The Integrated Form of a First-Order Kinetics Equation


Let us use the following chemical equation: A ---> products.

The decrease in the concentration of A over time can be written as: - d[A] / dt = k [A]

Rearrangement yields the following: d[A] / [A] = - k dt

Integrate the equation, which yields: ln [A] = - kt + C

Evaluate the value of C (the constant of integration) by using boundry conditions. Specifically, when t = 0, [A] = [A]o. [A]o is the original starting concentration of A.

Substituting into the equation, we obtain: ln [A]o = - k (0) + C.

Therefore, C = ln [A]o

We now can write the integrated form for first-order kinetics, as follows:

ln [A] = - kt + ln [A]o

This last equation can be rearranged into several formats, such as:

ln ([A] / [A]o) = - kt

[A] / [A]o = exp (- kt)

[A] = [A]o exp (-kt)

Remember the exp notation means the natural constant e raised to the power of whatever follows. Usually what follows is enclosed in parenthesis, but not always.

How do we determine the value of k for a given first-order reaction?

The integrated first-order equation is the equation of a straight line. (Remember the general equation of a straight line is y = mx + b.) In this case the y-value is ln [A], m equals negative k, the x-value is t, and the y-intercept is ln [A]o. A plot of ln [A] versus t will yield a line with slope equal to negative k.