Rate Laws and Integrated Rate Laws

The rate law describes how the reaction rate depends on theconcentrations of the reacting species.

For a general reaction

$$aA+bB\rightarrow\text{products},$$

the reaction rate is defined as

$$v = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt}.$$

An experimentally determined rate law may have the form

$$v=k[A]^\alpha[B]^\beta,$$

where $\alpha$ and $\beta$ are the reaction orders with respect to $A$ and $B$. The overall reaction order is

$$n=\alpha+\beta.$$

Reaction orders are determined experimentally. They do not generally have to be the same as the stoichiometric coefficients in the balanced equation. For an elementary reaction, however, the rate law follows directly from the molecularity of that elementary step.

Units of the Rate Constant

For an overall rate law

$$v=k[A]^n,$$

the units of rate are concentration per unit time:

$$[v]=\mathrm{mol\,L^{-1}\,s^{-1}}.$$

Therefore,

$$[k] = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}} {(\mathrm{mol\,L^{-1}})^n},$$

which gives

$$\boxed{ [k] = (\mathrm{mol\,L^{-1}})^{1-n}\mathrm{s^{-1}}. }$$

For a zero-order reaction,

$$[k]=\mathrm{mol\,L^{-1}\,s^{-1}}.$$

For a first-order reaction,

$$[k]=\mathrm{s^{-1}}.$$

For a second-order reaction,

$$[k]=\mathrm{L\,mol^{-1}\,s^{-1}}.$$

For a third-order reaction,

$$[k]=\mathrm{L^2\,mol^{-2}\,s^{-1}}.$$

General Integrated Rate Law

Consider a reaction described by

$$-\frac{d[A]}{dt}=k[A]^n.$$

To obtain concentration as a function of time, separate the concentration and time variables:

$$\frac{d[A]}{[A]^n}=-k\,dt.$$

or

$$[A]^{-n}d[A]=-k\,dt.$$

Integrate from the initial concentration $[A]_0$ at $t=0$ to the concentration $[A]$ at time $t$:

$$\int_{[A]_0}^{[A]}[A]^{-n}d[A] = -k\int_0^t dt.$$

For $n\neq1$,

$$\int [A]^{-n}d[A] = \frac{[A]^{1-n}}{1-n}.$$

Therefore,

$$\left[ \frac{[A]^{1-n}}{1-n} \right]_{[A]_0}^{[A]} = -kt.$$

Expanding the limits,

$$\frac{[A]^{1-n}-[A]_0^{1-n}}{1-n} = -kt.$$

Multiplying by $1-n$ gives

$$[A]^{1-n}-[A]_0^{1-n} = (n-1)kt.$$

Hence,

$$\boxed{ [A]^{1-n} = [A]_0^{1-n}+(n-1)kt }$$

for $n\neq1$.

The first-order case cannot be obtained by directly substituting $n=1$ into this expression because the integration then involves $[A]^{-1}$ and gives a logarithm.

Zero-Order Reactions

For a zero-order reaction,

$$-\frac{d[A]}{dt}=k.$$

Rearranging,

$$d[A]=-k\,dt.$$

Integrating from $[A]_0$ at $t=0$ to $[A]$ at time $t$,

$$\int_{[A]_0}^{[A]}d[A] = -k\int_0^t dt.$$

This gives

$$[A]-[A]_0=-kt,$$

and therefore

$$\boxed{ [A]=[A]_0-kt. }$$

The concentration decreases linearly with time, so a plot of $[A]$ against $t$ has slope $-k$.

For the half-life,

$$[A]=\frac{[A]_0}{2}.$$

Substituting this into the integrated rate law,

$$\frac{[A]_0}{2} = [A]_0-kt_{1/2}.$$

Thus,

$$kt_{1/2} = \frac{[A]_0}{2},$$

and

$$\boxed{ t_{1/2} = \frac{[A]_0}{2k}. }$$

The half-life of a zero-order reaction therefore depends on the initial concentration.

