Home
Class 12
CHEMISTRY
The conversion of molecules of A to B fo...

The conversion of molecules of A to B follows second order kineitcs. Doubling the concentration of A will increase the rate of formation of B by a factor of?

Text Solution

Verified by Experts

The correct Answer is:
4
`2AtoB,r=K.[A]^(2),` If [A] is doubled `r_(B)=4` times
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    AAKASH SERIES|Exercise PRACTICE SHEET-3 (SINGLE OR MORE THAN ONE OPTION QUESTIONS)|16 Videos

  • CHEMICAL KINETICS

    AAKASH SERIES|Exercise PRACTICE SHEET-3 (LINKED COMPREHENSION TYPE QUESTIONS)|6 Videos

  • CHEMICAL KINETICS

    AAKASH SERIES|Exercise PRACTICE SHEET-2 (MATCH THE FOLLOWING QUESTIONS)|2 Videos

  • CHEMICAL KINETCS

    AAKASH SERIES|Exercise EXERCISE - 3.2|45 Videos

  • CHEMICAL THERMODYNAMICS

    AAKASH SERIES|Exercise Additional Practice Exercise|54 Videos

Similar Questions

Explore conceptually related problems

The conversion of A to B follows second order kinetics. Doubling the conc. of A will increase the rate of formation of B by factor

The convertion of molecules X to Y follows second order kinetics. If concentration of X is increased by three times, how will it affect the rate of formation of Y.

For the reaction A+B+CtoD . The following observation were made (i) When the concentrations of A was doubled, the rate of formation of D was doubled (ii) When the concentration of B was halved, the rate of formation of d becomes one fourth (iii) Doubling the concentration of C and no effect on rate Select the correct statement (s).

For the reaction 2A+Bto Products, it is found that doubling the concentration of both reactants increases the rate by a factor of 8. But doubling the concentration of B alone, only doubles the rate. What is the order of the reaction w.r.t to A?

For the reaction A+B products, it is observed that (a) On doubling the initial concentration of A only, the rate of reaction is also doubled (b) On doubling the initial concentration of both A and B, there is a change by a factor of 8 in the rate of the reaction. The rate of this reaction is given by

Decompositon of NH_(3) on gold surface follows zero order kinetics. If rate constant is 5xx10^(-4)Ms^(-1) , rate of formation of N_(2) will be