Question

Consider the reaction A+2B⇌C whose rate at 25 ∘C was measured using three different sets of initial concentrations as listed in the following table: Trial [A] (M) [B] (M) Rate (M/s) 1 0.20 0.010 4.8×10−4 2 0.20 0.020 9.6×10−4 3 0.40 0.010 1.9×10−3 What is the rate law for this reaction? Express the rate law symbolically in terms of k, [A], and [B].

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.

Answer 1

A+2B⇌C

Let the rate law be , r = k[A]^m[B]ⁿ      --------(1)

Where

m is the order of the reaction with respective to A

n is the order of the reaction with respective to B

Let us apply the first data values to the above rate law,then it becomes,

for the first set of values 4.8×10⁻⁴ = k[0.20]^m[0.010]ⁿ      -------- (2)

for the second set of values 9.6×10⁻⁴ = k[0.20]^m[0.020]ⁿ      -------- (3)

for the third set of values      1.9×10⁻³ = k[0.40]^m[0.010]ⁿ      -------- (4)

Eqn(3) / Eqn(2) gives     (0.02/0.01)ⁿ = (9.6×10⁻⁴ ) / (4.8×10⁻⁴ )

                                                2ⁿ = 2

                                                 n = 1

Eqn(4) / Eqn(2) gives     (0.40/0.20)^m = ( 1.9×10⁻³) / (4.8×10⁻⁴ )

                                                2^m = 2²

                                                 m = 2

Plug the values of m& n in Equ (1) we get r = k[A]²[B]¹

Therefore the rate law be r = k[A]²[B]¹