1. How to solve

2.  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.

3. A + 2B 3C + D [A] [B] Initial Rate.240.4808.00.240.1202.00.360.2409.00.120.1200.500.240.06001.00.01401.35?

4. Rate = k[A] x [B] y Take any two experiments where the [ ] of one species is held constant. 8.00 = k[.240] x [.480] y 2.00 = k[.240] x [.120] y 4.00 = 4 y therefore, y = 1

5. Now repeat the process, holding the other [ ] constant. 2.00 = k[.240] x [.120] y.500 = k[.120] x [.120] y 4.00 = 2 x therefore, x = 2

6. 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

7.  For the reaction BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 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

8. Initial concentrations (M) Rate (M/s) BrO 3 - Br - H+H+H+H+ 0.100.100.108.0 x 10 -4 0.100.100.108.0 x 10 -4 0.200.100.101.6 x 10 -3 0.200.100.101.6 x 10 -3 0.200.200.103.2 x 10 -3 0.200.200.103.2 x 10 -3 0.100.100.203.2 x 10 -3 0.100.100.203.2 x 10 -3 Now we have to see how the rate changes with concentration

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