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
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A. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\text{k}} + {{\text{k}}_{\text{g}}}$$
B. $${{\text{k}}_{\text{e}}}{\text{ff}} = \frac{{{\text{k}} + {{\text{k}}_{\text{g}}}}}{2}$$
C. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\left( {{\text{k}}{{\text{k}}_{\text{g}}}} \right)^{\frac{1}{2}}}$$
D. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$
The half life period of a first order reaction is given by (where, K = rate constant. )
A. 1.5 K
B. 2.5 K
C. $$\frac{{0.693}}{{\text{K}}}$$
D. 6.93 K
Catalyst is a substance, which __________ chemical reaction.
A. Increases the speed of a
B. Decreases the speed of a
C. Can either increase or decrease the speed of a
D. Alters the value of equilibrium constant in a reversible
A. $$ \propto {\text{CA}}$$
B. $$ \propto \frac{1}{{{\text{CA}}}}$$
C. Independent of temperature
D. None of these
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?