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	0.50	0.010	3.0×10−3
2	0.50	0.020	6.0×10−3
3	1.00	0.010	1.2×10−2

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M. Express your answer to two significant figures and include the appropriate units.

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

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

(3*10^-3)/(6*10^-3) = (0.5/0.5)^a * (0.01/0.02)^b

0.5 = 1 * (0.5) ^b

b = 1

so it is 1st order with respect to B

now,

for B:

choose point 1 and 3

r1 / r3 = ([A]1/[A3])^a * ([B]1/[B]3)^b

(3*!0^-3)/(1.2*10^-2) = (0.5/1)^a (0.01/0.01)

0.25 = 0.5^a

ln(0.25)/ln(0.5) = a

a = 2

it is 2nd order with respec tto A

so

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

then, we need K

choose any point, (I will choose 1)

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

3*10^-3 = k*(0.5^2)(0.01)

k = (3*10^-3 ) / ((0.5^2)(0.01)) = 1.2 1/M2-s

now,

Rate = 1.2*[A]^2[B]

substitute units given

Rate = 1.2*[A]^2[B]

Rate = 1.2*(0.5^2)(0.075) = 0.0225 M/s