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Chapter 13: Problem 113

Does a substance that increases the rate of a reaction also increase the rate of the reverse reaction?

Short Answer

Expert verified
Explain your answer. Answer: Yes, a substance that increases the rate of a reaction (catalyst) can also increase the rate of the reverse reaction. This is because the catalyst reduces the activation energy for both forward and reverse reactions without changing the position of the equilibrium, providing an alternate pathway with lower activation energy for both reactions.

Step by step solution

01

Understand the role of a catalyst

A substance that increases the rate of a reaction is often called a catalyst. A catalyst is a substance that increases the rate of a chemical reaction by reducing the energy required for the reaction to occur, without being consumed in the process.
02

Identify what affects reaction rates

Reaction rates are influenced by various factors, such as temperature, pressure, concentration, and the presence of a catalyst. In this exercise, we will focus on the role of a catalyst in affecting reaction rates.
03

Consider the impact of a catalyst on forward and reverse reactions

A catalyst can increase the rate of a forward reaction by providing an alternate pathway with lower activation energy. However, it is crucial to remember that a catalyst does not change the overall energy balance of the reaction. In other words, the position of equilibrium remains unchanged.
04

Analyze the catalyst's effect on the reverse reaction

Since a catalyst provides an alternate pathway with lower activation energy for both forward and reverse reactions, and it does not change the overall energy balance of the reaction, it means that it can potentially increase the rate of the reverse reaction as well.
05

Conclusion

A substance that increases the rate of a reaction (catalyst) can also increase the rate of the reverse reaction. This is because the catalyst reduces the activation energy for both forward and reverse reactions without changing the position of the equilibrium.

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Most popular questions from this chapter

Under what circumstances is the activation energy of a reaction proceeding in the forward direction greater than the activation energy of it happening in reverse?

In the following mechanism for NO formation, oxygen atoms are produced by breaking \(\mathrm{O}=\mathrm{O}\) bonds at high temperature in a fast reversible reaction. If \(\Delta[\mathrm{NO}] / \Delta t=k\left[\mathrm{N}_{2}\right]\left[\mathrm{O}_{2}\right]^{1 / 2},\) which step in the mechanism is the rate-determining step? $$\begin{aligned} (1)\quad\quad\quad\quad\quad\mathrm{O}_{2}(g) & \rightleftharpoons 2 \mathrm{O}(g) \\ (2)\quad\quad\mathrm{O}(g)+\mathrm{N}_{2}(g) & \rightarrow \mathrm{NO}(g)+\mathrm{N}(g) \\ (3)\quad\quad\mathrm{N}(g)+\mathrm{O}(g) & \rightarrow \mathrm{NO}(g) \\ overall \quad \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \rightarrow 2 \mathrm{NO}(g) \end{aligned}$$

Each of the following reactions is first order in each reactant and second order overall. Which reaction is fastest if the initial concentrations of all the reactants are the same? a. \(\mathrm{ClO}_{2}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO}_{3}(g)+\mathrm{O}_{2}(g)\) \(k=3.0 \times 10^{-19} \mathrm{cm}^{3} /(\text { molecule } \cdot \mathrm{s})\) b. \(\mathrm{ClO}_{2}(g)+\mathrm{NO}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{ClO}(g)\) \(k=3.4 \times 10^{-13} \mathrm{cm}^{3} /(\text { molecule } \cdot \mathrm{s})\) c. \(\mathrm{ClO}(g)+\mathrm{NO}(g) \rightarrow \overline{\mathrm{C} 1(g)}+\mathrm{NO}_{2}(g)\) \(k=1.7 \times 10^{-11} \mathrm{cm}^{3} /(\text { molecule } \cdot \mathrm{s})\) d. \(\mathrm{ClO}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO}_{2}(g)+\mathrm{O}_{2}(g)\) \(k=1.5 \times 10^{-17} \mathrm{cm}^{3} /(\text { molecule } \cdot \mathrm{s})\)

Can the half-life of a second-order reaction have the same units as the half- life of a first-order reaction?

A proposed mechanism for the gas phase decomposition of hydrogen peroxide at an elevated temperature consists of three elementary steps: $$\begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(g) & \rightarrow 2 \mathrm{OH}(g) \\ \mathrm{H}_{2} \mathrm{O}_{2}(g)+\mathrm{OH}(g) & \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{HO}_{2}(g) \\ \mathrm{HO}_{2}(g)+\mathrm{OH}(g) & \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) \end{aligned}$$ If the rate law for the reaction is first order in \(\mathrm{H}_{2} \mathrm{O}_{2},\) which step in the mechanism is the rate-determining step?
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