Question

a first order reaction completes 60% in 20 minutes .the time period reqhired for completion if 90% of the reaction is approximate​

Answer 1

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

Answer 2

Answer:

this is your answer

Explanation:

k=t2.303loga−xa

The reaction is 60% complete in 20 minutes.

k=202.303log100−60100.....(1)

The reaction is 84% complete

k=t2.303log100−84100.....(2)

But (1) = (2),

202.303log100−60100=t2.303log100−84100

201log100−60100=t1log100−84100

t=log100−6010020log100−84100

t=0.397920×0.79588

t=40 minutes.

The reaction will take 40 minutes to be 84% complete