Chemical Kinetics Question 264
Question: The rate law for a reaction between the substances A and B is given by rate $ =k{{[A]}^{n}}{{[B]}^{m}} $ On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as
Options:
A) $ ( m+n ) $
B) $ ( n-m ) $
C) $ {2^{( n+m )}} $
D) $ \frac{1}{{2^{( m+n )}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ Rate_1=k{{[A]}^{n}}{{[B]}^{m}};Rate_2=k{{[2A]}^{n}}{{[{\scriptstyle{}^{1}/ _2}B]}^{m}} $
$ \therefore \frac{Rate_2}{Rate_1}=\frac{k{{[2A]}^{n}}{{[{\scriptstyle{}^{1}/ _2}B]}^{m}}}{k{{[A]}^{n}}{{[B]}^{m}}} $
$ ={{[2]}^{n}}{{[{\scriptstyle{}^{1}/ _2}]}^{m}}=2^{n}.,{2^{-m}}={2^{n-m}} $
