Ncert Exercises & Solutions

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

NEETprep Answer:

(i) The differential rate equation will be

$- \frac{d [R]}{d t} = K [A] {[B]}^{2}$

(ii) If the concentration of B is increased three times, then

$- \frac{d [R]}{d t} = K [A] {[3 B]}^{2}$
$= 9 . k [A] {[B]}^{2}$

Therefore, the rate of reaction will increase 9 times.
(iii) When the concentrations of both A and B are doubled,

$- \frac{d [R]}{d t} = K [A] {[B]}^{2}$
$= K [2 A] {[2 B]}^{2}$
$= 8 . K [A] {[B]}^{2}$

Therefore, the rate of reaction will increase 8 times.