Integrated Rate Law Method

Submitted by jwmoore on Thu, 02/17/2011 - 18:52

Integrated Rate LawAn equation which describes the rate of a reaction as a function of the rate constant and the concentrations of reactants (and any other substances that affect the rate, such as products or catalysts); also called rate equation.

For a reaction where the chemical equationA representation of a chemical reaction in which chemical symbols represent reactants on the left side and products on the right side. has the form

A   ->     products

and where the rate law is assumed to be of the form

Rate = - d[A]/dt = k [A]m

where m is the order of the reaction with respect to substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. A, three important cases can be treated: m = 0, m = 1, and m =2. These are called zeroth order, first order, and second order, respectively. The method for finding the rate law involves assuming that the reaction is first order, graphing the concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated.-time data in a way that should give a straight line, and rejecting the hypothesis that the reaction is first order if the graph is not linear. Then the process is repeated for second order and zeroth order.

First-Order ReactionA reaction for which the rate is directly proportional to the concentration of one reactant., m = 1

In this case the rate law is

- d[A]/dt = k [A].

This equation can be rearranged to

d[A]/[A] = - k t

and then integrated using the methods of the calculus (if you know how to integrate, you can verify the result below). The integration gives

ln[A]t = - k t + ln[A]0

y       =  m x   +    b

where ln represents taking a natural logarithm, t represents a particular time after the reaction began, and [A]t and [A]0 represent the concentration of A at time t and time zero. (It is assumed that the reaction is timed from whenever the reactants were mixed, so time zero, t = 0, is the start of the reaction.) This equation is in the form y = m x + b that represents a straight line. Thus, if the reaction is first order, a graph of the natural logarithm of the concentration of A (on the y axis) versus the time from the start of the reaction (on the x axis) should be a straight line. (If this graph is not straight, the reaction is not first order; perhaps it is second order or zeroth order.) If the graph is linear, then the slope of the graph is - k, the negative of the rate constantIn a differential rate equation, the proportionality constant that relates the rate with the concentrations of reactants and other species that affect the rate. The rate constant is the rate of reaction when all concentrations are 1 M., and the intercept is [A]0, the initial concentration of substance A.

Second-Order ReactionA reaction for which the rate is directly proportional to the square of the concentration of one species or to the product of the concentrations of two species., m = 2

For the same reaction, A  ->  products, but where the rate depends on the square of the concentration of A, the rate law becomes

Rate = - d[A]/dt =  k [A]2

As was done with the first-order rate equationAn equation that describes the rate of a reaction as a function of the rate constant and the concentrations of reactants (and any other substances that affect the rate, such as products or catalysts); also called rate law., this one can be rearranged and integrated. The result is

1/[A]t = k t + 1/[A]0

This equation is also in the form of a straight line, but this time the slope is k, the rate constant, and the intercept is the reciprocal of the initial concentration of substance A. If a plot of the reciprocal of the concentration of A on the vertical axis versus time on the horizontal axis gives a straight line, then it is reasonable to conclude that the reaction is second order. The rate constant can be calculated from the slope.

Zeroth-Order Reaction, m = 0

For a few reactions,  A   ->  products, the rate does not depend on the concentration of the reactant, A. In such a case the rate is always equal to the rate constant, as shown in the next equation

Rate = - d[A]/dt = k [A]0 =  k

This equation says that the slope of a graph of [A]t on the y-axis and t on the x-axis will be constant (equal to the rate constant, k). A constant slope means that a graph of [A]t versus t will be a straight line. You could also integrate this equation to give

[A]t  =  - k t  +  [A]0

This makes it quite clear that the slope of such a graph is the negative of the rate constant.

To review, the process for finding the order of the reaction (and hence the rate law) and the rate constant involves making up to three graphs. As soon as one of the graphs is found to be a straight line, the order of the reaction is known and the rate constant can be calculated from the slope of the straight line. The units of the rate constant are obtained from the units of the quantities plotted on the graph. For example, for the first-order plot, the y-axis is ln[A]t and the x-axis is t. A logarithm is a number and has no units; t has units of time, usually seconds. The slope is calculated as Δyx so there is a pure number in the numerator and time units in the denominator. This gives the rate constant units of s-1, reciprocal seconds.

Calculating Concentrations from Rate Laws

Once a rate law has been determined for a reaction, it is possible to use the integrated rate equationIn chemical kinetics, an equation that describes the concentrations of reactants (and products) as a function of time. to calculate the concentration of a reactant or product given the initial concentration of reactant and the time elapsed. For example, in an earlier section the rate law and rate constant were determined for decomposition of a dye at 80 °C. The rate law found was first order

Rate = k [dye]                        k = 0.073 s-1 (at 80 °C)

so the integrated rate law has this form

ln[dye]t  =  - k t  +  ln[A]0            or             ln([dye]t /[dye]0)  =  - k t

Suppose that you started with 0.100 M dye and allowed it to decompose as described earlier. What concentration of dye would be left after 25 s? You can calculate the concentration of dye after 25 s by substituting into the integrated rate equation:

[dye]t /(0.100 M) = exp(- 0.073 s-1 × 25 s)  and [dye]t  = 0.016 M

You can also use the integrated rate law to determine how long it would take for the concentration of the dye to drop to half its initial value, as

ln([dye]t /[dye]0)  =  - k t            and            ln{(0.050 M)/(0.100 M) = - 0.073 s-1 × t

This gives t = 9.5 s, which is in agreement with the figure showing that the concentration had dropped from 1.0 M to 0.49 M (a little less than half the starting concentration)in 10 s.

The time for the concentration to drop to half its initial value is called the half-lifeIn chemical kinetics, the time it takes for one half of the limiting reactant to be consumed. In nuclear chemistry, the time for half of a sample to undergo radioactive decay. and is represented by t1/2 . For a first-order reaction the half-life is the same regardless of the initial concentration, but this is not true for zeroth-order or second-order reactions. The half-life of a first-order reaction is related to the first-order rate constant and either can be determined if the other is known. The relationship is

ln[A]t1/2  -  ln[A]0  =  - k t1/2

ln{[A]0/2} -  ln[A]0  =  ln[A]0  -  ln2  -  ln[A]0  =  - ln2  =  - k t1/2

t1/2  = ( - ln2)/(- k)  =  0.693/k