1.

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

Answer»

SOLUTION :For the first order reaction
`k=((2.303)/(t))log((a)/(a-x))`
When `x=((40)/(100))a=0.4 a`
`t = 50 m`
`THEREFORE k = ((2.303)/(50))log((a)/(a-0.4a))`
`k=((2.303)/(50))log((1)/(0.6))`
`=0.010216 "min"^(-1)`
t = ? When x = 0.8 a
From above `K = 0.010216 "min"^(-1)`
`therefore t ((2.303)/(0.010216))log((a)/(a-0.8a))`
`((2.303)/(0.010216))log((1)/(0.2))=157.58` min
The time at which the reaction will be 80% COMPLETE is 157.58 min.


Discussion

No Comment Found

Related InterviewSolutions