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. One method for making this determination is to experimentally measure how the concentration of a reactant or product varies with time and then make characteristic kinetics plots. Another strategy for determining the rate law is to use the method of initial rates.

The Method of Initial Rates involves measuring the rate of reaction, r, at very short times before any significant changes in concentration occur. Suppose 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 following form:

r = k [A]a [B]b

The initial concentrations of A and B are known; 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.

Consider the following set of data:

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
= 0.100a

0.150a

0.4446 = 0.6667a

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

ln[0.4446] = ln[0.6667a]

-0.8106 = -0.4054a

The value of a may now be readily determined.

a = -0.8106

-0.4054
= 1.9995

In most cases, the exponents are integers (or less commonly fractions such as 1/2). In this case the reaction is second order in A (a = 2). A similar strategy can be employed to determine the value of b. (Actually, it should be obvious from inspection of trials 1 and 3 that the reaction is zero-order in B.) 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

Consider the following chemical reaction:

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

The experimental conditions are:

The graph shows the variation in the concentration of triiodide ion with time. The slope of this plot can be used to determine the rate of reaction (do not forget the stoichiometric coefficient for the triiodide ion when calculating the rate constant). To obtain the initial rate, the reaction must be studied as a relatively short time so that no curvature is present in the [I3-] vs t plot. The line-of-best-fit for the [I3-] vs t data can be calculated and displayed on the graph.

It is extremely difficult in practice to determine the initial rate, because as soon as the reaction begins, the rate slows (unless the reaction is zero order). Unfortunately, the shorter the time period (and there are experimental limitations here), the lower [I3-] and thus the more problematic the error in [I3-]. Try varying the time over which measurements are made and compare the rate to see how precise the rates are.

Select various volume ratios in order to obtain various initial concentrations of H2SeO3 and I- in the reaction solution. For each trial, measure the initial rate of reaction and use this information to determine the rate law. Hint: In this reaction, the exponents for H2SeO3 and I- are both integers.

Unlike the example above and examples in your textbook, it will not be possible to keep one concentration constant while varying the other concentration, owing to the design of the stopped flow apparatus. It will be necessary to construct two equations of the form

ln r1

r2
  =   a ln [H2SeO3]1

[H2SeO3]2
  +   b ln [I-]1

[I-]2

and solve the equations simultaneously to determine a and b. The exponents in the rate law are integers; be sure to round the results to the nearest integer.

Once the rate law has been determined, use the data for each trial to calculate a rate constant and report the average value of the rate constant. Be sure to work out the correct units for the rate constant and do not forget that the reaction is second-order in hydrogen ion.


   



Volumes Delivered

H2SeO3 Soln.: mL
I- Soln.: mL



final time = msec
time interval = msec





slope (mole L-1 sec-1) =

intercept (mole L-1) =


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