JEE Main2012ChemistryChemical KineticsActual
Reaction rate between two substance A and B is expressed as following: rate =k[A]^n[B]^m If the concentration of A is doubled and concentration of B is made half of initial concentration, the ratio of the new rate to the earlier rate will be:
Options
- Am+n
- Bn-m
- C1 .2^ (m+n )
- D.2^ (n-m )
Correct answer
D. .2^ (n-m )
Step-by-step solution
aligned Rate ₁ & =k[A]^n[B]^m Rate ₂ & =k[2 A]^n [ 1 2 B ]^m Rate ₂ Rate ₁ & = k[2 ~A ]^n [ 1 2 ~B ]^m k[ ~A ]^n[ ~B ]^m =(2)^n ( 1 2 )^m & =2^n (2)^ -m =2^ n-m aligned