Question

The following results have been obtained during the kinetic studies of the reaction. 2A + B → C + D

Experiment	[A] mol L-1	[B] mol L-1	Initial rate of formation of D (molL1min1)
I	0.1	0.1	6.00 x 10-3
II	0.3	0.2	7.20 x 10-2
III	0.3	0.4	2.88 x 10-1
IV	0.4	0.5	2.40 x-10-2

What is the rate law ? what is the order with respect to each reactant and the overall order ? Also calculate rate constant and write its unit.

Answer 1

Let the equation is,
Rate = k [A]^x [B]^y
Then According to given data.
(rate)_I = 6.0 x 10^-3 = k(0.1)^x (0.1)^y.
(rate)_II = 7.2 x 10^-2 = k(0.3)^x (0.2)^y
(rate)_III = 2.88 x 10^-1 = k(0.3)^x (0.4)^y
(rate)_IV = 2.40 x 10^-2 = k(0.4)^x (0.1) ^y
From equation (2) and (3)

y = 2
From equation (1) and (4)

x = 1
Hence, rate = k [A] [B]²
because x = 1, y = 2.
Rate law is
rate = k [A] [B]²
Hence calculation of rate constant with the help of eq (1)
rate = k [A] [B]²
6.0 x 10^-3 = k(0.1)(0.1)²
k = \(\frac { 6.0\times 10^{ -3 } }{ 10^{ -3 } } \)
k = 6.0 mol^-2 L² min^-1.