Chemical Kinetics Question 62
Question: The rate law of a reaction $ A+B\to $ Product is rate = $ k{{[A]}^{n}} $
$ {{[B]}^{n}} $ On doubling the concentration of A and halving the concentration of B, the ratio of new rate to the earlier rate of reaction will be
Options:
A) $ n-m $
B) $ {2^{n-m}} $
C) $ \frac{1}{{2^{m+n}}} $
D) $ m+n $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ Rat{e_1}=K{{[A]}^{n}}{{[B]}^{m}};Rat{e_2}=K{{[2A]}^{n}}{{[ \frac{1}{2}B ]}^{m}} $
$ \therefore \frac{Rate_2}{Rate_1}=\frac{K{{[2A]}^{n}}{{[ \frac{1}{2}B ]}^{m}}}{K{{[A]}^{n}}{{[B]}^{m}}} $
$ ={{(2)}^{n}}{{( \frac{1}{2} )}^{m}}=2^{n}\cdot {2^{-m}}={2^{n-m}} $
