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 [
Rate2 = 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
At a certain temperature, the first order rate constant k1 is found to be smaller than the second order rate constant k2...When the temperature of a reaction system is increased by 10°C (from 25°C to 35°C) at constant volume, the reaction rate...Which of the following statements are true for chemisorption...Rates of hydration of the following alkenes are :-(P) CH3-O-CH = CH2(Q) F-CH = CH2(R) CH3 -CH = CH2(S) CH = CH2...How many coulomb of electricity would be required to reduce the iron in 32.9 g of potassium hexa cyano ferrate (III) K3F...