Question

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?

Answer 1

Answer:

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..

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