RATE LAWS

 

• Consider the general reaction: aA + bB à products where A and B are the reactants and a and b are the coefficients

 

• The values of the exponents in the rate law equation establish the order of the reaction

 

• Look at this theoretical reaction: 2X + 2Y + 3Z à products
• Experimental data gives us rate = k [X]¹ [Y]² [Z]⁰
• The order of reaction with respect to X is 1
• The order of reaction with respect to Y is 2
• The order of reaction with respect to Z is 0
• The overall order of reaction is 3 (1 + 2 + 0)

 

• Because rate α [X]¹:

If initial [X] doubles; rate is multiplied by 2

If initial [X] increases by 5, rate is multiplied by 5

• Because rate α [Y]²:

If initial [Y] doubles; rate is multiplied by 4 (2²)

If initial [Y] increases by 5; rate is multiplied by 25 (5²)

• Because rate α [Z]⁰:

If initial [Z] doubles; rate is multiplied by 1 (2⁰) = rate remains unchanged

 

• A first order reaction has an overall order of 1
• 2N₂O₅(g) à 4NO₂(g) + 5O₂(g)

                        rate = k[N₂O₅]¹

 

• (CH₃)₃CBr(aq) + H₂O(l) à (CH₃)₃OH(aq) + H⁺¹(aq) + Br^-1(aq)

rate = k [(CH₃)₃CBr]¹[H₂O]⁰

 

 

 

 

 

 

 

 

 

 

 

 

• a second order reaction has an overall reaction order of 2
• 2HI(g) à H₂(g) + I₂(g)

rate = k[HI]²

 

• NO(g) + O₃(g) à NO₂(g) + O₂(g)

Rate = k [NO]¹ [O₃]¹

 

 

 

 

 

 

 

 

 

 

 

 

• The rate constant and exponents in a rate law can only be determined experimentally

 

To find the rate law:

 

1. Determine the exponents for each reactant

(if there is more than one reactant, you must determine one exponent at a time by making sure the other [reactant] doesn’t change)

 

eg1)           2N₂O₅(g) à 4NO₂(g) + 5O₂(g)

 

Experiment	Initial [N2O5] (mol/L)	Initial rate (mo/L*s)
1	0.01	4.8*10-6
2	0.02	9.6*10-6
3	0.03	1.5*10-5

 

General rate law: rate = k[N₂O₅]^m   

 

rate2 = k[0.02]^m = 9.6*10^-6               2^m = 2             therefore m = 1

rate1 = k[0.01]^m = 4.8*10^-6

 

rate = k[N₂O₅]¹

 

 

eg2)           2ClO₂(aq) + 2OH^-1(aq) à ClO₃^-1(aq) + ClO₂^-1(aq) + H₂O(l)

 

Experiment	Initial [ClO2] (mol/L)	Initial [OH-1] (mol/L)	Initial Rate (mol/L*s)
1	0.015	0.025	1.3*10-3
2	0.015	0.05	2.6*10-3
3	0.045	0.025	1.16*10-2

 

General rate law: rate = k[ClO₂]^m [OH^-1]ⁿ

 

Rate2 = k[0.015]^m [0.05]ⁿ =   2.6*10^-3         2ⁿ = 2              n = 1

Rate1 = k[0.015]^m [0.025]ⁿ = 1.3*10^-3

 

Rate3 = k[0.045]^m [0.025]ⁿ = 1.16*10^-2      3^m = 9             m = 2

Rate1 = k[0.015]^m [0.025]ⁿ = 1.3*10^-3

 

rate = k[ClO₂]² [OH^-1]¹

 

 

2. Find the value of the rate constant by substituting data from any of the experiments into the rate law equation.

 

Eg1)    rate = k[N₂O₅]¹, sub in exp. #1                     Eg2)    rate = k[ClO₂]² [OH^-1]¹, sub in exp #1

            4.8*10^-6 = k[0.01]¹                                                     1.3*10^-3 = k[0.015]² [0.025]¹

            k = 4.8*10^-4 s^-1                                                           k = 231 L²/mol²*s

 

            rate = 4.8*10^-4 s^-1 [N₂O₅]¹                                       rate = 231 L²/mol²*s [ClO₂]² [OH^-1]¹