Answer :

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

Time, t = 40 mins

Let, [R_{0}] be the initial concentration.

[R] at 75% completion = [R_{0}] - [R_{0}] = =

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. [R_{0}]. 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. [R_{0}]. We can say it is a __zero order reaction__ from the following formula-

t_{1/2} = i.e. t_{1/2} [R_{0}]

**OR**

2 NO + O_{2}→2 NO_{2}

(a) let the rate law be

rate = k[NO]^{x}[O_{2}]^{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 O_{2} is 1 and 2 respectively.

(b) Rate law, rate = k[NO]^{1}[O_{2}]^{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}

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