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)
Given, doubling the concentration of A and halving the concentration of B
then Rate2 = K[2A]n [
]mRate2 = K[A]n [B]mx 2(n-m)
Create a free account to view solution
View Solution FreeTopic: Chemical Kinetics and Nuclear Chemistry·Practice all Chemical Kinetics and Nuclear Chemistry questions
More Chemical Kinetics and Nuclear Chemistry Questions
The correct increasing order of basic strength for the following compounds is :...What is the final product in the disintegrations series of starting from 92U235 :...Which of the following quantity/quantities is/are temperature independent?...A reaction involving two different reactants can never be...T50 of first - order rection is 10 min. Starting with 10 mol-1 rate after 20 min is...