2.3: First-Order Reactions - Chemistry LibreTexts

The Integral Representation

First, write the differential form of the rate law.

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

Rearrange to give:

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

Second, integrate both sides of the equation.

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

Recall from calculus that:

\[ \int \dfrac{1}{x} = \ln(x) \nonumber \]

Upon integration,

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

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

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

This can be rearranged to:

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

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

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

The equation is a straight line with slope m:

\[mx=-kt \nonumber \]

and y-intercept b:

\[b=\ln [A]_o \nonumber \]

Now, recall from the laws of logarithms that

\[ \ln {\left(\dfrac{[A]_t}{ [A]_o}\right)}= -kt \nonumber \]

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

Because the logarithms of numbers do not have any units, the product \(-kt\) also lacks units. This concludes that unit of \(k\) in a first order of reaction must be time-1. Examples of time-1 include s-1 or min-1. Thus, the equation of a straight line is applicable:

\[ \ln [A] = -kt + \ln [A]_o.\label{15} \]

To test if it the reaction is a first-order reaction, plot the natural logarithm of a reactant concentration versus time and see whether the graph is linear. If the graph is linear and has a negative slope, the reaction must be a first-order reaction.

To create another form of the rate law, raise each side of the previous equation to the exponent, \(e\):

\[ \large e^{\ln[A]} = e^{\ln[A]_o - kt} \label{16} \]

Simplifying gives the second form of the rate law:

\[ [A] = [A]_{o}e^{- kt}\label{17} \]

The integrated forms of the rate law can be used to find the population of reactant at any time after the start of the reaction. Plotting \(\ln[A]\) with respect to time for a first-order reaction gives a straight line with the slope of the line equal to \(-k\). More information can be found in the article on rate laws.

This general relationship, in which a quantity changes at a rate that depends on its instantaneous value, is said to follow an exponential law. Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms. The reason that the exponential function \(y=e^x\) so efficiently describes such changes is that dy/dx = ex; that is, ex is its own derivative, making the rate of change of \(y\) identical to its value at any point.