a first order reaction is 15% complete in 20 min.how long will it take to b 60% complete????
for first order reaction
t = 2.303 / K (Log a / a-x)
Let a = 100
initially reaction is 15% completed, so x = 15 , t = 20 mn.
so, 20 = 2.303 / K ( log 100 / 100-15 )
K = 2.303 / 20 ( log 100 / 85 )
K = 2.303 / 20 (log 1.176 )
K = 2.303 / 20 ( 0.706 )
K = 0.11515 (0.706 )
K = 0.00813
now reaction is 60% completed
so, t = 2.303 / K (log a / a-x )
t = 2.303 / 0.00813 ( log 100 / 100-60 ) ( putting value of K )
t = 283.27 ( log 100 / 40)
t = 283.27 ( log 2.25)
t = 283.27( 0.3979 )
t = 112.71 min.