ConcTime Relation HalfLife Order 0 Ao 2k 0693k In A0 A k

Conc-Time Relation Half-Life Order 0 [A]o /2k 0.693/k In ( [A]0 / [A] ) = kt 2

Solution

A   +   B     ----->     products

So, rate, r = k [A]^m [B]ⁿ

Considering the expt. (1) and (2); Keeping the concentration [B] same and increasing the concentration of [A] by five times the rate increases by 25 times.

So, 25 x rate = k [5A]^m [B]ⁿ

25 x rate = k 5^m [A]^m [B]ⁿ

25 = 5^m

25 = 5²

m = 2

So, A is of second order.

Considering the expt. (1) and (3); Keeping the concentration [A] same and increasing the concentration of [B] by five times the rate remains constant.

So, rate = k [A]^m [5B]ⁿ

rate = k 5ⁿ [A]^m [B]ⁿ

1 = 5ⁿ

1 = 5⁰

n = 0

So, B is of zero order

Now, the overall order of the reaction = 2 + 0 = 2.

Hence, the overall rate equation is expressed as

r = k [A]^m [B]ⁿ

rate, r = k [A]² [B]⁰

rate, r = k [A]²

Now, substituting the values of expt.(2) in the overall rate equation, we get;

r = k [A]²

2.0 x 10^-1 = k (0.500)²

Or, 0.20 = k (0.25)

Or, k = 0.20 / 0.25

Or, k = 0.8

When the [A] = 0.300 M and [B] = 0.190 M

The rate, r = k [A]²

r = 0.8 x (0.300)²

r = 0.072