given:
a=100 ,x=60, time taken for 60percent =20min
time taken for 90 percent=?
we know that for first order ,
k=2.303/t(log([a]/[a-x])
k=2.303/20(log[100/40])
k=2.303/20((log5)-(log2)
we know that k is same throughout the reaction
t=2.303/k(log[100/100-99]
2.303/20(log5-log2)=2.303/t(log[100/10])
log5-log2)/20=log10/t
t=20/(0.6991-0.3010)
t=50.26