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it takes 1 hour for first order reaction to go to 50% completion. total time required for same reaction to reach 87.5 completion will be
By using simple unitary method..we get that in 1hour 50% reaction is completed so in 1/50hr 1% reaction will take place..so 87.5% reaction will take place in 87.5/50hour I.e equal to 1.75hr50% > 1hr1% > 1/50hr87.5% >87.5/50= 1.75 hr
For first order reactionKt =2.303 log a/a-x Where "a" is intial concentration of reactant and "a-x" is concentration at t time.So when reaction 50% completes K t(50%) = 2.303 log a÷a/2 ____(1)When reaction 87.5% completes K t(87.5%) = 2.303 log a÷ a/8 _____(2)Because when reaction 87.5 % completes 87.5% reactant changed into product and only 12.5% (which is a-x) remained. (12.5% = a/8)Now devide both equation t(50%)÷ t(87.5%) = log 2 ÷ log 8By solving this we get t(87.5%) = 3 t(50%)t(87.5%) = 3h
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