Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided. a + b yields products trial [a] [b] rate 1 0.30 m 0.25 m 1.2 × 10-2 m/min 2 0.30 m 0.50 m 4.8 × 10-2 m/min 3 0.60 m 0.50 m 9.6 × 10-2 m/min

Rate = (0.64 M²/min) [A]¹[B]²

Explanation:

1) Determination of the orders of A & 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:

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:

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.

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

r just means there's a number of carbon in teh chain. for example rcooh means that after x number of carbon, the chain ends with cooh.

explanation: