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-2L2 min-1.