Consider a certain reaction A → Products with k `= 2.0 × 10^(–2)s^(–1)`. Calculate
the concentration of A remaining after 100 s if the initial concentration
of A is `1.0 mol L^(–1)`.

Answer

`0.135 mol e L^(-1)`

Explanation

The unit of `k(i.e.,s^(-1))` shows that the reaction is of first order. Hence, `k=(2.303)/(t)log``([A]_(i nitial))/([A]_(l eft))`
`=:. 2.0xx10^(-2)s^(-1)=(2.303)/(100s)log``(1.0mol L^(-1))/([A]_(l eft))`
`or log[A]_(remai ni ng or l eft)=-0.8684`
`:.[A]=Antilog(-0.8684)`
`=Antilog(--0.8684+1-1)`
`=Antilogbar(1).1316=0.1354mol L^(-1)`