Answer:
Rate = (0.64 M²/min) [A]¹[B]²
Step-by-step explanation:
1) Determination of the orders of A & B:
- The rate law of the reaction = k [A]ᵃ[B]ᵇ
where, k is the rate law constant,
a is the order of the reaction with respect to reactant A,
b is the order of the reaction with respect to reactant B.
This is initial rate method problem:
From trial 1 & 2:
- Reactant [A] has the same concentration in both trials, but [B] has different concentrations and the rate of the reaction changes, so the reaction rate depends on [B].
- From trial 1, Rate₁ = k [A₁]ᵃ[B₁]ᵇ, [1.2 × 10⁻² M/min] = k [0.30 M]ᵃ[0.25 M]ᵇ
- From trial 2, Rate₂ = k [A₂]ᵃ[B₂]ᵇ, [4.8 × 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ
By dividing Rate₁ / Rate₂:
Rate₁ / Rate₂ = k [A₁]ᵃ[B₁]ᵇ / k [A₂]ᵃ[B₂]ᵇ
[1.2 × 10⁻² M/min] / [4.8 × 10⁻² M/min] = k [0.30 M]ᵃ[0.25]ᵇ / k [0.30 M]ᵃ[0.50 M]ᵇ
0.25 = [0.50]ᵇ
Taking log for both sides; log(0.25) = b log(0.5)
b = log(0.25) / log(0.5) = 2.
The reaction is second order with respect to reactant B.
By the same way for reactant A:
From trial 2 & 3:
- Reactant [B] has the same concentration in both trials, but [A] has different concentrations and the rate of the reaction changes, so the reaction rate depends on [A].
- From trial 2, Rate₂ = k [A₂]ᵃ[B₂]ᵇ, [4.8 × 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ
- From trial 2, Rate₃ = k [A₃]ᵃ[B₃]ᵇ, [9.6 × 10⁻² M/min] = k [0.60 M]ᵃ[0.50 M]ᵇ
By dividing Rate₂ / Rate₃:
Rate₂ / Rate₃ = k [A₂]ᵃ[B₂]ᵇ / k [A₃]ᵃ[B₃]ᵇ
[4.8 × 10⁻² M/min] / [9.6 × 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ / k [0.60 M]ᵃ[0.50 M]ᵇ
0.50 = [0.50]ᵃ
Taking log for both sides; log(0.50) = a log(0.5)
a = log(0.50) / log(0.50) = 1.
The reaction is first order with respect to reactant A.
- The rate law of the reaction will be: Rate = k [A]¹[B]²
- The overall order of the reaction is third order reaction.
2) Determining the rate law constant:
Rate = k [A]¹[B]²
[1.2 × 10⁻² M/min] = k [0.30 M]¹[0.25 M]²
k = [1.2 × 10⁻² M/min] / [0.30 M]¹[0.25 M]² = 0.64 M²/min.
So, the rate law will be: Rate = (0.64 M²/min) [A]¹[B]²