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	0.20	0.050	2.4×10−3
2	0.20	0.100	4.8×10−3
3	0.40	0.050	9.6×10−3

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.

Please help!

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

(2.4)/(4.8) = 0.2/0.2^a * (0.05/0.10)^b

0.5 = 1 * 0.5^b

b = 1

choose point 1 and 3 so B cancel

2.4/9.6 = (0.2/0.4)^a * (0.05/0.05)^b

0.25 = 0.5^a

ln(0.25) / ln(0.5) = a

a = 2

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

get k fom ay point:

2.4*10^-3 = k*(0.2^2)(0.05)

k = (2.4*10^-3 ) / ((0.2^2)(0.05))

k = 1.2

substitute values

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

Rate = 1.2*(0.2^2)(0.05)

Rate = 0.0024 M/s