Question

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:

What is the value of the rate constant k for this reaction?

Trial	[A] (M)	[B] (M)	[C] (M)	Initial rate (M/s)
1	0.10	0.10	0.10	3.0×10−5
2	0.10	0.10	0.30	9.0×10−5
3	0.20	0.10	0.10	1.2×10−4
4	0.20	0.20	0.10	1.2×10−4

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 and B cancels

(3*10^-5)/(9*10^-5) = (0.1/0.1)^a * (0.1/01)^b * (0.1/0.3)^c

0.333 = 0.33^c

c = 1,

choose point1 and 3, so b and c ancels

(3*10^-5)/(1.2*10^-4) = (0.1/0.2)^a * (0.1/0.1)^b * (0.1/0.1)^c

0.25 = 0.5^a

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

now... chose point 3 and 4 so a and C cancel

(1.2*10^-4)/(1.2*10^-4) = (0.2/0.2)^a * (0.1/0.2)^b * (0.1/0.1) ^c

1 = 0.5^b

b = 0

then,

Rate = k*[A]^2 *[B]^0 *[C]^1

choose any point for k

k = RAte /([A]^2 * [C])

k = (3*10^-5) / (0.1^2 * 0.1)

k = 0.03

units --> 1/M2s