Kinetics Made Simple 2N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) Example 1:

Kinetics Made Simple

2N2O5 (aq) 4NO2 (aq) + O2 (g)

Example 1:

Experiment # [2N2O5] (mol/L) Time (s)

1 1.00 0

2 0.88 200

3 0.78 400

4 0.69 600

5 0.61 800

6 0.54 1000

7 0.48 1200

8 0.44 1400

9 0.38 1600

10 0.34 1800

2N2O5 (aq) 4NO2 (aq) + O2 (g)

Rates of disappearance and appearance

Rate = - 1[N2O5] = 1[NO2] = [O2]

2 t 4 t t

Rate = - [A] = k[A]n

t

Rate = - [N2O5] = k[N2O5]n

t

k is rate constant for this reaction at this temperature and pressure

n is the rate order

Exp # [2N2O5] (mol/L) Time (s) Rate (mol/Ls)

2 0.88 200 5.4x10-4

8 0.44 1400 2.7x10-4

5.4x10-4 (mol/Ls)

2.7x10-4 (mol/Ls)

[2N

2O5]

(m

ol/

L)

Time (s)

(200,0.88)

(1400,0.44)

rate2 = [N2O5]2 = 0.88 = 2 = 5.4x10-4 = 2 also

rate8 [N2O5]8 0.44 2.7x10-4

Note that the reaction rate at [N2O5] = 0.88M is twice that at [N2O5] = 0.44M

A doubling of concentration is a doubling of rate in a first order reaction.

2(rate) = 2[N2O5]1

Example 2:

NH4+(aq) + NO2

-(aq) N2 (g) + H2O (l)

Exp # [NH4]+(mol/L) [NO2]-(mol/L) Rate (M/s)

1 0.100M 0.005M 1.35x10-7

2 0.100M 0.010M 2.70x10-7

3 0.200M 0.010M 5.40x10-7

NH4 (aq) + NO2 (aq) N2 (g) + H2O (l)

General Rate Law: Rate = k[NH4+]n[NO2

-]m

Rate relationship of appearance & disappearance

Rate = -[NH4+] = -[NO2

-] = [N2] = [H2O]

t t t t

If we examine data from experiments 2 and 3 where the [NO2

-] does not change, then any change in rate must be caused by [NH4

+].

A doubling in concentration caused a doubling in rate.

So, first order in terms of [NH4+].

Exp # [NH4+] (mol/L) [NO2

-] (mol/L) Rate (M/s)

1 0.100M 0.005M 1.35x10-7

2 0.100M 0.010M 2.70x10-7

3 0.200M 0.010M 5.40x10-7

If you need to see the math, then:

Rate3 = 5.40x10-7 M/s = k(0.200M)n (0.01M)m

Rate2 = 2.70x10-7 M/s = k(0.100M)n (0.01M)m

So,

Rate3 = 5.40x10-7 M/s = k(0.200M)n (0.01M)m

Rate2 = 2.70x10-7 M/s = k(0.100M)n (0.01M)m

= (0.200M)n = 2.00 = (2.00)n so n=1 First order

(0.100M)n

If we examine data from experiments 1 and 2 where the [NH4

+]. does not change, then any change in rate must be caused by [NO2

-] .

A doubling in concentration caused a doubling in rate.

So, first order in terms of [NO2-] .



1 0.100M 0.005M 1.35x10-7

2 0.100M 0.010M 2.70x10-7

3 0.200M 0.010M 5.40x10-7

If you need to see the math, then:

Rate2 = 2.70x10-7 M/s = k(0.100M)n (0.01M)m

Rate1 = 1.35x10-7 M/s = k(0.100M)n (0.005M)m

So,

Rate2 = 2.70x10-7 M/s = k(0.100M)n (0.01M)m

Rate1 1.35x10-7 M/s k(0.100M)n (0.005M)m

= (0. 010M)n = 2.00 = (2.00)m so n=1 First order

(0.005M)n

The rate law is first order for both [NH4+] and

[NO2-] . Note that it is merely coincidence

that n and m have the same values as the stoichiometric coefficients

The overall reaction order is the sum of the n and m orders. n+m=2. The reaction is second order overall. The value of k can now be determined using any of the experimental data.

