Question

for a first order reaction 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 the content of 0.9 mole of a to produce 0.675 mole of A
pls answer it fast guys

Answer 1

Chemistry 

 Best Answer

for the reaction 
A ---> B 

the rate of the reaction can be considered 
rate = -d[A] / dt 

where 
[ ] means concentration 
the "-" is because the concentration of [A] is dropping 
and d[A] / dt is change in concentration of A per time 

make sense? 

also.. we know that rate is proportional to the amount of A present. Actually to the concentration of A. The higher the concentration.. the faster the rate... 
ie.. 
rate α [A] 
and if we put in a constant 
rate = k x [A] 

and because sometimes the mechanisms require two molecules of A to collide and sometimes 1 molecule can convert by itself.. there's an exponent on that [A] 
rate = k x [A]^n 

putting those two equations together.. 
rate = -d[A] / dt = k x [A]^n 

agreed? 

Answer 2

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

  n 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