1. Rate Law 5-2 an expression which relates the rate to the concentrations and a specific rate constant

2. For a general reaction between reactant A and B at a constant temperature the reaction can be represented by: aA + bB -------> products Rate law is expressed as: Rate = k[A] m [B] n

3. The rate law equation expresses the relationship between the concentration of the reactants and the rate of the equation.

4. Rate constant (k) Exponents m and n 1Varies with temperature Constant with temperature 2Constant under constant conditions Can only be determined experimentally

5. Exponents and Orders of reactions. The values of exponents (m or n) can only be determined through experimentation and may or may not be the values of the coefficients in the balanced chemical equation.

6. The value of the exponents determines the order of the reaction. If a reaction has a single reactant and the value of the exponent is one then it is first order. If the exponent were two then it is second order. If more than one reactant is present in a reaction, the sum of the exponents (m + n) is called the overall reaction order.

7. 2 N 2 O 5 (g) → 4 NO 2 (g)+ O 2 (g) Experimentally the rate was found to be first order for N 2 O 5 (g). Rate = k [N 2 O 5 (g)] 1

8. The value of exponent is not the same as the balance for N 2 O 5 in the equation. The overall reaction order would be first order.

9. NO 2 (g) + CO (g) → NO (g) + CO 2 (g) Experimentally the rate was found to be second order for NO 2 (g) and zero order for CO (g). The rate law can be written: Rate = k [NO 2 ] 2 [CO] 0 or Rate = k [NO 2 ] 2

10. The value of the exponents is not the same as the coefficients. The overall reaction order is second order ( 2 + 0 ).

11. 2 HI (g) → H 2 (g) + I 2 (g) Experimentally the rate was found to be second order for HI (g). Rate = k [HI] 2

12. Assignment #1: Rate Law Equations ReactionRate depende nt on Rate lawReacti on order 1A + B -----> products [A] 2 [B] 1 22NO(g) +O 2 (g) ----> NO 2 (g) [NO] 1 [O 2 ] 1 =k[A] 2 [B] 1 3 =k[NO] 1 [O 2 ] 1 2

13. 32NO(g) + 2H 2 (g) ----> N 2 (g) + 2 H 2 O(g) [NO] 2 [H 2 ] 1 42NO 2 (g) +Cl 2 (g) ------> 2NO 2 Cl(g) [NO 2 ] 1 [Cl 2 ] 1 52ClO 2 (aq)+ 2OH - ------> ClO 3 - (aq) + ClO 2 - (aq) + H 2 O (l) [ClO 2 ] 0 [OH] 1 6C + D ------> products [C] 2 [D] 0 =k[NO] 2 [H 2 ] 1 3 =k[NO] 1 [Cl 2 ] 1 2 =k[ClO] 0 [OH] 1 1 =k[C] 2 [D] 0 2

14. Determination of Rate Exponents using Initial rates of reactions. Chemists can examine the change in the initial rates of the reaction when concentrations of reactants are changed to determine the order of reactants (for a specific equation).

15. Order of Reaction

16. Order of reactants Change in intial rate when concentration is doubled First order Doubles the rate of reaction second order Quadruples the rate of the reaction zero order No change in the rate of the reaction

17. In the following questions determine the order of each reactant and the overall rate the value of the rate constant All the experiments were conducted under conditions of constant temperature.

18. a.. Single reactant A -----> B + C the following data was collected ExpInitial [A] (mol/L) Initial rate (mol/L·s) 10.014.8 x 10 -6 20.029.6 x 10 -6 30.031.4 x 10 -5

19. Finding the rate law and the constant Choose the data from the table to solve for the exponent and the constant. Rate law = k [ A ] m Choose two experimental data and then divide one by the other to find the exponent Calulate K by putting the experimental data into the rate law including the value of the expnent and solve for K.

20. Rate law = k [ A ] m Exp1: 4.8 x 10 -6 = k [0.01] m Exp 2: 9.6 x 10 -6 =k [0.02] m Exp1: 1 = [1] m Exp 2: 2 = [2] m – m = 1

21. Finding k Chose one set of data and solve Exp 1 Exp1: 4.8 x 10 -6 = k [0.01] 1 4.8 x 10 -6 = k [0.01] 1 0.01 0.01 4.8 x 10 -4 = k

22. Rate law = 4.8 x 10 -4 [A] 1

23. b. Two reactants expInitial [A]Initial [B]Initial rate 10.010.032.4 x 10 -4 20.03 7.2 x 10 -4 30.010.062.4 x 10 -4

24. Choose the data such that the concentration of one of the reactants, A, is the same in both the experimental data. In this case exp 1 and exp 2

25. Rate law = k [A] m [B] n Exp1: 2.4 x 10 -4 = k [0.01] m [0.03] n Exp 3: 2.4 x 10 -4 =k [0.01] m [0.06] n 1 = ( 1/2 ) n n = 0

26. Solve for m now Exp1: 2.4 x 10 -4 = k [0.01] m [0.03] n Exp 2: 7.2 x 10 -4 =k [0.03] m [0.03] n 1/3 = ( 1/3 ) m m = 1

27. Rate law = k [A] 1 [B] 1 Solve for k Exp1: 2.4 x 10 -4 = k [0.01] 1 [0.03] 0 0.01 0.01 2.4 x 10 -2 = k Rate law = 2.4 x 10 -2 [A] 1 [B] 0

28. 2 ICl + H 2 → I 2 + 2 HCl exp[ ICl ][ H 2 ]Initial rate 10.100.010.002 20.200.010.004 30.100.040.008

29. Rate law = k [ICl] m [H 2 ] n Exp1: 0.002 = k [0.10] m [0.01] n Exp 2: 0.004 = k [0.20] m [0.01] n 1/2 = (1/2) m m = 1

30. Exp 1: 0.002 = k [0.10] m [0.01] n Exp 3: 0.008 = k [0.10] m [0.04] n 1/4 = (1/4) n n = 1

31. Solve for k Exp1: 0.002 = k [0.10] 1 [0.01] 1 0.001 0.001 2 = k

32. Rate law = 2 [ICl] 1 [H 2 ] 1

33. 1. For the general reaction: A + B -----> C the following data was collected Exp[A] (mol/L) [B] (mol/L) Initial rate (mol/L·s) 10.0020.052 x 10 -5 20.0040.058 x 10 -5 30.0020.102 x 10 -5

34. Rate law = 5 [A] 2 [B] 0 Answer

35. 2. For the reaction: H 2 O 2 (aq)+2HI (aq)→ 2H 2 O (l)+I 2 (aq)the following data was collected exp[H 2 O 2 ] (mol/L [HI] (mol/L) Initial rate (mol/L·s) 10.05 0.002 20.050.100.004 30.100.050.004

36. answer Rate law = 0.8 [H 2 O 2 ] 1 [H I ] 1

37. 3. For the reaction: (CH 3 ) 3 Br + OH - →(CH 3 ) 3 COH +Br - the following data was collected exp[(CH 3)3 CBr] (mol/L) [OH - ] (mol/L) Initial rate (mol/L·s) 10.030.041.2 x 10 -3 20.030.081.2 x 10 -3 30.060.042.4 x 10 -3

38. answer Rate law = 0.04 [(CH 3 ) 3 CBr] 1 [OH] 0