A first order rate law would take the form

\[ \dfrac{d[A]}{dt} = k[A]\]

Again, separating the variables by placing all of the concentration terms on the left and all of the time terms on the right yields

\[ \dfrac{d[A]}{[A]} =-k\,dt\]

This expression is also easily integrated as before

\[ \int_{[A]=0}^{[A]} \dfrac{d[A]}{[A]} =-k \int_{t=0}^{t=t}\,dt\]

Noting that

\[ \dfrac{dx}{x} = d (\ln x)\]

The form of the integrated rate law becomes

\[ \ln [A] - \ln [A]_o = kt\]

or

\[ \ln [A] = \ln [A]_o - kt \label{In1}\]

This form implies that a plot of the natural logarithm of the concentration is a linear function of the time. And so a plot of ln[A] as a function of time should produce a linear plot, the slope of which is -k, and the intercept of which is ln[A]_{0}.

Example \(\PageIndex{1}\):

Consider the following kinetic data. Use a graph to demonstrate that the data are consistent with first order kinetics. Also, if the data are first order, determine the value of the rate constant for the reaction.

**Time (s)** |
0 |
10 |
20 |
50 |
100 |
150 |
200 |
250 |
300 |

**[A] (M)** |
0.873 |
0.752 |
0.648 |
0.414 |
0.196 |
0.093 |
0.044 |
0.021 |
0.010 |

**Solution**:

The plot looks as follows:

From this plot, it can be seen that the rate constant is 0.0149 s^{-1}. The concentration at time \(t = 0\) can also be inferred from the intercept.

It should also be noted that the integrated rate law (Equation \ref{In1}) can be expressed in exponential form:

\[ [A] = [A]_o e^{-kt}\]

Because of this functional form, 1^{st} order kinetics are sometimes referred to as exponential decay kinetics. Many processes, including radioactive decay of nuclides follow this type of rate law.

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