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Alok Singh Grade: 12
        

the reaction A->B follows a first order kinetics. the time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour . what is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B? plzz tell how did u find the answer

6 years ago

Answers : (2)

SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,


The integrated rate law for a 1st order reaction is:


ln([A]0) - ln([A]) = kt

One interesting thing about first order reactions is that the time required for 1/2 of the beginning reagent to react is independant of the conentration of the reagent. This can easily be seen by combining the two logs in the above integrated rate expression


ln([A]0/[A]) = kt























































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Sagar Singh


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6 years ago
vikas askiitian expert
510 Points
										

for first order reaction ,


t = 2.303log(a/a-x) / k                    ....................1


k is rate constant ..


when .8 mol of A produces .6mol of B then .8-.6 = .2 mol of A is left ...


final moles A = a-x = .2


initial moles = a = .8


time taken = 1hr


by plugging these values in eq 1 , we get


k = 2.303log4 ...............2


 


in second case , .9mol of A produces .675mols of B so remaining moles


of A are .9-.675 = .225


final moles of A = a-x = .225


initial mole = .9


now againg using eq 1


t = 2.303log(.9/.225)/k


  =2.303log4/k


k = 2.303log4 , so


t = 1hr


 


therefore time in second part will be same 1 hr...


 


appr0ve if u like this ans

6 years ago
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