Zero-order behavior can occur when another part of the reaction system is saturated. Once all available catalytic sites are occupied, for example, increasing the reactant concentration may no longer increase the rate.

First-Order Reactions

For a first-order reaction,

$$-\frac{d[A]}{dt}=k[A].$$

Separate the variables:

$$\frac{d[A]}{[A]}=-k\,dt.$$

Integrating between the initial and final conditions,

$$\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k\int_0^t dt.$$

Since

$$\int\frac{d[A]}{[A]}=\ln[A],$$

we obtain

$$\ln[A]-\ln[A]_0=-kt.$$

Using

$$\ln x-\ln y=\ln\frac{x}{y},$$

the integrated rate law becomes

$$\boxed{ \ln\frac{[A]}{[A]_0}=-kt. }$$

Exponentiating both sides,

$$\frac{[A]}{[A]_0}=e^{-kt},$$

so

$$\boxed{ [A]=[A]_0e^{-kt}. }$$

The logarithmic form can also be written as

$$\ln[A]=\ln[A]_0-kt.$$

Thus a plot of $\ln[A]$ against $t$ is linear with slope $-k$.

For the half-life,

$$[A]=\frac{[A]_0}{2}.$$

Substitution gives

$$\ln \left( \frac{[A]_0/2}{[A]_0} \right) = -kt_{1/2}.$$

Therefore,

$$\ln\frac12=-kt_{1/2}.$$

Since

$$\ln\frac12=-\ln2,$$

we obtain

$$\boxed{ t_{1/2}=\frac{\ln2}{k}. }$$

The initial concentration disappears from the result, so the half-life of a first-order reaction is constant.

If $x$ is the fraction of reactant converted,

$$[A]=[A]_0(1-x).$$

Using the integrated first-order law,

$$\ln \left( \frac{[A]_0(1-x)}{[A]_0} \right) = -kt,$$

so

$$\ln(1-x)=-kt.$$

Hence,

$$\boxed{ t=-\frac{1}{k}\ln(1-x). }$$

For $90\%$ conversion,

$$x=0.90,$$

so

$$t_{90} = -\frac{1}{k}\ln(0.10) = \frac{\ln10}{k}.$$

For $99\%$ conversion,

$$t_{99} = -\frac{1}{k}\ln(0.01) = \frac{\ln100}{k} = 2t_{90}.$$

Second-Order Reactions

For a second-order reaction involving one reactant,

$$-\frac{d[A]}{dt}=k[A]^2.$$

Separate the variables:

$$\frac{d[A]}{[A]^2}=-k\,dt.$$

or

$$[A]^{-2}d[A]=-k\,dt.$$

Integrating,

$$\int_{[A]_0}^{[A]} [A]^{-2}d[A] = -k\int_0^t dt.$$

Since

$$\int[A]^{-2}d[A] = -\frac{1}{[A]},$$

we obtain

$$\left[ -\frac{1}{[A]} \right]_{[A]_0}^{[A]} = -kt.$$

Therefore,

$$-\frac{1}{[A]} + \frac{1}{[A]_0} = -kt.$$

Multiplying by $-1$,

$$\boxed{ \frac{1}{[A]} = \frac{1}{[A]_0}+kt. }$$

A plot of $1/[A]$ against $t$ is linear with slope $k$.

For the half-life,

$$[A]=\frac{[A]_0}{2}.$$

Substitution gives

$$\frac{2}{[A]_0} = \frac{1}{[A]_0} + kt_{1/2}.$$

Therefore,

$$kt_{1/2} = \frac{1}{[A]_0},$$

and

$$\boxed{ t_{1/2} = \frac{1}{k[A]_0}. }$$

Unlike first-order kinetics, the half-life increases as the concentration decreases.

