Lecture Notes: Friday, October 5, 2001

For the reaction

aA(g) + bB(g) ----> cC(g) + dD(g)

The rate law for the reaction is;

 

rate = k[A]m[B]n

The rate law expression tells us how the rate of the reaction changes with concentration. However, from a more practical stand point we are more interested in the relationship between time and concentration.

If were studying a reaction we would be interested in knowing how much reactant remained after 5 minutes, 1 day or several years. Our rate expression

rate = k[A]m[B]n

tells us how the rate of the reaction changes with concentration. It is possible to derive algebraic expression relating the concentration to time. The form of these equations maybe very simple or quite complex depending on the order of the reaction and the number of reactants. We will discuss the simpler concentration/time equations. Those for first and second order reactions involving a single reactant.

A simple reaction is defined as a reaction of a single reactant and has the following form;

A(g) ---> products

we can write the differential form of the rate law (which is useful in describing how the rate of disappearance of reactant is proportional to the concentration of the reactant) as;

which can be rearranged to

When this relationship is integrated from t = 0â to t = t we obtain the relationship

This integrated form of the rate law for a first order reaction is handy because it provides us with a way to determine the relationship between concentration of the reacting species and time. Be sure to note that the [A]t listed in the equation represents the concentration of A remaining at time 't'.

Sample Problem 1:

The decomposition of H2O2 to H2O and O2 follows first order kinetics with a rate constant of 0.0410 min-1.

H2O2(l) ---> H2O(l) + O2(g)

Calculate the [H2O2] after 10 mins if [H2O2]o is 0.200 M.

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Answer

 

If we rearrange the relationship

we find an interesting new form

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

The equation has the form of a straight line. Plotting ln [A] vs. time gives a straight line with a slope equal to -k. Remember that the rate constant, k, must be positive, therefore, the line must have a negative slope.

Sample Problem 2:

Time

[A]

Time

[A]

0

0.310

3000

0.116

600

0.254

3600

0.0964

1200

0.208

4200

0.0812

1800

0.172

4800

0.0669

2400

0.141

6000

0.0464

If we plot this data [A] versus time and ln [A] versus time we see interesting results.

Plot of [A] versus time

Plot of ln [A] versus time

The rate constant can be determined from the slope of the line.

slope = -0.000326 sec-1 = -k = 3.26 x 10-4 sec-1

 

Another value that is frequently used to indicate how rapid a reaction is, is the half-life of the reaction, called t1/2. The half-life occurs when the initial concentration, [A]o, has fallen to half it value. The time required for this to occur is called the half-life of the reaction.

We can obtain a half-life relationship for first order reactions by recognizing that

[A]t1/2 = .5[A]o

For a simple reaction following first order kinetics;

Substituting;

ln 0.5 = -kt1/2

-0.693 = -kt1/2

0.693 = kt1/2

So we can calculate the half-life for a reaction that follows first order kinetics if we know the rate constant for the reaction.

Sample Problem 3:

How long would it take for half of the H2O2 (from Sample Problem 1) to decompose?

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Answer

For a simple reaction

A(g) ---> products

following second order kinetics. We can write the differential form of the rate law (which is useful in describing how the rate of disappearance of reactant is proportional to the concentration of the reactant) as;

which can be rearranged to

When this relationship is integrated from t = 0 to t = t we obtain the relationship

This integrated form of the rate law for a first order reaction is handy because it provides us with a way to determine the relationship between concentration of the reacting species and time. Be sure to note that the [A]t listed in the equation represents the concentration of A remaining at time 't'.

This integrated form of the rate law for a second order reaction is handy because it provides us with a way to determine the relationship between concentration of the reacting species and time.

Here are a few sample problems using the second order integrated rate law.

Sample Problem 4:

The decomposition of NOCl(g) follows the reaction

2NOCl(g) ---> 2NO(g) + Cl2(g)

is a second order reaction with a rate constant of 0.0480 M-1sec-1 for the reaction at 200 degrees C. In an experiment at 200 degrees C, the initial concentration of NOCl was 0.400 M. What is the concentration of NOCl after 15.0 min have elasped?

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Answer

 

Sample Problem 5:

How many minutes will it take for the concentration of NOCl(g) to drop to 0.150 M?

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Answer

The integrated form of the rate law for a second order reaction is handy because it provides us with a way to determine the relationship between concentration of the reacting species and time. We rearrange the relationship;

to the form of a straight-line equation;

The equation has the form of a straight line. Plotting [A]-1 vs. time gives a straight line with a slope equal to k.

Let's look at a sample problem where we must plot the data to see how well the reaction follows second order kinetics.

The following data were obtained for the gasÐphase decomposition of nitrogen dioxide at 300 degrees C.

Time

[NO2] (M)

0

0.0100

25

0.0088

50

0.0079

75

0.0071

100

0.0065

150

0.0055

175

0.0051

200

0.0048

250

0.0042

300

0.0038

 

If we plot this data [A] versus time and ln [A] versus time and [A]-1 vs. time we see interesting results.

Plot of conc versus time

Plot of [A]-1 versus time

Plot of ln [A] versus time

From this data the rate constant is 0.543 M-1sec-1

The rate constant can be determined from the slope of the line of the [A]-1 vs. time, which is 0.543 M-1sec-1

 

If we rearrange this relationship

Arrhenius Plot