Question

Considering the following reaction:
BrO3- (aq) + 5Br- (aq) + 6H+(aq) → 3Br2(aq) + 3H2O (1)

Using the data in the following table, find the rate law equation, k and the order of the reaction with

respect to each reactant as well as the overall order. (Use whole numbers or common fractions for

order; e.g. 1.11 would be considered first order and 0.333 would be 1 )

Exp. Number	(BrO3-)(Mol/L)	(Br-)(mol/L)	(H+)(mol/L)	Initial Rate (10-3 mol/Ls)
1	0.10	0.10	0.10	1.2
2	0.20	0.10	0.10	2.4
3	0.10	0.30	0.10	3.5
4	0.20	0.10	0.15	5.4

Answer 1

BrO₃^- (aq) + 5Br^- (aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O

Let the rate law be r = k[BrO₃^- ]^m [Br^- ]ⁿ [H⁺]^o     ------ (1)

Where m = order of the reaction with respect to BrO₃^-

        n = order of the reaction with respect to Br^-

        o = order of the reaction with respect to H⁺

Apply the values in the tabular form in the above equation we have

1^st values ----> 1.2 = k (0.10)^m (0.10)ⁿ (0.10)^o      ---- (2)

2^nd values ----> 2.4 = k (0.20)^m (0.10)ⁿ (0.10)^o      ---- (3)

3^rd values ----> 3.5 = k (0.10)^m (0.30)ⁿ (0.10)^o      ---- (4)

4^th values ----> 5.4 = k (0.20)^m (0.10)ⁿ (0.15)^o      ---- (5)

Eqn(3) / Eqn(2) gives    2^m = 2

                                    m = 1

Eqn(4) / Eqn(2) gives    3ⁿ = 3

                                    n = 1

Eqn(5) / Eqn(3) gives    1.5^o = 2.25

                                    o = 2

Therefore the rate law becomes r = k[BrO₃^- ]¹ [Br^- ]¹ [H⁺]²

                                             r = k[BrO₃^- ] [Br^- ] [H⁺]²

Order of the reaction is = sum of the powers of concentration terms in the rate law

                                  = 1+1+2

                                  = 4

Substitute the values of m , n , o in Eqn(2) we get ,

1.2 = k (0.10)¹ (0.10)¹ (0.10)²

k = 12000 (mol/L)^-3 s^-1