How to solve

 This method requires that a reaction be run several times.  The initial concentrations of the reactants are varied.  The reaction rate is measured just after the reactants are mixed.  Eliminates the effect of the reverse reaction.

A + 2B 3C + D [A] [B] Initial Rate ?

Rate = k[A] x [B] y Take any two experiments where the [ ] of one species is held constant = k[.240] x [.480] y 2.00 = k[.240] x [.120] y 4.00 = 4 y therefore, y = 1

Now repeat the process, holding the other [ ] constant = k[.240] x [.120] y.500 = k[.120] x [.120] y 4.00 = 2 x therefore, x = 2

Rate = k[A] 2 [B] Determine the value of k with units: 8.00 = k[.240] 2 [.480] 290 L 2 /mol 2 sec = k Determine the initial rate for the last experiment rate = 290[.0140] 2 [1.35] rate =.0767 mol/l sec

 For the reaction BrO Br - + 6H + 3Br H 2 O  The general form of the Rate Law is Rate = k[BrO 3 - ] n [Br - ] m [H + ] p  We use experimental data to determine the values of n, m, and p

Initial concentrations (M) Rate (M/s) BrO 3 - Br - H+H+H+H x x x x x x x x Now we have to see how the rate changes with concentration

x = 1 y = 1 z = 2 Rate Law = k[BrO 3 - ][Br - ][H + ] 2