Q. For the chemical change $A \rightarrow B$ it is found that the rate of reaction doubles when the concentration is increased by 4 times. The order of the reaction is
A) One
B) Two
C) Half
D) None of these
Explanation
(c)
$A \rightarrow B$
$\frac{dx}{dt} = k\lbrack A\rbrack^{n}$
$\ \frac{dx^{'}}{dt} = k\lbrack 4A\rbrack^{n}$
$\frac{dx^{'}/dt}{dx/dt} = \left( \frac{4}{1} \right)^{n}$
$2 = \left( \frac{4}{1} \right)^{n}$
$n = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.