Question

A first order reaction has `k=1.5 xx 10^(-6) s^(-1)` at `200^(@)`C. If the reaction is allowed to run for 10 hours, what percentage of initial concentration would have changed in products?

Answer 1

For the first order reaction,
`t=2.303/k log a/(a-x)`
`loga/(a-x) = (kxxt)/(2.303)=((1.5 xx 10^(-6)s^(-1)) xx (10 xx 60 xx 60s))/(2.303)=0.0235`
`a/(a-x) ="Antilog" 0.0235=1.055`
or `100=105.5 -1.055 x or x=(5.5)/(1.055) = 5.2=5.2%`