Home
Class 12
CHEMISTRY
A first order reaction is 50% completed ...

A first order reaction is 50% completed in 40 minutes at 300K and in 20 minutes at 320K. Calculate the activation energy of the reaction.
(Given:`log2=0.3010," "log4=0.6021, " " R=8.314J.K^(-1)."mol"^(-1))`

Text Solution

Verified by Experts

`k_(1)=(0.693)/(40)=0.017325" min"^(-1)`
`k_(2)=(0.693)/(20)=0.03465" min"^(-1)`
`thereforelog ((k_(2))/(k_(1)))=(E_(a))/(2.303xx8.314)xx(320-300)/(320xx300)`
or, `log .((0.03465)/(0.017325))=(E_(a))/(2.303xx8.314)xx(20)/(320xx300)`
or, `E_(a)=(log2xx2.303xx8.314xx320xx300)/(20)`
`=27666.55 " J.mol"^(-1)=27.66655"kJ.mol"^(-1)`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    CHHAYA PUBLICATION|Exercise SOLVED NCERT TEXTBOOK PROBLEMS (NCERT INTEXT QUESTIONS)|21 Videos

  • CHEMICAL KINETICS

    CHHAYA PUBLICATION|Exercise SOLVED NCERT TEXTBOOK PROBLEMS (NCERT EXERCISE QUESTIONS)|44 Videos

  • CHEMICAL KINETICS

    CHHAYA PUBLICATION|Exercise SOLVED CBSE SCANNER (ALL INDIA-2017)|1 Videos

  • BIOMOLECULES

    CHHAYA PUBLICATION|Exercise Practic Set 29|1 Videos

  • CHEMICAL THERMODYNAMICS

    CHHAYA PUBLICATION|Exercise Practice Set 6|15 Videos

Similar Questions

Explore conceptually related problems

A first order reaction takes 20 minutes for 25% decomposition . Calculate the time when 75% of the reaction will be completed. (Given : log 2 = 0.3010, log3 = 0.4771 log4 = 0.6021 )

A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction . (log2=0.3010)

A first order reactions 50% completed in 1.26xx10^4 sec. How much time would take for 100% completion.

Specific reaction rates for a reaction at 300K and 310K are 3xx10^(-6) S^(-1) and 2.8xx10^(-5 S^(-1 . Calculate the activation enrgy for that reaction.

At 300K , frequency factor (A) of a first order reaction is 1000 times its rate constant . The activation energy of the reaction ("in kJ. Mol"^(-1)) is -

The rate constant of a first order reaction becomes 6 times when the temperature is raised from 350 K to 410K. Calculate the energy of activation for the reaction.

The rate of a reaction doubles when its temperature changes from 300K to 310K . Activation energy of such a reaction will be - (R=8.314"J.K"^(-1)"mol"^(-1) and log2=0.310)

For increasing temperature from 300K to 320K the rate of reaction increases by 4 times calculate energy of activation for that reaction.

The specific reaction rates of a chemical reaction are 2.45xx10^(-5)sec^(-1)" at "273K and 16.2 xx10^(-4) sec^(-1) at 303 K. Calculate the activation energy.