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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

```
3 years ago

Nitin jha
8 Points
```							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
```
3 years ago
Tarun Kumar
16 Points
```							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
```
3 years ago
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