Third-Order Reactions

For a third-order reaction involving one reactant,

$$-\frac{d[A]}{dt}=k[A]^3.$$

Separate the variables:

$$[A]^{-3}d[A]=-k\,dt.$$

Integrating,

$$\int_{[A]_0}^{[A]} [A]^{-3}d[A] = -k\int_0^t dt.$$

Since

$$\int[A]^{-3}d[A] = -\frac{1}{2[A]^2},$$

we obtain

$$\left[ -\frac{1}{2[A]^2} \right]_{[A]_0}^{[A]} = -kt.$$

Thus,

$$-\frac{1}{2[A]^2} + \frac{1}{2[A]_0^2} = -kt.$$

Multiplying by $-2$ gives

$$\boxed{ \frac{1}{[A]^2} = \frac{1}{[A]_0^2}+2kt. }$$

For the half-life,

$$[A]=\frac{[A]_0}{2}.$$

Therefore,

$$\frac{4}{[A]_0^2} = \frac{1}{[A]_0^2} + 2kt_{1/2}.$$

Hence,

$$\frac{3}{[A]_0^2} = 2kt_{1/2},$$

and

$$\boxed{ t_{1/2} = \frac{3}{2k[A]_0^2}. }$$

Half-Life for a General Reaction Order

For $n\neq1$,

$$[A]^{1-n} = [A]_0^{1-n}+(n-1)kt.$$

At the half-life,

$$[A]=\frac{[A]_0}{2}.$$

Substitution gives

$$\left( \frac{[A]_0}{2} \right)^{1-n} = [A]_0^{1-n} + (n-1)kt_{1/2}.$$

Since

$$\left( \frac{[A]_0}{2} \right)^{1-n} = [A]_0^{1-n}2^{n-1},$$

we have

$$[A]_0^{1-n}2^{n-1} - [A]_0^{1-n} = (n-1)kt_{1/2}.$$

Factorizing,

$$[A]_0^{1-n} \left( 2^{n-1}-1 \right) = (n-1)kt_{1/2}.$$

Therefore,

$$\boxed{ t_{1/2} = \frac{2^{n-1}-1} {(n-1)k[A]_0^{n-1}}. }$$

For $n=1$, the separate first-order result applies:

$$\boxed{ t_{1/2}=\frac{\ln2}{k}. }$$

Fractional and Negative Reaction Orders

Reaction orders do not have to be positive integers.

For example,

$$v=k[A]^{1/2}$$

is half-order in $A$, while

$$v=k[A][B]^{-1}$$

is negative first-order in $B$.

Such rate laws usually arise from multistep mechanisms. Intermediate concentrations, adsorption equilibria, inhibition, and radical mechanisms can all produce fractional or negative concentration dependences.

A fractional reaction order does not imply that a fractional number of molecules participates in an elementary step. Reaction order is an experimentally observed concentration dependence, whereas molecularity describes an elementary event.

Determining Reaction Order from Initial Rates

Suppose

$$v=k[A]^m[B]^n.$$

For two experiments in which only $[A]$ changes,

$$v_1=k[A]_1^m[B]^n$$

and

$$v_2=k[A]_2^m[B]^n.$$

Dividing the second equation by the first eliminates $k$ and $[B]$:

$$\frac{v_2}{v_1} = \frac{k[A]_2^m[B]^n} {k[A]_1^m[B]^n}.$$

Therefore,

$$\frac{v_2}{v_1} = \left( \frac{[A]_2}{[A]_1} \right)^m.$$

Taking logarithms,

$$\ln\frac{v_2}{v_1} = m\ln\frac{[A]_2}{[A]_1}.$$

Hence,

$$\boxed{ m= \frac{\ln(v_2/v_1)} {\ln([A]_2/[A]_1)}. }$$

The order with respect to $B$ can be found in the same way:

$$\boxed{ n= \frac{\ln(v_2/v_1)} {\ln([B]_2/[B]_1)}. }$$

Pseudo-First-Order Kinetics

Consider

$$v=k[A][B].$$

If $B$ is present in large excess, its concentration changes very little during the reaction:

$$[B]\approx[B]_0.$$

The rate law becomes

$$v=k[A][B]_0.$$

Since $k$ and $[B]_0$ are both constant, they can be combined into a new observed rate constant:

$$k_{\mathrm{obs}}=k[B]_0.$$

Therefore,

$$\boxed{ v=k_{\mathrm{obs}}[A]. }$$

The reaction is second order according to its complete rate law, but under these conditions the observed kinetics are first order in $A$.

