1. Determining the Form of the Rate Law

2. Rate Law
The mathematical expression that allows calculations of reaction rate as a function of reactant concentration.
Must be determined experimentally by altering the initial concentrations of all reactants and observing the effect on initial rate.

3. How Data is created

4. k, m and n must be experimentally determined
Rate Law Equation
Since the concentrations within a reaction change the instant the reaction starts, chemists look at the effect of the initial concentrations.
For any general reaction:
aA + bB → products
The rate law equation may be expressed as the product of the initial concentrations raised to some exponential value.
Rate = k[A]m[B]n
k, m and n must be experimentally determined

5. Method of Initial Rates
Used to find the form of the rate law
Choose one reactant to start with
Find two experiments where the concentration of that reactant changes but all other reactants stay the same
Write the rate laws for both experiments
Divide the two rate laws
Solve for the order
Follow the same technique for other reactants

6. Example Choose one reactant to start with NH4+
Find two experiments where the concentration of that reactant changes but all other reactants stay the same
Exp 2 & 3

7. Write the rate laws for both experiments
Exp 2: 2.70x10-7 = k(0.100)x(0.010)y
Exp 3: 5.40x10-7 = k(0.200)x(0.010)y
Divide the two rate laws
0.50 = 0.50x
Use log rules to solve for the order
x = 1 so the order for NH4+ is one

8. Follow the same technique for other reactants
NO2-: Exp 1 & 2
Exp 1: 1.35x10-7 = k(0.100)1(0.0050)y
Exp 2: 2.70x10-7 = k(0.100)1(0.010)y
0.5 = 0.5y
y = 1
So Rate = k[NH4+]1[NO2-]1
Overall Reaction Order – sum of orders of reactants

9. Finding k We can find k using values from any of the experiments given
Units will be different for k depending on order of reactants

10. Example BrO3- : Exp 1 & 2 Exp 1: 8.0x10-4 = k(0.10)x(0.10)y(0.10)z

11. Example Br- : Exp 2 & 3 Exp 2: 1.6x10-3 = k(0.20)1(0.10)y(0.10)z

12. Example H+ : Exp 1 & 4 Exp 1: 8.0x10-4 = k(0.10)1(0.10)1(0.10)z
0.25 = 0.50z or log0.25=zlog0.50
z = 2

13. Example So Rate = k[BrO3-]1[Br-]1[H+]2 Solve for rate constant, k
Overall order of reaction = 4
Solve for rate constant, k

14. Homework
Page 380 #1-5
Monday we will determine the rate law equation for the iodine clock reaction, please read investigation6.5.1 on page 392