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We have seen that the unit of \(k\) depends on the overall order of a reaction. Derive a general expression for the units of \(k\) for a reaction of any overall order, based on the order of the reaction (o) and the units of concentration (M) and time (s).

Short Answer

Expert verified
The general expression for the unit of the rate constant 'k' in terms of the overall order 'o' of the reaction, the units of concentration (M) and time (s) is \(k = \frac{1}{(M)^{o-1}*s}\).

Step by step solution

01

Knowing the Rate Law

Remember the rate law for a reaction: \(Rate = k * [A]^n\), where 'Rate' is the rate of reaction, 'k' is the rate constant, '[A]' is the molarity (M), 'n' is the order with respect to 'A'.
02

Rearranging the Rate Law for 'k'

Rearrange the rate law to isolate 'k', so the equation becomes \(k = \frac{Rate}{[A]^n}\).
03

Recognizing Units

Recognize the units of the parts of the equation. The 'Rate' generally has units of M/s, '[A]' is in Molarity (M), and 'n' is the overall order without unit.
04

Inserting Units into the Equation

Plug the units in place of the variables in the equation as follows: \(k = \frac{M/s}{(M)^n}\).
05

Simplifying the Units

Simplify the equation. The result is the unit of 'k', which is \(k = \frac{1}{(M)^{n-1}*s}\). The minus one comes from multiplying by the 'M' in the rate numerator;
06

General Expression

To express it more generally, consider the order 'o' of the reaction, the unit for 'k' derived for a reaction of order 'o' will be \(k = \frac{1}{(M)^{o-1}*s}\).

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Key Concepts

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

Understanding Chemical Kinetics
Chemical kinetics is the study of the speed at which chemical reactions occur and the factors that influence this rate. It's an essential area within physical chemistry that helps us understand reaction mechanisms and conditions under which chemical processes can be optimized. In kinetics, we examine variables like temperature, pressure, and concentration to determine how they affect the rate of a reaction.

To fully grasp chemical kinetics, we must begin with the rate of reaction, which refers to the change in concentration of a reactant or product over time. It can be expressed as the decrease in the concentration of reactants per unit time or the increase in concentration of products per unit time. This rate can be affected by various factors like temperature, catalysts, and the physical state of the reactants. For educational purposes, it's useful to understand methods to measure reaction rates and how to interpret kinetic data in graphical form.
Deciphering the Rate Law
The rate law is a mathematical expression that links the rate of a chemical reaction to the concentration of the reactants. It takes the form of an equation: \(Rate = k * [A]^n\), where 'Rate' is the rate of reaction, 'k' is the rate constant, '[A]' is the molarity of reactant A, and 'n' represents the reaction order with respect to A. The rate constant 'k' is a proportionality factor that is specific to the given reaction at a particular temperature.

Determining the Rate Law

To find the rate law, one typically conducts experiments to measure the reaction rate at different concentrations of reactants. The rate constant can be determined from these experimental data, and its units depend on the reaction order, as we will explore next.
Exploring Reaction Order and Rate Constant Units
Reaction order is a key concept in chemical kinetics that specifies how the rate is affected by the concentration of each reactant. It is usually derived from experimental data and can be an integer, zero, or even a fraction. The overall order of a reaction is the sum of the orders with respect to each reactant involved.

Rate Constant Units

Based on the overall reaction order, the units for the rate constant 'k' can be derived. For a reaction of order 'o', the unit for 'k' is given by \(k = \frac{1}{(M)^{o-1}*s}\). This shows that the rate constant units vary and are inversely related to the order of the reaction minus one, reflecting changes in the concentration's impact on the rate. For example, a first-order reaction (o=1) will have 'k' units of \(s^{-1}\), while a second-order reaction (o=2) would have 'k' units of \(M^{-1}s^{-1}\), indicating more complex relationships between concentration changes and the reaction rate.

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Most popular questions from this chapter

The half-life of the radioactive isotope phosphorus- 32 is 14.3 days. How long does it take for a sample of phosphorus-32 to lose \(99 \%\) of its radioactivity?

The following substrate concentration [S] versus time data were obtained during an enzyme-catalyzed reaction: \(t=0 \min ,[\mathrm{S}]=1.00 \mathrm{M} ; 20 \mathrm{min}, 0.90 \mathrm{M}; 60 \min , 0.70 \mathrm{M} ; 100 \mathrm{min}, 0.50 \mathrm{M} ; 160 \mathrm{min}, 0.20 \mathrm{M}.\) What is the order of this reaction with respect to \(\mathrm{S}\) in the concentration range studied?

Three different sets of data of \([\mathrm{A}]\) versus time are giv the following table for the reaction \(A \longrightarrow\) prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ What is the approximate half-life of the first-order reaction?

A kinetic study of the reaction \(A \longrightarrow\) products yields the data: \(t=0 \mathrm{s},[\mathrm{A}]=2.00 \mathrm{M} ; 500 \mathrm{s}, 1.00 \mathrm{M}; 1500 \mathrm{s}, 0.50 \mathrm{M} ; 3500 \mathrm{s}, 0.25 \mathrm{M} .\) Without performing detailed calculations, determine the order of this reaction and indicate your method of reasoning.

A simplified rate law for the reaction \(2 \mathrm{O}_{3}(\mathrm{g}) \longrightarrow\) \(3 \mathrm{O}_{2}(\mathrm{g})\) is $$\text { rate }=k=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]}$$ For this reaction, propose a two-step mechanism that consists of a fast, reversible first step, followed by a slow second step.

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