Chemical Kinetics and Nuclear ChemistryHard

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.1/2m + n
B.(m + n)
C.(n - m)
D.2(n - m)

Solution

Rate = K[A]n [B]m
Given, doubling the concentration of A and halving the concentration of B
then  Rate2  = K[2A]n []m
Rate2 = K[A]n [B]mx 2(n-m)

Create a free account to view solution

View Solution Free
Topic: Chemical Kinetics and Nuclear Chemistry·Practice all Chemical Kinetics and Nuclear Chemistry questions

More Chemical Kinetics and Nuclear Chemistry Questions