11.1 Reaction Rate 2

11.1 11.1 REACTION RATEREACTION RATE

C H E M I S T RY U N I TC H E M I S T RY U N I T

S K 0 2 7S K 0 2 7

OBJECTIVES:OBJECTIVES:

C H A P T E R 1 1 : R E A C T ION K INE T IC SC H A P T E R 1 1 : R E A C T ION K INE T IC S

11.1 REACTION RATE11.1 REACTION RATE

1. define rate laws, order of reaction and

rate constant

2. write the rate law / rate equation

S K 0 2 7 C H E M I S T RYS K 0 2 7 C H E M I S T RY

2. write the rate law / rate equation

3. calculate the order with respect to a

certain reactant and the overall order of

a reaction from experimental data.

4. calculate the value and determine the

units of rate constants, k.



THE RATE LAWTHE RATE LAW

�The rate law for a chemical reaction relates the rate of reaction to the concentrations of reactants.

�At constant T & P, rate of chemical reaction directly proportional to [reactants].

Rate = k [A]m[B]n

directly proportional to [reactants].

Rate ∝ [reactants]

Consider this equation,

aA + bB → cC

Rate ∝ [A]m [B]n

k = rate constant

n = reaction order wrt B

m = reaction order with

respect to A

m + n= overall rxn order




� The exponents in a rate law / reaction order

must be determined by experiment.

They are not derived from the stoichiometric

coefficients in an overall chemical equation.

They are generally small, positive integers They are generally small, positive integers

(0,1,2…).

� The rate constant, k and its value depends on

specific rxn, T & the presence of catalyst.

The larger k, the faster the reaction proceeds.

Units of k depend on the values of m, n, …




Example 1:

Determining Reaction Order from Rate Laws

2NO(g) + O2(g) 2NO2(g)

rate = k [NO]2 [O ]

The overall reaction order = 3

The reaction order with respect to NO = 2

The reaction order with respect to O2 = 1

rate = k [NO]2 [O2]

Solution:




S2O82- + 3I- 2SO4

2- + I3-

The above reaction is first order with respect to Iodide ion and

to thiosulphate ion.

a) Write the rate equation for the reaction

Example 2 :

a) Write the rate equation for the reaction

b) What is the unit of rate constant, k?

Solution :

a) Rate = k [S2O82-]1[I-]1

b) Rate = k [S2O82-]1[I-]1

k = rate

[S2O82-]1[I-]1

=Ms-1

M2= M-1s-1




Determine rate law: Method Of Initial Rates

• The method of initial rates involves a series of

expt in which the initial [ ] of some reactants are

held constant and others are varied in order to

determine the rate law for that reaction.determine the rate law for that reaction.

• The effects of doubling one initial concentration:

– For zero-order reactions, no effect on rate.

– For first-order reactions, the rate doubles.

– For second-order reactions, the rate quadruples.

– For third-order reactions, the rate increases

eightfold.S K 0 2 7 C H E M I S T RYS K 0 2 7 C H E M I S T RY



Method Of Initial Rates

Initial Rates of the Reaction:

2NO(g) + Cl2(g) → 2NOCl(g)

Exp. Initial [NO] Initial [Cl2] Initial rate, Ms-1

1 0.0125 0.0255 2.27 × 10-5

2 0.0125 0.0510 4.55 × 10-5

3 0.0250 0.0255 9.08 × 10-5




Method Of Initial Rates (continue..)

Compare 2 experiments, of [one reactant] varies and the

[ other reactant] constant.

2NO(g) + Cl2(g) → 2NOCl(g)

The rate law in general is; Rate = k [NO]m [Cl2]n

[ other reactant] constant.

=exp1 rate

exp2 rate

nm

nm

(0.0255)k(0.0125)

(0.0510)k(0.0125)

n

exp12

m

exp1

n

exp22

m

exp2

][Clk[NO]

][Cl k[NO]

=×

×

5-

-5

102.27

104.55

2 = 2n Thus, n, order of rxn wrt Cl2 = 1S K 0 2 7 C H E M I S T RYS K 0 2 7 C H E M I S T RY



Method Of Initial Rates (continue..)

