1.

A first order reaction is 60% complete in 20 minutes . How long will the reaction take to be 84% complete ?

Answer»

54 MINS
68 mins
40 mins
76 mins

SOLUTION :If `[A]_(0) = 100` then `[A] = 100 - 60 = 40 ` in 20 minutes
Rate constant (k) = `(2.303)/(t) "log" ([A]_(0))/([A])`
`k = (2.303)/(20) "log" (100)/(40)`
`k = (2.303)/(20) (1- 0.6020) =(2.303)/(20) xx 0.397 "min"^(-1) … (1)`
The time for 84 % completion of the REACTION is `t_(84 %) = (2.303)/(k) "log" (100)/(100 - 84)`
substituting the VALUE of k from eq. (1) we get `t_(84 %) = (2.303 xx 20)/(2.303 xx 0.397) "log" (100)/(16)`
`t_(84 %) = (20)/(0.397) ( 2 - 1.20)`
`t_(84 %) = 50.377 xx 0.795 = 40.09` min


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