react when they collide. High activation energy barriers can make it less likely for molecules to react when they collide, resulting in a slower reaction rate. By lowering the activation energy barrier, a catalyst increases the probability of successful reactions during molecular collisions, thus accelerating the reaction rate.
03

Molecular Mechanism of a Catalyst

The molecular mechanism of a catalyst involves creating an alternative reaction pathway that requires a lower activation energy. This is achieved by forming temporary intermediate compounds that are more reactive and have weaker bonds, which are then broken more easily to form the desired products. The catalyst then returns to its original state, allowing it to be used repeatedly throughout the reaction.
04

Catalyst Effect on Time to Reach Equilibrium

The time it takes for a reaction to reach equilibrium is the time it takes for the rate of the forward reaction to equal the rate of the reverse reaction. By increasing the reaction rate, a catalyst causes the system to approach this state of equilibrium more quickly, shortening the time it takes to reach equilibrium. It's important to note that while a catalyst affects the rate at which equilibrium is reached, it does not alter the final equilibrium concentrations or the position of the equilibrium.
05

Summary

In summary, a catalyst accelerates a chemical reaction and shortens the time it takes to reach equilibrium by lowering the activation energy barrier and providing an alternative reaction pathway. This increases the collision success rate for the reacting molecules and helps the system reach a state of equilibrium faster. However, the catalyst does not affect the final concentrations or position of the equilibrium.

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Most popular questions from this chapter

Write the expression for \(K_{\text {eq }}\) for the reaction \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

Indicate with an arrow the direction of the equilibrium shift and predict what will happen to the amount of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (increases, decreases, unchanged, need more information) when the following stresses are applied to the following exothermic reaction: \(2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g)\) The reaction is cooled. \(\mathrm{H}_{2}\) gas is added. \(\mathrm{H}_{2} \mathrm{O}\) is removed. Volume is reduced. A catalyst is added. Fe is added while the reaction temperature is increased.

Consider the reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{PCl}_{5}(g)\) Use forward and reverse rate meters to represent the forward and reverse reaction rates for: (a) The initial equilibrium (b) Moments after you disturb the equilibrium by adding \(\mathrm{PCl}_{5}\) (c) The restored equilibrium (d) Which way did the reaction shift to get from the initial equilibrium, part (a), to the restored equilibrium, part (c)?

As noted in the chapter, the value of \(K_{\mathrm{eq}}\) for the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) is \(0.0017\) at \(2027^{\circ} \mathrm{C}\) and \(2.3 \times 10^{-9}\) at \(25^{\circ} \mathrm{C}\). (a) Judging from the values of \(K_{\text {eq }}\), does this reaction shift to the left or to the right when the reaction mixture is heated? Explain your answer. (b) Is this reaction endothermic or exothermic? Explain your answer.

For the reaction \(4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g)\) at \(25^{\circ} \mathrm{C}, K_{\mathrm{eq}}=0.150 .\) What is the equilibrium concentration of \(\mathrm{NO}_{2}(g)\) if \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.300 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=1.20 \mathrm{M} ?\)
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