Answer:
rate = k [Br⁻][H⁺]²[BrO₃⁻]
Explanation:
Here we are going to determine the rate law for the reaction
5 Br-(aq) + BrO3-(aq) + H+(aq) → 3 Br2(aq) + 3 H2O(l)
by performing experiments in which we vary concentration of the reactants and determining the effect this has on the initial rate.
Exp [ Br− ] [BrO3− ] [ H+] (Δ[BrO3− ]/Δt)0 M·s−1
I 0.10 0.10 0.10 6.8 x 10-4
II 0.15 0.10 1.0 10-3
III 0.10 0.20 0.10 1.4 x 10-3
IV 0.10 0.10 0.25 4.3 10-3
The way to do the comparison is by taking experiments in which we keep constant the concentration of two of the reactants and vary the third and study the effect this change has on the initial rate of the reaction.
Comparing experiments I and III we see that the initial reaction doubled when we doubled the [BrO3− ] while keeping the other two the same. Thus the reaction rate is of order 1 respect to [BrO3− ].
In experiments I and IV we increased the concentration of H⁺ by 2.5 times, and the rate of the reaction increased by a factor of 6.3 which is 2.5 squared . If you do not see it, lets try using logarithms
(0.25/ 0.10)^ x = 4.3 x 10⁻³ / 6.8 x 10⁻⁴
2.5 ^ x = 6.3235
x log 2.5 = log 6.3235
0.40 x = 0.80 ∴ x = 2
Therefore the rate is second order respect to [H⁺].
Comparing experiments I and II we see that increasing the concentration of Br⁻ by a factor of 1.5, the initial rate also went up by a factor of 1.5 ( 1.0 x 10⁻³ / 6.8 x 10⁻⁴ =1.5. Thus the rate is first order respect to [ Br⁻ ].
Then our rate law is
rate = k [Br⁻][H⁺]²[BrO₃⁻]