=exp1 rate

exp3 raten

exp12

m

exp1

n

exp32

m

exp3

][Cl k[NO]

][Cl k[NO]


nm

nm

(0.0255)k(0.0125)

(0.0255)k(0.0250)=

×

×

−

−

5

5

1027.2

1008.9

4 = 2m

Thus, m, order of rxn wrt NO = 2

Overall order of reaction = m + n = 2 + 1 = 3

Thus, rate law is Rate = k [NO]2 [Cl2]



The order of reaction

For Reaction,

A Products

Rate = k [A]xRate = k [A]x

Rate is not dependent on [A]

Rate = k [A]0

Rate = k

Thus, this reaction is zero order with respect to A

i) If x = 0





ii) Rate = k [A]1

A Products

Rate = k [A]x

If x = 1ii) Rate = k [A]

If the [A] is doubled, A 2A

Double the [A] will double the rate of reaction.

Thus, this reaction is first order with respect to A

Rate = k[2A]1

Rate = 2k[A]1

If x = 1





A Products

iii) Rate = k[A]2If x = 2

Rate = k [A]x

iii) Rate = k[A]2

If the [A] is doubled, A 2A

Rate = 4k[A]2

Doubling the [A], the rate increase by a factor of 4.

Thus, the reaction is second order with respect to A

Rate = k[2A]2

If x = 2




The results of the kinetic studies are given below.

Exercise:

exp [ClO2] / M [OH-] / M Initial rate / Ms-1

1 0.0421 0.0185 8.21 ×10-3

2 0.0522 0.0185 1.26 ×10-2

ClO2(aq) + 2OH- (aq) → products

2 0.0522 0.0185 1.26 ×10-2

3 0.0421 0.0285 1.26 ×10-2

a) Explain what is meant by the order of reaction

b) Referring to the data determine

(i) rate law /rate equation

(ii) rate constant, k

(iii) the reaction rate if the concentration of both ClO2and OH- = 0.05 M

rate = k [ClO2]2[OH-]

k = 250 M-2s-1

rate = 3.12 ×10-2 M/sS K 0 2 7 C H E M I S T RYS K 0 2 7 C H E M I S T RY



Exercise:

i) When [A] is double, rate also double. But double the

[B] has no effect on rate.

A + B → C

Write rate law for this equation,

Rate = k[A][B] has no effect on rate.

ii) When [A] is increase 3x, rate increases 3x, and

increase of [B] 3x causes the rate to increase 9x.

iii) Reduce [A] by half has no effect on the rate, but

reduce [B] by half causing the rate to be half of the

initial rate.

Rate = k[A]

Rate = k[A] [B]2

Rate = k[B]




Exercise:The results of the kinetic studies are given below.

exp Initial [ ] (M) Time

interval

(Min)

The change in

concentration

of C (M)[C] [D]

1 0.10 1.0 30 2.5 ×10-3

C + D → E

1 0.10 1.0 30 2.5 ×10-3

2 0.10 2.0 30 1.0 ×10-2

3 0.05 1.0 120 5.0 ×10-3

a) Calculate the rate of reaction for each experiment

a) Determine the order of reaction with respect to C

and D and write the rate law. rate = k [C] [D]2

8.33 ×10-5 ,3.33 × 10-4, 4.17×10-5 M min-1




Exercise:The results of the kinetic studies are given below.

exp Initial [ ] (M) Time

interval

(Min)

The change in

concentration

of C (M)[C] [D]

1 0.10 1.0 30 2.5 ×10-3

C + D → E

c) State the effect on the reaction rate if the

concentration of D is doubled but the concentration

of C remains constant. rate increase by a factor of 4;

1 0.10 1.0 30 2.5 ×10-3

2 0.10 2.0 30 1.0 ×10-2

3 0.05 1.0 120 5.0 ×10-3


11.1 Reaction Rate 2

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