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Rate Laws

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Example: Determine the rate law for the following reaction given the data below.

H2O2(aq) + 3 I-(aq) + 2H+(aq) ïƒ I3-(aq) + H2O (l)

[H2O2] [I-] [H+]Initial

Expt # (M) (M) (M)Rate (M /s)

10.0100.0100.000501.15 x 10-6

20.0200.0100.000502.30 x 10-6

30.0100.0200.00050 2.30 x 10-6

40.0100.0100.001001.15 x 10-6

For the reaction:

5Br- (aq) + BrO3- (aq) + 6 H+ (aq) ïƒ 3 Br2 (aq) + 3 H20 (l)

the rate law was determined experimentally to be:

Rate = k[Br-] [BrO3-] [H+]2

- The reaction is first order with respect to Br-, first order with respect to BrO3-, and second order with respect to H+.

The previous reaction is fourth order overall.

The overall reaction order is the sum of all the exponents in the rate law.

Note: In most rate laws, the reaction orders are 0, 1, or 2.

Reaction orders can be fractional or negative.

The value of the rate constant can be determined from the initial rate data that is used to determine the rate law.

Select one set of conditions.

Substitute the initial rate and the concentrations into the rate law.

Solve the rate law for the rate constant, k.

The value of the rate constant, k, depends only on temperature.

It does not depend on the concentration of reactants.

Consequently, all sets of data should give the same rate constant (within experimental error).

Example: The following data was used to determine the rate law for the reaction:

A + B ïƒ C

Calculate the value of the rate constant if the rate law is:

Rate = k [A]2[B]

Expt #[A] (M)[B] (M)Initial rate (M /s)

10.1000.1004.0 x 10-5

20.1000.2008.0 x 10-5

30.2000.100 16.0 x 10-5

Select a set of conditions:

Substitute data into the rate law:

Solve for k

Important:

The units of the rate constant will depend on the overall order of the reaction.

You must be able to report your calculated rate constant using the correct units.

A first order reaction is one whose rate depends on the concentration of a single reactant raised to the first power:

Rate = k[A]

Calculus (integration) gives us the integrated form of the rate law which allows us to predict the concentration of a reactant after a given amount of time has elapsed:

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

To predict the concentration of a reactant after a given time has elapsed during a first order reaction:

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

where ln = natural logarithm (not log)

t =time (units depend on k)

[A]t = conc.or amount of A at time t

[A]0 = initial concentration or amâ€™t of A

k = rate constant

Example: A certain pesticide decomposes in water via a first order reaction with a rate constant of 1.45 yr-1. What will the concentration of the pesticide be after 0.50 years for a solution whose initial concentration was 5.0 x 10-4 g/mL?

The time required for the concentration of a reactant to drop to one half of its original value is called the half-life of the reaction.

t1/2

After one half life has elapsed, the concentration of the reactant will be:

[A]t = Â½ [A]0

For a first order reaction:

t1/2 = 0.693

k

Â½

Given the half life of a first order reaction and the initial concentration of the reactant, you can calculate the concentration of the reactant at any time in the reaction OR the amount of time needed for the concentration to decrease to a given value.

Use t1/2 = 0.693/k to find the rate constant

Substitute k and other given values into the integrated rate law.

- The integrated rate law can also be re-written in terms of the half life of the reaction:
[A]t = [A]0 x (0.5)

- This equation is easier to use when solving for [A]t or [A]0.
- The original integrated rate law is easier to use if you need to find telapsed.

telapsed

t1/2

Example: A 25.0 mass % solution of a certain drug in water has a half life of 14 days. Calculate the concentration of the drug present in the solution after 30. days.

Example: A certain pesticide has a half life of 0.500 yr. How many years will it take for the concentration of a 2.30 x 10-3 M solution of the pesticide to decrease to 1.15 x 10-5M.

Notice that the integrated rate law for first order reactions follows the general formula for a straight line:

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

y = mx + b

Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k.

CH3NC

CH3CN

Consider the process in which methyl isonitrile is converted to acetonitrile.

CH3NC

CH3CN

This data was collected for the reaction at 198.9Â°C.

Remember that the partial pressure of the gas will be directly proportional to the number of moles of that gas

- When ln P is plotted as a function of time, a straight line is obtained.
- Therefore, the process is first-order.

A reaction that is second order with respect to A has the following rate law:

Rate = k[A]2

The integrated form of the second order rate law follows the general formula for a straight line:

1

[A]t

1

[A]0

= kt +

For a reaction that is second order with respect to A, a plot of 1/[A] versus time will give a straight line.

slope of line = k

1

[A]

- A zero order reaction is one whose rate is independent of the reactant concentration.
Rate = k

- The integrated form of this rate law follows the general form for a straight line:
[A]t = -kt + [A]o

- Plotting [A] vs. time gives a straight line for a zero order reaction.

- On your exam, you must be able to use experimental data to graphically determine if a reaction is zero, first, or second order with respect to a given reactant.
- Zero Order:
- A plot of [A] vs. time is a straight line.

- First Order:
- A plot of ln[A] vs. time is a straight line.

- Second Order:
- A plot of 1/[A] vs. time is a straight line.

Example: The concentration of A in the reaction: A ïƒ 2 B was monitored as a function of time. Graphically determine if the reaction is zero, first, or second order with respect to A.

The data obtained:

A plot of [A] vs. time is a straight line so the reaction is zero order with respect to A.

Notice that the plots of ln[A] vs. time and 1/[A] vs. time are both curves instead of straight lines! This means that the reaction is not first or second order with respect to [A].

Time (s)

[NO2], M

0.0

0.01000

50.0

0.00787

100.0

0.00649

200.0

0.00481

300.0

0.00380

Example: The decomposition of NO2 at 300Â°C is described by the equation:

NO2 (g) ïƒ

NO (g) + 1/2 O2 (g)

and yields the data below. Determine if the reaction is zero, first, or second order with respect to NO2?

- A plot of [NO2] vs. time is a curve instead of a straight line.
- The reaction is not zero order with respect to [NO2]

Time (s)

[NO2], M

ln [NO2]

0.0

0.01000

âˆ’4.610

50.0

0.00787

âˆ’4.845

100.0

0.00649

âˆ’5.038

200.0

0.00481

âˆ’5.337

300.0

0.00380

âˆ’5.573

- A graph of ln [NO2] vs.t is a curve instead of a straight line.
- Reaction is not first order in [NO2].

Time (s)

[NO2], M

1/[NO2]

0.0

0.01000

100

50.0

0.00787

127

100.0

0.00649

154

200.0

0.00481

208

300.0

0.00380

263

- Graphing 1/[NO2] vs. t gives a straight line.

- Therefore, the process is second-order in [NO2].