Determining the Form of the Rate Law

How Data is created

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

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

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

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

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

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

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

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

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