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	.50	.30	9.0x10^-3
2	.50	.60	1.8x10^-2
3	.100	.30	3.6x10^-2

What is the rate law for this reaction? Express the rate law symbolically in terms of k, [A], and [B].

Answer 1

In order to calculate the rate law expression for a A+B reaction, we need to apply Initial Rates Method.

Note that the generic formula goes as follows:

r = k [A]^a [B]^b

Note that if we got at least 3 sets of point, in which we have A and B constant, then we could use:

r1 / r2 = (k1 [A]1^a [B]1^b) / (k2 [A]2^a [B]2^b)

If we assume K1 and K2 are constant, then K1= K2 cancel each other

r1 / r2 = ([A]1^a [B]1^b) / ( [A]2^a [B]2^b)

Then, order according to [A] and [B]

r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b

If we get two points in which A1 = A2, then we could get B, and vise versa for A...

From the data shown in YOUR table

choose point 1 and 2 so [A] cancels

(9*10^-3)/(1.8*10^-2) = (0.3/0.6)^b

b = ln((9*10^-3)/(1.8*10^-2) ) / ln((0.3/0.6))

b = 1

now, choose point 1 and 3 so b cancels

(9*10^-3)/(3.6*10^-2) = (0.5/0.100)^a

0.25 = 0.5^a

a = ln(0.25)/ln(0.5) = 2

then

Rate = k*[A]^2 *[B]

for k

choose nay point

9*10^-3 = k* (0.5^2)(0.3)

k = (9*10^-3 ) / ((0.5^2)(0.3)) =

k = 0.12 1/(M2-s)

Rate = 0.12 *[A]^2 *[B]