Kinetics Part II:Rate Laws & Order of Reaction

Dr. C. Yau

Spring 201411

Jespersen Chap. 14 Sec 3

Rate of Rxn vs. Rate Law

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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!!

The Rate Law Depends On The Concentrations Used

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.

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A + B products

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."

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H2SeO3 + 6I- +4H+ Se + 2I3- + 3H2O

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 order with respect to H2SeO3,

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

second order with 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?

5Do Practice Exercises 7, 8 & 9 on p.648.

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?

6rate = 0.0027 M/srate = 0.045 M-1 s-1 [0.2M][0.3M]

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.

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

A. 0.02 M-1 /s

B. 0.02 M/s

C. 0.02 s-

D. None of these

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2.3x102 M-2 s-1

Orders…

• are indicated for each reactant,

• the overall reaction 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.

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Determining The Rate Law:

• 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.

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Use Rate Laws To Determine Orders :

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

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exp [NO] [O2] RNO2 M/s

1 .015 .015 .048

2 .030 .015 .192

3 .015 .030 .096

2NO(g) + O2(g) → 2NO2(g)

2. bemust x

2.0 .004

[0.015MNO]

]MNO030.0[ 4.00

[0.015MNO]

]MNO030.0[ 4.00

]MO015.0[]k[0.015MNO

]MO015.0[]MNO030.0[k

s/M048.0

s/M192.0

x

x

x

x

y2

x

y2

x

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

Use Rate Laws To Determine Orders :

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

(O2) changes.

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exp [NO] [O2] RNO2 M/s

1 .015 .015 .048

2 .030 .015 .192

3 .015 .030 .096

x=2, y = 1

so…..

Rate = k [NO]2[O2]

2NO(g) + O2(g) → 2NO2(g)

1. bemust y

2.0 .002

][0.015MO

]MO030.0[ 2.00

]MO015.0[

]MO030.0[ 2.00

]MO015.0[]k[0.015MNO

]MO030.0[]MNO015.0[k

s/M048.0

s/M096.0

y

y

2

2

y2

y2

y2

x

y2

x

Determining The Value Of k

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

exp [NO] [O2] RNO2

M/s

1 .015 .015 .048

2 .030 .015 .192

3 .015 .030 .096

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rate = k[NO]2[O2]

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

1.4×104 M-2s-1 =kDo Example 14.5, 14.6 p.651, Exercises 10 thru 14p.651+.

Determine The Rate Law From Given Data

[A] [B] [C]Rate

M/s

0.01 0.02 0.15 0.0002

0.02 0.02 0.15 0.0004

0.01 0.01 0.15 0.0001

0.01 0.02 0.3 0.0002

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z=0x=1 y=1

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.”

]15.0[]02[.]M01[.k

]15.0[]02[.]M02[.k

s/M0002.0

s/M0004.0zyx

zyx

]15.0[]01[.]M01[.k

]15.0[]02[.]M01[.k

s/M0001.0

s/M0002.0zyx

zyx

]3.0[]02[.]M01[.k

]15.0[]02[.]M01[.k

s/M0002.0

s/M0002.0zyx

zyx

Effect of Order of Rxn on Rate

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 14

Visual Determination of Reaction Order

• 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.)

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What is the rate law?

Rate = k[A]?[B]?

Rate = k[A]1[B]2 16

p. 648

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

2 NO2(g) + F2(g)→ 2 NO2F(g):

Exp [NO2] [F2] Rate NO2 disappearance (M/s)

1 0.001 0.005 2 x10-4

2 0.002 0.005 4 x10-4

3 0.006 0.002 4.8 x10-4

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A. 0 B. 1 C. 2 D. 3 E. not enough information given

When Visual Determination Fails...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!

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