What is the value of 'n' if the reaction rate of the chemical reaction A → B, is proportional to CAn and it is found that the reaction rate triples, when the concentration of 'A' is increased 9 times?
A. $$\frac{1}{2}$$
B. $$\frac{1}{3}$$
C. $$\frac{1}{9}$$
D. 3
Answer: Option A

r1=kCa1^n. r2=kCa2^n
r2=3r1. Ca2=9Ca1
r1/3r1=Ca1^n/(9Ca1) ^n
n log 9=log 3
n=0.5
how did we get 1/2?