Question

Experiment shows that the reaction 6I−(aq) + BrO3 −(aq) + 6H+(aq) ⟶ 3I2 (aq) + Br−(aq) + 3H2O(l) obeys this rate law: rate = k[I- ][BrO3 ][H+] 2 .

(a) What is the reaction order in each reactant and the overall reaction order?

(b)By what factor does the rate change if [I−] and [BrO3 −] are tripled and [H+] is doubled?

Answer 1

The rate law (or rate equation) of a reaction gives the relationship between the rate of a reaction and the concentration of the reactants. It is usually written in the form

where [A], [B], [C] are the concentration of the reactants.

k= rate constant

m= order of the reaction with respect to A

n = order of the reaction with respect to B and so on.

Overall order of the reaction is the sum of the individual order of reactions.

Overall order of reaction = m+n+p+...

The given reaction is -

The rate law for this reaction is -

a) Comparing with the general form, we get

Order of the reaction with respect to I^- = 1

Order of the reaction with respect to BrO₃^- = 1

Order of the reaction with respect to H⁺ = 2

Overall order of reaction = 1+1+2 = 4

b) Case I

Let the concentrations of the reactants be [I^-], [BrO₃^-] and [H⁺].

Case II

Let the concentrations of the reactants be [I^-]', [BrO₃^-]' and [H⁺]',

where [I^-]' = 3[I^-]

[BrO₃^-]' = 3[BrO₃^-]

and [H⁺]' = 2[H⁺]

( [I^-] and [BrO₃^-] are tripled and [H⁺] is doubled)

so the new rate is

(since the reactant concentrations change, the rate will also change. Rate constant is independent on reactant concentrations, so it will remain the same.)

So the rate changes by a factor of 36.