May 15, 2017

Solve the mass balance

#### Explanation:

For the first order reaction of $A \setminus \rightarrow P r o \mathrm{du} c t s$

$\frac{{\mathrm{dC}}_{A}}{\mathrm{dt}} = - k {C}_{A}$
$t = - \frac{1}{k} \left[\ln \left({C}_{A}\right)\right] {|}_{{C}_{A} 0}^{{C}_{A}}$
$t = - \frac{1}{k} \ln \left(\frac{{C}_{A}}{{C}_{A 0}}\right)$

The given half time is when there is 50% of initial concentration present. Hence we have

${C}_{A} = 0.5 {C}_{A 0}$

$3 = - \frac{1}{k} \ln \left(0.5\right)$
$k = \ln \frac{2}{3}$
$k = 0.231 h {r}^{- 1}$

When 70 % of A is consumed the remaining concentration is

${C}_{A} = \left(1 - 0.7\right) {C}_{A 0}$
${C}_{A} = 0.3 {C}_{A 0}$
$t = - \frac{1}{0.231} \ln \left(0.3\right)$
$t = 5.212$ hours.