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:

What is the rate law for this reaction?

Express the rate law symbolically in terms of k , [A] , and [B] .

Trial	[A] (M )	[B] (M )	Rate (M/s )
1	0.15	0.010	2.7×10−4
2	0.15	0.020	5.4×10−4
3	0.30	0.010	1.1×10−3

Answer 1

Ans. Step 1: Order of reaction with respect to [B]:

Compare trial 1 and 2 – [A] remains constant while [B] varies

So, order of reaction with respect to B [Note: [B] varies while [A] kept constant] is given by-

            ([B]_T2 / [B]_T1)ⁿ = (r_T2 / r_T1) - equation 1

            Where, T1 and T2 represent trial 1 and 2 respectively.

                        r = rate of reaction.

                        [B] = concentration of B

                        n = order of reaction

Putting the values in equation 1-

            (0.02 / 0.01)ⁿ = (5.4 x 10^-4) / (2.7 x 10^-4)

            Or, (2)ⁿ = 2

            Hence, n = 1

Therefore, order of reaction with respect to [B] = 1      

Step 2: Order of reaction with respect to [A]:

Compare trial 1 and 3 – [A] varies while [B] remains constant.

So, order of reaction with respect to A is given by-

            ([A]_T3 / [B]_T1)ⁿ = (r_T3 / r_T1) - equation 1

Putting the values in equation 1-

            (0.03 / 0.15)ⁿ = (1.10 x 10^-3) / (2.7 x 10^-4)

            Or, (2)ⁿ = 4

            Hence, n = 2

Therefore, order of reaction with respect to [A] = 2

Step 3: Rate law is-

            r = k [A]² [B]