Question

Consider the reaction and data shown below

Experiment	Initial Rate	[A]	[B]
1	5.00	0.100	0.100
2	45.00	0.300	0.100
3	10.0	0.100	0.200
4	90.0	0.300	0.200

a. What is the rate law? b. What is the rate constant?

Answer 1

The rate law in general just relates the rate r(t) with the rate constant k and the concentration [A] of reactant A and [B] of reactant B:

r(t)=k[A]^m[B]ⁿ

where m is the order of reactant A and n is the order of reactant B. We do not know m or n yet, so we must find those to finish writing the rate law.

RATE LAW ORDERS, AND RATE LAW

To make our lives easier, let us set [B] as a constant in the rate law. That way, when [A] changes, we know it must influence r(t). Thus, we focus on experiments 1 and 2 and the change in the initial A concentration in relation to the change in initial rate:

r_i,₁(t)=k[A]^m_i,₁[B]ⁿ_i,₁
r_i,₂(t)=k[A]^m_i,₂[B]ⁿ_i,₂

But since [B] is constant, [B]_i,₂=[B]i,1.

r_i,₂(t) / r_i,₁(t) = (kA]^m_i,₂[B]ⁿ_i,₂)   /   (k[A]^m_i,₁[B]ⁿ_i,₁)

r_i,₂(t) / r_i,₁(t) = [A]^m_i,₂ / [A]^m_i,₁

45 / 5 =( 0.3 / 0.1)^m

If you do the math, you'd get a comparison:

9=(3)^m

Thus m=3, and A is third order, or the reaction is "third order with respect to A".

For B, we set [A] constant and follow the same process, comparing experiments 1 and 3, such that [A]_i,₁=[A]_i,₃:

r_i,₃(t) / r_i,₁(t)=[B]ⁿ_i,₃ / [B]ⁿ_i,₁

10/5 = (0.1/0.1)^m

If you do the math, you'd get a comparison:

2=(1)^m

Thus, n=0, and B is zero order, or the reaction is "zero order with respect to B".

Therefore, the overall rate law is:

r(t)=k[A]³[B]⁰

RATE CONSTANT

The rate constant is specific to the reaction, not to the experiment. That means we can pick any rate from any trial, and find the rate constant for the entire reaction.

ri,1(t)=k1[A]2i [B]i,1

45 = K₁ (0.3)³ (0.1)⁰

K₁ 0.027 = 45

K₁= 45/0.027 =1666 M−2⋅s−1