Question

what fraction of reaction in first order reaction remains after 40 mins if t1/2 is 20min

Answer 1

for first order reaction
t1/2 = 0.693 / K
so
K = 0.693/20  1/sec

as we know first order reaction K x t = 2.303 log(a/a-x)

and (0.693/20) x 40 = 2.303 log(a/a-x)
log(a/a-x) = 0.6
and when we take antilog of a/a-x 
so a/a-x = 4
a = 4a-4x
x=(3/4)a
then remain is a-x = a- (3/4)a = a/4

regards