# Integrated Rate Laws

Synopsis

It would be very helpful to have the equation that governs the curves like the one above on the decomposition of sucrose. If we had such an equation we could then calculate the concentrations of the reactant(s) at any given time and would then know a lot more about the travel of reactants to products. It is possible to obtain such equations, but the methods used to obtain the equations involve the use of differential and integral calculus. For the reaction, R --> P, which is first order, we can write the rate law as

r = -[R]/t = k[R] and then we can transform this into a differential equation: r = -d[R]/dt = k[R]. Now we can separate variables and obtain the equation

-d[R]/[R] = kdt. Now we can integrate this equation to obtain the integrated rate law for a first order reaction:

ln([R]t/[R]o) = -kt ………………………………..(3)

with [R]t being the concentration of the reactant at any time t, and [R]o being the concentration of the reactant at time t = 0, the beginning of the reaction. The "ln" stands for the natural logarithm, which is explained in Appendix A. Many readers will not know calculus, and that is ok, but for those who do know calculus you will appreciate how easy this derivation is. This is a very important equation. If we experimentally measure the concentration of reactant at some time, we can easily calculate the rate constant. If we remember (Appendix A) that the ln(a/b) = lna - lnb, then we can easily rearrange the above equation into the equation for a straight line:

ln[R]t = ln[R]o -kt ……………………………….(4)

If we take the antilog of each side we obtain

[R]t = [R]o exp(-kt) ………………………………(5)

which shows there is an exponential decrease in the concentration of reactant with the passage of time. Many radioactive decays are first-order reactions.

If the reaction R --> P is a second order reaction then the rate law is r = k[R]2 and the integrated rate law is

1/[R]t = 1/[R]o + kt ………………………………(6)

Sometimes the rate of a reaction does not depend upon the concentration of the reactant at all; the reaction is a zeroth-order reaction. If the reaction R à P is a zeroth-order reaction then the rate law is r = k[R]o =k and the integrated rate law is

[R]t = [R]o - kt …………………………………..(7)

We have written the three integrated rate laws all in the equation of a straight line (eq 4, 6, 7) which suggests a method for determining the rate constants. We can plot [R]t vs. t and the slope will be the value of the rate constant.

The half-life of a reaction, t1/2, is the time required for the concentration of the reactant to decrease to ½ of the initial value (at t = t1/2, [R]t = ½[Ro]). We can make this substitution into each of the integrated rate laws and obtain the following equations for the half-life of a reaction (R à P) for each of the orders we have examined (zeroth, first, and second):

 Reaction Order Half-life equation 0 [R]o/(2k) 1 0.693/k 2 1/(k[R]o)

The number 0.693 is the value of ln2. Note that it is only for the first-order reaction that the half-life does not depend upon the initial concentration at all. We can use these equations to obtain the order of the reaction and the rate constant. We can measure the half-life experimentally, double the initial concentration and measure the half-life again. If the half-life is the same for both experiments then the reaction is first-order, and we can use the corresponding half-life equation to calculate k. If the half-life also doubled, then we know that the reaction is zeroth-order, and we can use that half-life equation to calculate k. If the half-life was ½ of the initial value, then we know that the reaction is second order, and we can use that half-life equation to calculate k.

Review Questions

1. Derive the equations for the half-lifes in the above table.
2. Outline the methods that we now have for experimentally determining the orders and rate constants for chemical reactions.
3. Work Exercise 15.9 on page 622.

Web Author: Dr. Leon L. Combs