Question

What is the rate law and the value of K?

2MnO4 +5ClO3 +6H ---> 2Mn+5 ClO4 +3H2O

Exp. # [MnO₄^-]       [ClO₃^-]         [H⁺]           Initial Rate (M/s)

1          0.100 M        0.100 M      0.100 M        5.20 x 10^-3

2          0.250 M        0.100 M      0.100 M        3.25 x 10^-2

3           0.100 M        0.300 M      0.100 M        1.56 x 10^-2

4           0.100 M        0.100 M      0.200 M        7.35 x 10^-3

  

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 ClO3- and H+ cancels

(5.2*10^-3)/(3.25*10^-2) = (0.1/0.25)^a

a = ln(0.16) / ln(0.4) = 2 with respect to MnO4-

now.. choose points:

1 and 4, so Mn4- an dClO3- cnacels

(5.2*10^-3)/(7.35*10^-3) = (0.1/0.2)^c

c = ln(0.707) / ln(0.5) = 0.5

0.5 order with respect to H+

choose point 1 and 3 so ClO3- remains

(5.2*10^-3)/(1.56*10^-2) = (0.1/0.3)^b

b = ln(0.333) / ln(0.333) = 1

1st order with respec to ClO3-

Rate = k*[MnO4-]^2 * [ClO3-] * [H+]^0.5

for k

choose any point

Rate = k*[MnO4-]^2 * [ClO3-] * [H+]^0.5

(5.2*10^-3) = k*(0.1^2)(0.1)(0.1^0.5)

k = (5.2*10^-3) / ((0.1^2)(0.1)(0.1^0.5)) = 16.443