Loading in 5 sec....

Kinetics Part II: Rate Laws & Order of ReactionPowerPoint Presentation

Kinetics Part II: Rate Laws & Order of Reaction

- 153 Views
- Uploaded on
- Presentation posted in: General

Kinetics Part II: Rate Laws & Order of Reaction

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Kinetics Part II:Rate Laws & Order of Reaction

Jespersen Chap. 14 Sec 3

Dr. C. Yau

Spring 2014

1

Rate of reaction is based on one component (reactant or product) of the reaction: disappearance of a reactant or formation of a product.

Rate law is a rate expression that includes all reactants.

LEARN THESE TERMS SO YOU KNOW WHAT IS BEING ASKED FOR!!

2

Rate= k [reactant]order

- k is a reaction rate constant, a measure of time efficiency(not to be confused with “Rate”).
- High values of k mean high efficiency.(Reaction goes fast.)
- k must be determined experimentally.
- Each experiment has its own rate law.
- Rate law must be determined experimentally.

Rate = k [A]m[B]n

where m and n are the "orders of reaction" and are found by experiment, NOT based on the coefficients of the chemical equation,

and k is the "rate constant."

This expression is called the "rate law."

Rate = 5.0x105 L5mol-5 s-1 [H2SeO3][I-]3[H+]2

5.0x105 mol-5 s-1 is the rate constant (k).

We speak of the reaction as being…

first orderwith respect to H2SeO3,

third orderwith respect to I-(Nothing to do with 6 in eqn)

second orderwith respect to H+, and

the overall order of reaction is 6 (sum of all the orders).

Learn this terminology!

What is the unit of Rate in the equation shown above?

Do Practice Exercises 7, 8 & 9 on p.648.

What is a rate law used for?

Rate changes with concentrations. The rate law allows us to determine the rate for various concentrations of the reactants.

Example:

The rate law for the reaction 2A +B→3C is

Rate = 0.045M-1s-1 [A][B]

If the concentration of A is 0.2M and that of B is 0.3M, what will be the reaction rate?

rate = 0.045 M-1 s-1 [0.2M][0.3M]

rate = 0.0027 M/s

Do Practice Exercises 5 & 6 p.646

Chlorine dioxide, ClO2, is a reddish-yellow gas that is soluble in water. In basic solution it gives ClO3- and ClO2- ions.

ClO2(aq) + OH(aq) ClO3(aq) + ClO2(aq) + H2O (l)

The rate law is Rate=k[ClO2]2[OH-]. What is the value of the rate constant given that when [ClO2]=0.060M, [OH-] = 0.030M, the reaction rate is 0.0248 M/s

- 0.02 M-1 /s
- 0.02 M/s
- 0.02 s-
- None of these

2.3x102 M-2 s-1

- are indicated for each reactant,
- the overallreaction order is the sum of individual reactant orders,
- may be zero, negative, fractional or integers, but in this course we will usually encounter positive integers, and
- must be determined from experimental data.

- Run reaction under the same conditions, varying only the concentrations of reactants (not the temperature).
- A ratio of rate laws for each experiment allows us to determine the order of each reactant.
- The rate law is unique to temperature and concentration conditions. Therefore, when a rate law is stated, it must include the temperature at which it is determined.

2NO(g) + O2(g)→ 2NO2(g)

Select 2 rate laws that vary in concentration for only one of the substances (NO).

Hint: Write the fractions with the larger R on top.

2NO(g) + O2(g)→ 2NO2(g)

Next choose 2 rate laws where the concentration for the other component

(O2) changes.

x=2, y = 1

so…..

Rate = k [NO]2[O2]

Finally we can solve for k. Use any rate law and the orders that we have determined.

rate = k[NO]2[O2]

0.048M/s =k [0.015M]2[0.015M]

1.4×104 M-2s-1 =k

Do Example 14.5, 14.6 p.651, Exercises 10 thru 14p.651+.

x=1

y=1

z=0

rate=k[A][B]

Note that changing the concentration of C had no effect on the rate. We say it is “zero order with respect to C.”

Consider Rate = k[A]n

If n = 0, change in conc has no effect on rate.

If n = 1, Rate = k[A]1 and when conc is 2x,

rate is 2x.

If n = 2, Rate = k[A]2 and when conc is 2x,

rate is 4x

If n = 2, when conc is tripled, rate is …?

rate is 9x

If n = 3, and conc is doubled, rate is…?

rate is 8x

- Once you understand how you can predict effect of a change in concentration on rates (as in the previous slide), you can often determine the rxn order visually without doing complicated calculations.
- HOWEVER, that is only if the conc were neatly doubled or tripled, etc.
(See next 2 slides.)

p. 648

What is the rate law?

Rate = k[A]?[B]?

Rate = k[A]1[B]2

For the following data, determine the order of NO2 in the reaction at 25°C

2 NO2(g) + F2(g)→ 2 NO2F(g):

- 0
- 1
- 2
- 3
- not enough information given

We cannot always determine the rxn order visually.

For example, if we ended with

32.1=3.18x

How do we determine what x is?

In high-level chemistry courses, x might even be a fraction!