Question

A reaction is of first order with respect to A and of second order with respect to B. What will happen to the rate of reaction if (i) only the concentration of B is increased to three times, and (ii) A is taken in excess ?

Hint:

Write rate = k[A][B]^2. Tripling B multiplies the rate by 3 squared. With A in large excess, [A] stays constant and the reaction becomes pseudo second order.

Answer 1

Step 1: Write the rate law:
The order is 1 in A and 2 in B. So the rate law is:
\[ \text{Rate} = k[\mathrm{A}][\mathrm{B}]^2 \]
The overall order is 1 + 2 = 3.

Step 2: Part (i) - B is tripled:
Let the new rate be \( R' \). Put \( [\mathrm{B}] \) as \( 3[\mathrm{B}] \) and keep \( [\mathrm{A}] \) the same.
\[ R' = k[\mathrm{A}](3[\mathrm{B}])^2 = 9\,k[\mathrm{A}][\mathrm{B}]^2 = 9 \times \text{Rate} \]
So the rate becomes 9 times the original rate.

Step 3: Part (ii) - A is in excess:
When A is present in large excess, only a tiny part of it is used up. So \( [\mathrm{A}] \) stays almost constant.
Then \( k[\mathrm{A}] \) can be written as a new constant \( k' \):
\[ \text{Rate} = k'[\mathrm{B}]^2 \]
The rate now depends only on \( [\mathrm{B}] \), and the reaction behaves as a second order reaction (a pseudo second order reaction).

Final Answer:
(i) The rate becomes 9 times. (ii) The reaction becomes pseudo second order, with rate = \( k'[\mathrm{B}]^2 \).