Chemical Kinetics Question 191
Question: In a reaction, the concentration of reactant is increased two times and three times then the increases in rate of reaction were four times and nine times respectively, order of reaction is [UPSEAT 2003]
Options:
A) Zero
B) 1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: C
Solution:
$ R=K{{(A)}^{2}},R’=K{{(2A)}^{2}},,\therefore \frac{R’}{R}=4 $
$ R=K{{(A)}^{2}},R’=K{{(3A)}^{2}},\frac{R’}{R}=9 $
