1.

If 30% of a first order reaction is completed in 12 mins, what percentage will be completed in 65.33 mins?

Answer»

Solution :Give data : Time taken for COMPLETION of 30% of the reaction = 12 mins.
a = 100 , x = 30 , a - x = 70 and t = 12 min
Formula : `K =(2.303)/(t)"log"(a)/(a-x)`
Solution : For a first order reaction,
`=(2.303)/(12)"log"(100)/(70)`
`=(2.303)/(12)xx0.1.549=0.02972 "min"^(-1)`
`k=2.97xx10^(-2)"min"^(-1)`
If t = 65.33 minutes, x = ?
`k=(2.303)/(65.33)"log"(100)/(100-x)`
`0.02972 = (2.303)/(65.33)"log"(100)/(100-x)`
`"log"(100)/(100-x)=(0.02972xx65.33)/(2.303)=0.8430`
`(100)/(100-x)` = Antilog of 0.8430
`100 = 6.966 (100 -x)`
`100 = 696.6-6.966 x`
`-6.966 x = 100 - 696.6=596.6`
`x=(596.6)/(6.966)=85.62%`
The reaction completed in 65.33 minutes


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