Determining the Form of the Rate Law

1. Determining the Form of the Rate Law

2. How Data is created

3. 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

4. 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

5. Example • Write the rate laws for both experiments Exp 2: Rate = 2.70x10-7 = k(0.100)x(0.010)y Exp 3: Rate = 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

6. Example • Follow the same technique for other reactants NO2-: Exp 1 & 2 Exp 1: Rate = 1.35x10-7 = k(0.100)1(0.0050)y Exp 2: Rate = 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

7. 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

8. Example • BrO3- : Exp 1 & 2 Exp 1: Rate = 8.0x10-4 = k(0.10)x(0.10)y(0.10)z Exp 2: Rate = 1.6x10-3 = k(0.20)x(0.10)y(0.10)z 0.50 = 0.50x x = 1

9. Example • Br- : Exp 2 & 3 Exp 2: Rate = 1.6x10-3 = k(0.20)1(0.10)y(0.10)z Exp 3: Rate = 3.2x10-3 = k(0.20)1(0.20)y(0.10)z 0.50 = 0.50y y = 1

10. Example • H+ : Exp 1 & 4 Exp 1: Rate = 8.0x10-4 = k(0.10)1(0.10)1(0.10)z Exp 4: Rate = 3.2x10-3 = k(0.10)1(0.10)1(0.20)z 0.25 = 0.50z OR ¼ = (½)z z = 2

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