The reversible liquid phase reaction

$A+2B3C$

has forward and reverse kinetics given by

forward: $-(r_A{)}_1=k_1{C}_A^{0\mathrm{.}8}{C}_B^2$;$,k_1=0\mathrm{.}003(\frac{L}{\text{m}\text{o}\text{l}}{)}^{1\mathrm{.}8}\frac{1}{(s)}$

reverse: $-(r_C{)}_2=k_2{C}_C^{2\mathrm{.}6}$;$,k_2=0\mathrm{.}006(\frac{L}{\text{m}\text{o}\text{l}}{)}^{1\mathrm{.}6}\frac{1}{(s)}$

The same reactions now take place in a CSTR with constant liquid volume $V=20L($also under isothermal condition$).$ Initially, a stream of inert liquid runs through the reactor at a volumetric flow rate $Q_0=2\frac{L}{s}.$

At time $t=0,$ molecules of A and B are introduced into the feed stream. The concentrations of A and $B$ in the feed stream are given by

$C_{A0}(t)=\{\begin{array}{c}0\text{;}t<mn\; id="EditorElement\_2208170">0</mn>\end{array}$

$5[1-$exp$(-\frac{t}{8})]$;$,t>0$

$C_{B0}(t)=\{\begin{array}{c}0\text{;}t<mn\; id="EditorElement\_2208256">0</mn>\end{array}$

$5$;$t>0$

where $t$ is in s and the concentrations are in mol$/$L$.$ The volumetric flow rate through the reactor is maintained at $2\frac{L}{s}$ at all times.

Determine the concentrations of all three components in the effluent stream at $t=20s.$