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A Reaction is Second Order in a and First Order in B - Chemistry

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Question

A reaction is second order in A and first order in B.

(i) Write the differential rate equation.

(ii) How is the rate affected on increasing the concentration of A three times?

(iii) How is the rate affected when the concentrations of both A and B are doubled?

 

Solution

(i) A reaction is second order in A and first order in B.

Differential rate equation:− `Rate=(-d[R])/dt=K[A]^2[B]`

(ii) On increasing the concentration of A three times i.e. 3A:

`Rate=k[3A]^2[B]=9k[A]^2[B]=9k[A]^2[B]=9(Rate)` , i.e. 9 times the initial rate.

(iii) On increasing the concentration of A and B as 2A and 2B:

`Rate=k[2A],^2[2B]=k(4xx2) i.e. 8 times the initial rate.[A]^2[B]=8k[A]^2[B]=8(Rate)` i.e. 8 times the initial rate

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2012-2013 (March) Delhi Set 1

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