Considering the following reaction: BrO3- (aq) + 5Br- (aq) + 6H+(aq) 3Br2(aq) +
ID: 947748 • Letter: C
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 5.4Explanation / Answer
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
1st values ----> 1.2 = k (0.10)m (0.10)n (0.10)o ---- (2)
2nd values ----> 2.4 = k (0.20)m (0.10)n (0.10)o ---- (3)
3rd values ----> 3.5 = k (0.10)m (0.30)n (0.10)o ---- (4)
4th values ----> 5.4 = k (0.20)m (0.10)n (0.15)o ---- (5)
Eqn(3) / Eqn(2) gives 2m = 2
m = 1
Eqn(4) / Eqn(2) gives 3n = 3
n = 1
Eqn(5) / Eqn(3) gives 1.5o = 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
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