Question

20%of a first order reaction occur in 10 min. Calculate its t1/2

Answer 1

time=10min

a=100

a-x=100-20=80

so formula

k=2.303÷t log10(a÷a-x)

k =2.303/10 log10(100/80)

k= 2.303/10 log10*10-3log10*2

k=2.303/10(1-3(.3010)

k=2.303/10(1- •9630)

k=2•303/10(.0970)

k=2.303×0.097

k=0.223391

because

T1/2=0.6932/k

t/12=0.6932/0.223391

t/12=3.10 min