A reaction is first order in A and second order in B. 1. Write the differential rate equation. 2. How is the rate affected on increasing the concentration of B three times? 3. How is the rate affected when the concentrations of both A and B are doubled?

1. Differential rate equation of reaction is dx dt =k[A]^1[B]^2=k[A] [B]^2 2. When cone. of B is tripled, it means cone. of B becomes (3 × B) New rate of reaction, dx' dt =k[A] [3B]^2=9k[A] [B]^2= ( dx dt ) i.e., rate of reaction will become 9 times. 3. When cone. of A is doubled and that of B is also doubled, then cone. of A becomes [2A] and that of B becomes [2B] rate of reaction, dx'' dt =k[2A] [3B]^2=8k[A] [B]^2 i.e., the rate of reaction will become 8 times the rate as in (1).

A reaction is first order in A and second order in B. 1. Write th…

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Question

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

Write the differential rate equation.
How is the rate affected on increasing the concentration of B three times?
How is the rate affected when the concentrations of both A and B are doubled?

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Answer

Differential rate equation of reaction is

$\frac{\text{dx}}{\text{dt}}=\text{k}[\text{A}]^1[\text{B}]^2=\text{k}[\text{A}]\ [\text{B}]^2$

When cone. of B is tripled, it means cone. of B becomes (3 × B)

$\therefore$ New rate of reaction,
$\frac{\text{dx'}}{\text{dt}}=\text{k}[\text{A}]\ [\text{3B}]^2=9\text{k}[\text{A}]\ [\text{B}]^2=\bigg(\frac{\text{dx}}{\text{dt}}\bigg)$
i.e., rate of reaction will become 9 times.

When cone. of A is doubled and that of B is also doubled, then cone. of A becomes [2A] and that of B becomes [2B] rate of reaction,

$\frac{\text{dx''}}{\text{dt}}=\text{k}[\text{2A}]\ [\text{3B}]^2=8\text{k}[\text{A}]\ [\text{B}]^2$
i.e., the rate of reaction will become 8 times the rate as in (1).

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