Question

38. A first order reaction completes 60% in
20 minutes. The time required for the completion of
90% of the reaction is approx
(1) 30 minutes (2) 40 minutes
(3) 50 minutes
(4) 60 minutes

Answer 1

answer : option (3) 50 min

explanation : according to first order reaction,

$t=\frac{2.303}{k}log\left(\frac{a}{a-x}\right)$

case 1 : x = 60% of a = 60a/100 = 0.6a

and t = 20min

so, 20min = 2.303/k log[a/(a - 0.6a)]

or, 20= 2.303/k log[a/0.4a]

or, k = 2.303/20 log[2.5 ] .....(1)

case2 : x = 90% of a = 0.9a

then, t = 2.303/k log [a/(a - 0.9a)]

from equation (1),

t = 2.303/[2.303/20 log(2.5)] log[10]

= 20/log(2.5)

= 50.258 min ≈ 50 min

so, option (3) is correct choice.

Answer 2

Answer:

50min

Explanation:

according to first order reaction,

case 1 : x = 60% of a = 60a/100 = 0.6a

and t = 20min

so, 20min = 2.303/k log[a/(a - 0.6a)]

or, 20= 2.303/k log[a/0.4a]

or, k = 2.303/20 log[2.5 ] .....(1)

case2 : x = 90% of a = 0.9a

then, t = 2.303/k log [a/(a - 0.9a)]

from equation (1),

t = 2.303/[2.303/20 log(2.5)] log[10]

= 20/log(2.5)

= 50.258 min ≈ 50 min

option 3