Class 12th Chemistry 2017 Set3 Delhi Board Paper Solution

Question 18

Following data are obtained for the reaction:

N₂O₅ → 2NO₂ + 1/2 O₂

t/s 0 300 600

[N₂O₅] / mol L-1 1.6 x 10-2 0.8 x 10-2 0.4 x 10-2

(a) Show that it follows first order reaction.

(b) Calculate the half-life.

(Given: log 2 = 0.3010 , log 4 = 0.6021 )

Answer

For first order reaction:

k = 2.303 / t Log a₀ /a

where a₀= initial concentration

and a = remaining concentration

At t = 300 seconds

K₁ = 2.303/300 log 1.6 x 10-2 / 0.8 x 10-2

K₁ = 2.303/300 log 2

K₁ = 2.303/300 x 0.3010

K₁ = 0.00231 s-1

Now at t = 600 seconds

K₂ = 2.303/600 log 1.6 x 10-2 / 0.4 x 10-2

K₂ = 2.303/600 log 4

K₂ = 2.303/600 x 0.6021

K₂ = 0.00231 s-1

Since the value of K is contant in both the situation so it is a first order reaction.

K = 0.00231 s-1

t_1/2= 0.693 / 0.00231 = 300 seconds.

Q:-

In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B?

t/s	0	300	600
[N₂O₅] / mol L-1	1.6 x 10-2	0.8 x 10-2	0.4 x 10-2