Question

The inversion of compound B is experimentally determined to follow the rate law: d[B]/dt = -k[B]^a[H⁺]^b.
At 25 °C and a constant pH of 3.00, the reaction proceeds with a constant half-life of 40.0 min. At this same temperature, but at a constant pH of 1.00, the half-life is constant at 4.00 min. What are the exponents (orders) a and b in the above rate law?

Answer 1

We know that the pH is kept constant in the duration of the reaction. That means the rate law simplifies to pseudo a-th order rate law:
d[B]/dt = k'∙[B]^a
with
k' = k∙[H⁺]^b

This is the form of rate law, which is known as simple nth order reaction [1]
with n = a..
The half-life for this reaction is given by
(i) for a = 1
t_½ = ln(2)/k'
(ii) for a ≠ 1
t_½ = (2^(a-1) - 1) / ( (a - 1)∙k'∙[B]₀^(a-1) )

[B]₀ in the second formula denotes the reactant concentration at the beginning of the half-life period.
Since [B]₀ decreases from half-life to half-life period due to consumption of B the half-life changes as reaction proceeds. i.e., if and only if a=1 the half-life of the reaction is constant through out the whole reaction,

To find exponent b we can rewrite formula (i) in terms of k and [H⁺] as:
t_½ = ln(2)/(k∙[H⁺]^b)

We have measured
(t_½)₁ = 40.0 min
at
[H⁺]₁ = 10^(-pH₁) = 10^(-3.0) = 0.001
and
(t_½)₂ = 4.00 min
at
[H⁺]₂ = 10^(-pH₂) = 10^(-1.0) = 0.1 M

From the formula for the half-life follows the ratio of the half-life periods:
(t_½)₁ / (t_½)₂ = {ln(2)/(k∙[H⁺]₁^b)} / {ln(2)/(k∙[H⁺]₂^b) }
<=>
(t_½)₁ / (t_½)₂ = ( [H⁺]₂ / [H⁺]₁ )^b
<=>
40.0 min / 4.00 min = ( 0.1 M / 0.001 M )^b
<=>
10 = ( 100 )^b
=>
b = 1/2

Now the reaction orders with respect to B and H⁺ are
a = 1 Ans.
b = 1/2 Ans.