Question

Wkst #13 CHEM& 162 Name 1. Several experiments were run on the chemical reaction A + B + 2C → products, giving the following results: Reaction 1 2 Rate (M.s-1) 4.998x102 1.767x102 1.767x102 3.534x102 [A] (M) 0.100 0.200 0.200 0.200 [B] (M) 0.500 0.500 0.250 0.250 [C] (M) 0.100 0.100 0.100 0.200 ديا 4 What is the rate law for this reaction? Be sure to include a numerical value for the rate constant and units!.

Answer 1

Assuming the overall Rate Law = k[A]^a[B]^b[C]^c

where,

[A], [B], [C] are concentration of reactants

a, b,c are respective concentration orders(on which the rate depends) and are constants.

In Reaction 1:

Rate = 4.988 x 10² = k x (0.1)^a x (0.5)^b x (0.1)^c​​​​​​​ ----(1)

In Reaction 2:

Rate = 1.767 x 10² = k x (0.2)^a x (0.5)^b​​​​​​​ x (0.1)^c​​​​​​​ ----(2)

Dividing (1) / (2) :

4.988 / 1.767 = (0.5)^a => a = 0.45 / log(0.5) = - 1.5 =>

• a = -3/2

In Reaction 3:

Rate = 1.767 x 10² = k x (0.2)^a x (0.25)^b​​​​​​​ x (0.1)^c​​​​​​​ ----(3)

Dividing (2) / (3) :

1 = (0.5/0.25)^b =>

• b = 0

In Reaction 4:

Rate = 3.534 x 10² = k x (0.2)^a x (0.25)^b​​​​​​​ x (0.2)^c​​​​​​​ ----(4)

Dividing (4)/(3):

3.534 / 1.767 = (0.2/0.1)^c => 2 = 2^c =>

• c=1

So, the Rate Law = k [A]^-3/2 [B]⁰[C]¹ = k [A]^-3/2[C]¹

Rate constant, k = Rate / [A]^-3/2[C]¹

Taking Reaction 1 values, k = 4.998 x 10² / (0.1)^-3/2 x (0.1) = 4.998 x 100 x (0.1)^1/2 = 157.73 M^3/2 s^- (Rate constant value)

Overall order of the reaction = a + b + c = -3/2 + 0 + 1 = -1/2