1.

A first order reaction is 40% complete in 50 minutes . Calculate the value of the rate constant. In wha time the reaction be 80% complete ?

Answer»

SOLUTION :(i) For the first order reaction `k=(2.303)/tloga/((a-x))`
Assume , a = 100% x = 40% t = 50 MINUTES
Therefore , a-x = 100-40 = 60
k = (2.303/50)log (100/60)
`k = 0.010216 "min"^(-1)` .
Hence the value of the rate constant is `0.010216"min"^(-1)`
(II) t = ? , wheb x = 80%
Therefore , a - x = 100-80 = 20
From above , k = `0.010216"min"^(-1)`
`t=(2.303//0.010216)log (100//20)`
t = 157.58 min
The time at which the reaction will be 80% COMPLETE is 157.58 min.


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