More generally, if

$$v=k[A]^m[B]^n$$

and $B$ is in large excess,

$$k_{\mathrm{obs}}=k[B]_0^n$$

and

$$v=k_{\mathrm{obs}}[A]^m.$$

Second-Order Reaction with Two Different Reactants

Consider

$$A+B\rightarrow P$$

with

$$v=k[A][B].$$

Let the initial concentrations be

$$[A]_0=a$$

and

$$[B]_0=b.$$

If an amount $x$ has reacted,

$$[A]=a-x$$

and

$$[B]=b-x.$$

Therefore,

$$\frac{dx}{dt} = k(a-x)(b-x).$$

Separating variables,

$$\frac{dx}{(a-x)(b-x)} = k\,dt.$$

Using partial fractions,

$$\frac{1}{(a-x)(b-x)} = \frac{1}{b-a} \left( \frac{1}{a-x} - \frac{1}{b-x} \right).$$

Thus,

$$\frac{1}{b-a} \int_0^x \left( \frac{1}{a-x} - \frac{1}{b-x} \right)dx = k\int_0^t dt.$$

After integration,

$$\frac{1}{b-a} \left[ -\ln(a-x)+\ln(b-x) +\ln a-\ln b \right] = kt.$$

Combining the logarithms gives

$$\boxed{ kt = \frac{1}{b-a} \ln \left[ \frac{a(b-x)} {b(a-x)} \right]. }$$

If $a=b$, then $[A]=[B]$ throughout the reaction and

$$-\frac{d[A]}{dt}=k[A]^2,$$

so the ordinary second-order integrated rate law applies:

$$\frac{1}{[A]} = \frac{1}{[A]_0}+kt.$$

Determining Reaction Order from Concentration-Time Data

The integrated rate laws give different linear relationships.

For zero-order kinetics,

$$[A]=[A]_0-kt,$$

so $[A]$ plotted against $t$ is linear.

For first-order kinetics,

$$\ln[A]=\ln[A]_0-kt,$$

so $\ln[A]$ plotted against $t$ is linear.

For second-order kinetics,

$$\frac{1}{[A]} = \frac{1}{[A]_0}+kt,$$

so $1/[A]$ plotted against $t$ is linear.

For third-order kinetics,

$$\frac{1}{[A]^2} = \frac{1}{[A]_0^2}+2kt,$$

so $1/[A]^2$ plotted against $t$ is linear.

For a general order $n\neq1$,

$$[A]^{1-n} = [A]_0^{1-n}+(n-1)kt.$$

Reaction order can also be obtained from rate measurements. Starting from

$$v=k[A]^n,$$

take the logarithm of both sides:

$$\ln v = \ln(k[A]^n).$$

Using logarithm rules,

$$\ln v = \ln k+n\ln[A].$$

This has the form of a straight line,

$$y=b+mx,$$

with

$$y=\ln v,$$ $$x=\ln[A],$$

and

$$\boxed{ m=n. }$$

Thus the slope of a plot of $\ln v$ against $\ln[A]$ gives the reaction order.

References

• IUPAC Compendium of Chemical Terminology (Gold Book), entry on reaction order.
• OpenStax, Chemistry 2e, sections on rate laws and integrated rate laws.
• Atkins, P.; de Paula, J.; Keeler, J. Atkins’ Physical Chemistry.