Answer :

(a) [A]_{1}= 0.1 mol L^{-1} at time = 0 seconds.

[A]_{2} = 0.05 mol L^{-1} at time = 10 seconds.

The rate constant (K) for a first order reaction is given as,

K =

K =

K = 0.069 sec^{-1}

Now, at time = 20 s, [A]_{2} = 0.025 mol L^{-1}

At time = 10 s, [A]_{1}= 0.05 mol L^{-1}

K =

K =

K = 0.069 sec^{-1}

Since, the value of rate constant at any time is same i.e. 0.069sec^{-1}. Therefore, the given reaction follows pseudo first order reaction.

(b) Average rate of reaction between the time intervals 10 to 20 seconds is

= =

= 0.0025 mol L^{-1} sec^{-1}

**OR**

(a) For a reaction A + B → P, the rate is given by

Rate = k [A][B]^{2}

(i) If the concentration of B is doubled,

New rate of reaction, rate’ = k[A][2B]^{2}

= 4k[A][B]

Rate’ = 4rate

The rate of reaction will become 4 times if the concentration of B is doubled.

(ii) Overall order of reaction if A is present in large excess will be 2. As A is present in large excess therefore the reaction will be independent of the concentration of A and will depend only on the concentration of B.

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

t_{1/2} = 30 mins

K = = = 0.0231 min^{-1}

[R] at 90% completion = [R_{0}] - [R_{0}] =

Time required for 90% completion of this reaction,

K =

t =

=

= 99.69 min

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