Question

The initial rate of a reaction \[\ce{A + B -> products}\] is doubled when the initial concentration of A is doubled and increases eight fold when the initial concentration of both A and B are doubled. State the order of the reaction with respect to A and with respect to B. Write the rate equation.

Answer 1

\[\ce{A + B -> products}\]    ...(Given)

And the following experimental observations:

When [A] is doubled, the rate is doubled.

When [A] and [B] are both doubled, the rate increases 8-fold.

Observation 1: [A] Doubles, [B] constant

Let the rate law be Rate = k[A]^x[B]^y

Where:

x = order with respect to A

y = order with respect to B

Let original rate be Rate₁ ​= k[A]^x[B]^y

Now double [A], keeping [B] constant:

Rate₂ = k(2[A])^x[B]^y = 2^x⋅k[A]^x[B]^y = 2^x ⋅ Rate₁

Given: 

Rate₂​ = 2 × Rate_1​

So

2^x = 2 ⇒ x = 1

Observation 2: [A] and [B] both doubled

Rate₃​ = k(2[A])^x(2[B])^y = 2^x × 2^y × k[A]^x[B]^y = 2^x+y × Rate₁​

Rate₃​ = 8 × Rate_1​

We already found x = 1, so:

2^1+y = 8 = 2³

⇒ 1 + y = 3

⇒ y = 2

∴ Order w.r.t A = 1

Order w.r.t. B = 2

Overall order = 1 + 2 = 3

∴ The rate equation is k[A]¹[B]²