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Chapter 14: Problem 9

What are the units for the rate constants of first-order and second-order reactions?

Short Answer

Expert verified
The unit of the rate constant for a first-order reaction is \(s^{-1}\) and for a second-order reaction, it is \(L \cdot Mol^{-1} s^{-1}\)

Step by step solution

01

Understanding Rate of Reaction

The rate of a chemical reaction is given by the equation: \[Rate = k[A]^n\], where: 'Rate' is the rate of reaction, 'k' is the rate constant, '[A]' is the concentration of the reactants, and 'n' is the order of reaction. We know that the units of the rate of reaction are always \(Mol \cdot L^{-1} \cdot s^{-1}\).
02

First-Order Reaction

A first-order reaction is one where n=1. In this case, the rate of reaction equation simplifies to: \[Rate = k[A]\]. To work out the units of 'k', we rearrange the equation to solve for 'k': \[k = Rate / [A]\]. Given that the units of rate are always \(Mol \cdot L^{-1} \cdot s^{-1}\) and the units of concentration '[A]' are \(Mol \cdot L^{-1}\), it can be found that the units of 'k' for a first-order reaction will be \(s^{-1}\).
03

Second-Order Reaction

A second-order reaction is one where n=2. In this case, the rate of reaction equation simplifies to: \[Rate = k[A]^2\]. To work out the units of 'k', we rearrange the formula for 'k': \[k = Rate / [A]^2\]. Given that the units of rate are always \(Mol \cdot L^{-1} \cdot s^{-1}\) and the units of concentration '[A]' are \(Mol \cdot L^{-1}\), it can be found that the units of 'k' for a second-order reaction will be \(L \cdot Mol^{-1} \cdot s^{-1}\).

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Most popular questions from this chapter

Strontium-90, a radioactive isotope, is a major product of an atomic bomb explosion. It has a half-life of 28.1 yr. (a) Calculate the first-order rate constant for the nuclear decay. (b) Calculate the fraction of \({ }^{90} \mathrm{Sr}\) that remains after 10 half-lives. (c) Calculate the number of years required for 99.0 percent of \({ }^{90} \mathrm{Sr}\) to disappear.

Given the same concentrations, the reaction $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g) $$ at \(250^{\circ} \mathrm{C}\) is \(1.50 \times 10^{3}\) times as fast as the same reaction at \(150^{\circ} \mathrm{C}\). Calculate the energy of activation for this reaction. Assume that the frequency factor is constant.

Write an equation relating the concentration of a reactant \(\mathrm{A}\) at \(t=0\) to that at \(t=t\) for a first-order reaction. Define all the terms and give their units.

The rate law for the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\). Which of these changes will change the value of \(k ?\) (a) The pressure of \(\mathrm{NO}_{2}\) is doubled. (b) The reaction is run in an organic solvent. (c) The volume of the container is doubled. (d) The temperature is decreased. (e) A catalyst is added to the container.

Consider this mechanism for the enzyme-catalyzed reaction $$ \mathrm{E}+\mathrm{S} \stackrel{k_{1}}{\rightleftharpoons_{-1}} \mathrm{ES} \quad \text { (fast equilbrium) } $$ $$ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \quad(\text { slow }) $$ Derive an expression for the rate law of the reaction in terms of the concentrations of \(\mathrm{E}\) and \(\mathrm{S}\). (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.)
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