A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.

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The Differential Representation

Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements. The differential equation describing first-order kinetics is given below:

\< Rate = - \dfrac{d}{dt} = k^1 = k \label{1}\>

The "rate" is the reaction rate (in units of molar/time) and \(k\) is the reaction rate coefficient (in units of 1/time). However, the units of \(k\) vary for non-first-order reactions. These differential equations are separable, which simplifies the solutions as demonstrated below.

The Integral Representation

First, write the differential form of the rate law.

\< Rate = - \dfrac{d}{dt} = k \nonumber\>

Rearrange to give:

\< \dfrac{d}{} = - k\,dt \nonumber\>

Second, integrate both sides of the equation.

\< \begin{align*} \int_{_o}^{} \dfrac{d}{} &= -\int_{t_o}^{t} k\, dt \label{4a} \\<4pt>\int_{_{o}}^{} \dfrac{1}{} d &= -\int_{t_o}^{t} k\, dt \label{4b} \end{align*}\>

Recall from calculus that:

\< \int \dfrac{1}{x} = \ln(x) \nonumber\>

Upon integration,

\< \ln - \ln_o = -kt\nonumber\>

Rearrange to solve for to obtain one form of the rate law:

\< \ln = \ln_o - kt \nonumber\>

This can be rearranged to:

\< \ln = -kt + \ln _o\nonumber\>

This can further be arranged into y=mx +b form:

\< \ln = -kt + \ln _o\nonumber\>

The equation is a straight line with slope m:


and y-intercept b:


Now, recall from the laws of logarithms that

\< \ln {\left(\dfrac{_t}{ _o}\right)}= -kt \nonumber\>

where is the concentration at time \(t\) and \(_o\) is the concentration at time 0, and \(k\) is the first-order rate constant.

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Plotting \(\ln\) with respect to time for a first-order reaction gives a straight line with the slope of the line equal to \(-k\).


Half-lives of first order reactions

The half-life (\(t_{1/2}\)) is a timescale on which the initial population is decreased by half of its original value, represented by the following equation.

\< = \dfrac{1}{2} _o \>

After a period of one half-life, \(t = t_{1/2}\) and we can write

\<\dfrac{_{1/2}}{_o} = \dfrac{1}{2}=e^{-k\,t_{1/2}} \label{18}\>

Taking logarithms of both sides (remember that \(\ln e^x = x\)) yields

\< \ln 0.5 = -kt\label{19}\>

Solving for the half-life, we obtain the simple relation

\< t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{k}\label{20}\>

This indicates that the half-life of a first-order reaction is a constant.

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Notice that, for first-order reactions, the half-life is independent of the initial concentration of reactant, which is a unique aspect to first-order reactions. The practical implication of this is that it takes as much time for to decrease from 1 M to 0.5 M as it takes for to decrease from 0.1 M to 0.05 M. In addition, the rate constant and the half life of a first-order process are inversely related.