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Chapter 7: Problem 5

For the zeroth-order reaction: \(\mathrm{A} \rightarrow\) products, what will happen to the rate of reaction if the concentration A is doubled? a. The rate will be doubled b. The rate will be halved c. The rate will remain the same d. The rate will be quadrupled

Short Answer

Expert verified
c. The rate will remain the same

Step by step solution

01

Understand Zeroth-Order Reactions

In a zeroth-order reaction, the rate of reaction is independent of the concentration of the reactant. This means that the rate of reaction does not change with changes in the concentration of the reactant.
02

Apply the Zeroth-Order Rate Law

The rate law for a zeroth-order reaction is given by \[\text{Rate} = k\]where \(k\) is the rate constant. Notice that the concentration of reactant \([A]\) is not present in this expression, which supports the independence of the rate from the concentration of \(A\).
03

Analyze the Effect of Doubling Concentration

Since the rate law for a zeroth-order reaction is independent of the concentration of \(A\), doubling the concentration of \(A\) will not affect the rate. The rate remains constant regardless of the concentration change.
04

Conclusion

Based on the above analysis, doubling the concentration of \(A\) has no effect on the rate of a zeroth-order reaction. Therefore, the correct answer is that the rate will remain the same.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate of Reaction in Zeroth-Order Reactions
In the context of chemical kinetics, the term 'rate of reaction' refers to how fast reactants are converted into products. For many reactions, this rate can vary with changes in the concentration of the reactants. However, zeroth-order reactions behave uniquely.

For a zeroth-order reaction, the rate of reaction is constant, regardless of how the concentration of the reactant changes. Whether you have a large amount or a tiny bit of the reactant, you'll find that the rate at which the reaction proceeds remains the same. This might be quite surprising because it contrasts with other types of reactions where the rate typically changes with concentration.

In simpler terms, think of zeroth-order reactions as being on autopilot. No matter what you do to the amount of the reactant, the rate stays the same!
Understanding the Rate Law for Zeroth-Order Reactions
The rate law is an equation that relates the rate of a chemical reaction to the concentration of its reactants. For most types of reactions, the rate law will include terms for the concentration of reactants, often raised to a certain power.

However, and interestingly, in the case of zeroth-order reactions, the rate law is written as \[\text{Rate} = k\]where \(k\) is the rate constant. Notice the absence of the concentration of reactant \([A]\) in this expression!

This absence highlights that the rate is entirely determined by the rate constant \(k\), not the amount of reactant present. The rate constant \(k\) remains constant regardless of whatever concentration shenanigans you might perform! This is essential in maintaining the reaction's independence from concentration.
Concentration Independence in Zeroth-Order Reactions
The fascinating concept of concentration independence is what sets zeroth-order reactions apart from other reaction orders. In typical chemical reactions, changes in the concentration of a reactant will result in a change in the reaction rate. But zeroth-order reactions are an exception to this rule.

Consider it like a moving conveyor belt in a factory鈥攏o matter how many items (or concentration) you place on it, the speed of the belt (or rate of reaction) does not change. Doubling, halving, or even quadrupling the concentration of the reactant will not alter the rate of reaction. The process continues at a fixed pace due to the rate law, which is not dependent on reactant levels.

This unique property makes zeroth-order reactions particularly easy to study and predictable in their behavior, as they stay unaffected by concentration changes.

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Most popular questions from this chapter

The following set of data was obtained by the method of initial rates for the reaction: \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow\) \(2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq})\) What is the rate law for the reaction? \begin{tabular}{lll} \hline\(\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right], \mathrm{M}\) & {\(\left[\mathrm{I}^{-}\right], \mathrm{M}\)} & Initial rate, \(\mathrm{M} \mathrm{s}^{-1}\) \\ \hline \(0.25\) & \(0.10\) & \(9.00 \times 10^{-3}\) \\ \(0.10\) & \(0.10\) & \(3.60 \times 10^{-3}\) \\ \(0.20\) & \(0.30\) & \(2.16 \times 10^{-2}\) \\ \hline \end{tabular} a. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]^{2}\) b. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]^{2}\left[\mathrm{I}^{-}\right]\) c. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right]\left[\mathrm{I}^{-}\right]\) d. Rate \(=\mathrm{k}\left[\mathrm{S}, \mathrm{O}_{\mathrm{g}}^{2-}\right]\left[\mathrm{I}^{-}\right]^{5}\)

Consider a reaction \(\mathrm{aG}+\mathrm{bH} \rightarrow\) Products. When concentration of both the reactants \(\mathrm{G}\) and \(\mathrm{H}\) is doubled, the rate increases by eight times. However when concentration of \(\mathrm{G}\) is doubled keeping the concentration of \(\mathrm{H}\) fixed, the rate is doubled. The overall order of the reaction is a. 0 b. 1 c. 2 d. 3

The reaction of hydrogen and iodine monochloride is given as:\(\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{ICl}(\mathrm{g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~g})\) This reaction is of first order with respect to \(\mathrm{H}_{2}(\mathrm{~g})\) and ICl (g), following mechanisms were proposed: Mechanism (1): \(\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{ICl}(\mathrm{g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~g})\) Mechanism (2): \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{ICl}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{HI}_{2}(\mathrm{~g}) ;\) slow \(\mathrm{HI}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~g}) ;\) fast Which of the above mechanism(s) can be consistent with the given information about the reaction? a. 2 only b. Both 1 and 2 c. Neither 1 nor 2 d. I only

If the volume of the vessel in which the reaction \(2 \mathrm{NO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{NO}_{2}\) is occurring is diminished to \(1 / 3\) rd of its initial volume. The rate of the reaction will be increased by a. 5 times b. 8 times c. 27 times d. 35 times

Hydrogen iodide decomposes at \(800 \mathrm{~K}\) via a second order process to produce hydrogen and iodine according to the following chemical equation. \(2 \mathrm{HI}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\) At \(800 \mathrm{~K}\) it takes 142 seconds for the initial concentration of HI to decrease from \(6.75 \times 10^{-2} \mathrm{M}\) to \(3.50 \times 10^{-2} \mathrm{M}\). What is the rate constant for the reaction at this temperature?a. \(6.69 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) b. \(7.96 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) c. \(19.6 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) d. \(9.69 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\)
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