# Chemical Kinetics: Method of Initial Rates

## Concepts

One of the first steps in studying the kinetics of a chemical reaction is to determine the rate law for the reaction. The differential rate law describes how the rate of reaction depends upon the concentrations of the reactants. In most cases, one knows the initial concentrations of each reactant, because one knows how the reaction mixture was prepared. The initial concentrations are the concentrations at time t = 0. The initial rate, r0, is the rate of reaction at time t = 0. In the Method of Initial Rates, the differential rate law is determined by varying the initial concentrations of the various reactants and observing the effect on the initial rate.

One challenge in this method is to accurately determine the initial rate. It is necessary to make measurements at very short times, before the concentrations change significantly from their initial value. It is also possible to extrapolate the rate back to t = 0.

In order to illustrate the Method of Initial Rates, suppone one is studying a reaction with the following stoichiometry.

A + 2 B   →   3 C

While the form of the differential rate law might be very complicated, many reactions have a rate law of the form

r   =   k   [A] a   [B] b

The initial concentrations of A and B are known, because one knows how the reaction mixture was prepared. Therefore, if the initial reaction rate is measured, the only unknowns in the rate law are the rate constant, k, and the exponents a and b. One typically measures the initial rate for several different sets of concentrations and then compares the initial rates.

The table below contains experimental data for the initial rate of reaction for various initial concentrations of A and B.

Trial Rate
(mole L-1 sec-1)
Initial Concentration
of A (mole L-1)
Initial Concentration
of B (mole L-1)
1 2.73 0.100 0.100
2 6.14 0.150 0.100
3 2.71 0.100 0.200

If simple multiples are chosen for the concentrations and only one concentration is varied at a time, one can determine a and b by inspection. In this case the values of a and b may not be obvious. One can employ the following algebraic technique for determining the exponents.

First, write the ratio of the rate laws for two trials.

r1   r2
=
k   [A]1a   [B]1b   k   [A]2a   [B]2b

Next, substitute the numerical values into the equation.

2.73 mole L-1 sec-1   6.14 mole L-1 sec-1
=
k   (0.100 mole L-1) a   (0.100 mole L-1) b   k   (0.150 mole L-1) a   (0.100 mole L-1) b

Notice that the units for each quantity and the rate constant can be removed, and in this case the exponent b is removed when the concentrations of B divide. The equation simplifies to

2.73   6.14
=
0.100 a   0.150 a

0.4446   =   0.6667 a

To convert a from an exponent into a coefficient, take the logarithm of both sides of the equation.

ln( 0.4446 )   =   ln( 0.6667 a )   =   a ln( 0.6667 )

-0.8106   =   -0.4054 a

a   =
-0.8106   -0.4054
=   1.9995

In most cases, the kinetic exponents are integers (or less commonly fractions such as ½ ). In this case the reaction is second order in A,     a   =   2.     Note that the value 1.9995 has been rounded to exactly 2.

A similar strategy can be employed to determine the value of b. It should be obvious from inspection of Trials 1 and 3 that the reaction is zero-order in B,     b   =   0.     Once the exponents are known, the rate constant can be calculated. Because the data generally suffers from experimental error, it is best to calculate the rate constant for each trial and use the average value.

## Experiment

### Objectives

• Determine the rate law for a chemical reaction.
• Determine the rate constant for a chemical reaction.

Consider the following chemical reaction between selenious acid and iodide ion in acidic solution.

H2SeO3 (aq)   +   6 I- (aq)   +   4 H+ (aq)   →   Se (s)   +   2 I3- (aq)   +   3 H2O (l)

The experimental details are:

• The solution in the left syringe contains 0.100 mole/L H2SeO3.
• The solution in the right sryinge contains 0.100 mole/L KI.
• Both solutions are buffered at 0.020 mole/L H+. The reaction is known to be second order in H+.
• In each experiment, 10.0 mL of the 0.100 mole/L H2SeO3 solution is used. You may select the volume of the 0.100 mole/L KI solution that will be mixed with the 10.0 mL of 0.100 mole/L H2SeO3
• The only colored species is that tri-iodide ion: I3- . Consequently, the experimental data shows how [I3 -] varies with time. (The colloidal selenium formed during the reaction scatters some of the light, but this effect is disregarded in this simulation.)
• Data is only collected for the first percent or two of the reaction, because we are only interested in the initial rate of reaction.

To the right of the stopped flow apparatus is a graph showing the variation of [I3 -] with time. The slope of this plot can be used to determine the rate of reaction. Select the Plot Line button to show a blue line tangent to one of the experimental data points. Use the and buttons to move the tangent point to see how the rate of reaction (slope) varies with time.
To obtain the initial rate, move the tangent point to   t = 0

Select various volume ratios in order to obtain various initial concentrations of H2SeO3 and I- in the reaction solution. For each trial, determine the initial rate of reaction and use this information to determine the rate law.

The design of the stopped-flow apparatus does not permit the concentration of H2SeO3 or I- to be held constant. Therefore one will need to determine the rate law using a variation of the algebraic method illustrated above. For simplicity, let A = H2SeO3 and B = I-. Suppose two experiments are performed (Trials 1 and 2) using different concentrations of A and B. One can begin by writing the ratio of differential rate law for the two trials.

r1   r2
=
k   [A]1 a   [B]1 b   k   [A]2 a   [B]2 b

The rate constant divides out. Then take the logarithm of each side of the equation to obtain

ln
r1   r2
=   a   ln
[A]1   [A]2
+   b   ln
[B]1   [B]2

There are two unknowns in this equation: a and b. Performing an additional experiment allows a second equation to be written.

ln
r1   r3
=   a   ln
[A]1   [A]3
+   b   ln
[B]1   [B]3

Now there are two equations and two unknowns, which allows one to solve for a and b

Tips for performing the data analysis.

• In this reaction, the exponents for H2SeO3 and I- are both integers.
• The graph plots data for the concentration of the product I3-. The stoichiometric coefficient for I3- must be taken into account in calculating a value for the rate constant k
• Recall that H+ is also a reactant in the reaction. [H + ] = 0.0200 M and the reaction is second order in H +. Use this information in calculating the rate constant.
• Pay attention to the units for the rate constant !

mL KI Soln
Delivered

slope

mole L-1 sec-1

MethodOfInitialRates.html version 3.0
© 2000, 2014, 2023 David N. Blauch