General Rate Law: Rate = k[NH4+]1[NO2

-]1



1 0.100M 0.005M 1.35x10-7

NH4+ (aq) + NO2

- (aq) N2 (g) + H2O (l)

General Rate Law: Rate = k[NH4+]1[NO2

-]1

Rate1 = 1.35x10-7 M/s = k(0.100M)n (0.005M)m

k = 1.35x10-7 M/s = 2.7x10-4 L/mol s

(0.100M)(0.005M)

Now, we can calculate the rate for ANY combination of these reactants at this temperature and pressure. THAT IS POWER!

Rate = 2.7x10-4 L/mol s[NH4+]1[NO2

-]1

General Rate Law:

Rate = k[NH4+]1[NO2

-]1

Example 3:

BrO3- (aq) + 5Br- (aq) + 6H+ 3Br2 (l) + 3H2O (l)

Exp # [BrO3-] [Br-] [H+] Rate (mol/Ls)

1 0.10 0.10 0.10 8.0x10-4

2 0.20 0.10 0.10 16.0x10-4

3 0.20 0.20 0.10 32.0x10-4

4 0.10 0.10 0.20 32.0x10-4

BrO3- (aq) + 5Br - (aq) + 6H+ (aq) 3Br2 (l) + 3H2O (l)

General form of the rate law for this reaction?

Determine the orders of each reactant in this reaction.

Write the rate law for this reaction.

Determine the value of the rate constant k for this reaction.

Check the value of k.


1 0.10 0.10 0.10 8.0x10-4

2 0.20 0.10 0.10 16.0x10-4

3 0.20 0.20 0.10 32.0x10-4

4 0.10 0.10 0.20 32.0x10-4



Rate = k [BrO3-]m [Br-]n [H+]p

Find the orders of each of the reactants by eliminating the causes of change in the reaction rate.


1 0.10 0.10 0.10 8.0x10-4

2 0.20 0.10 0.10 16.0x10-4


Rate2 = 1.60x10-3 M/s = k(0.20M)n (0.10M)m (0.10M)p

Rate1 8.0x10-4 M/s k(0.100M)n (0.005M)m(0.10M)p

= (0.20M)n = 2.00n = (2.00) so n=1 First order (0.10M)


2 0.20 0.10 0.10 16.0x10-4

3 0.20 0.20 0.10 32.0x10-4


Rate3 = 3.20x10-3 M/s = k(0.20M)n (0.20M)m (0.10M)p

Rate2 1.60x10-3 M/s k(0.20M)n (0.10M)m(0.10M)p

= (0.20M)m = 2.00m = (2.00) so m=1 First order (0.20M)


1 0.10 0.10 0.10 8.0x10-4

4 0.10 0.10 0.20 32.0x10-4


Rate4 = 3.20x10-3 M/s = k(0.10M)n (0.10M)m (0.20M)p

Rate1 8.00x10-4 M/s k(0.10M)n (0.10M)m(0.10M)p

= (0. 20M)p = 4.00 = (2.00)p so p=2 second order (0.10M)

The rate of this reaction is first order in [BrO3-]

and [Br-] and second order in [H+].

The over all reaction order is n + m + p = 4.


Rate = k [BrO3-]m [Br-]n [H+]p

Determine the orders & write the rate law for this reaction.

Rate = k [BrO3-]1 [Br-]1 [H+]2

Determine the value of the rate constant k for this reaction.

Exp #

[BrO3-] [Br-] [H+] Rate (mol/Ls)

1 0.10 0.10 0.10 8.0x10-4

8.0x10-4 (mol/Ls) = k (0.10M)1 (0.10M)1 (0.10M)2

k = 8.0x10-4 (mol/Ls) = 8.0 L3/mol3s

1.0x10-4 (mol4/L4)


2 0.20 0.10 0.10 16.0x10-4


Rate2 = 8.0 L3/mol3s (0.20M)1 (0.10M)1 (0.10M)2

Rate2 =16.0x10-4 (mol/Ls) = 1.60x10-3 (mol/Ls)

Check is correct !!

Now we can calculate rate for any combination of reactants at this temperature and pressure.

Check the value of k.

Kinetics Made Simple 2N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) Example 1:

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