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(a) Given, reaction is of 1^st order,

Time, t = 40 mins

Let, [R₀] be the initial concentration.

[R] at 75% completion = [R₀] - [R₀] = =

For 1^st order reaction, K =

K =

=

= 0.0346 min^-1

Now, for a 1^st order reaction, K =

t_1/2 = = 20.02 mins 20 mins

(b)

(i)

From graph, it is clear that t_1/2 does not depend on initial concentration i.e. [R₀]. We can say it is a 1^st order reaction because t_1/2 for a 1^st order reaction depends only on value of K.

t_1/2 =

(ii)

From graph, it is clear that value of t_1/2 varies linearly with value of initial concentration i.e. [R₀]. We can say it is a zero order reaction from the following formula-

t_1/2 = i.e. t_1/2 [R₀]

OR

2 NO + O₂→2 NO₂

(a) let the rate law be

rate = k[NO]^x[O₂]^y

From dividing rate of experiment 1 and 3,

0.25 = (0.5)^y

y = 2

From dividing rate of experiment 2 and 4,

0.25 = (0.25)^x

x = 1

The order of reaction with respect to NO and O₂ is 1 and 2 respectively.

(b) Rate law, rate = k[NO]¹[O₂]²

And overall order of reaction = 1+2 = 3

(c) k =

Substituting values in above equation from experiment 1,

k =

k = 0.06 mol^-1/2 l^1/2 s^-1