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 above reaction is carried out in a batch reactor of constant liquid volume and temperature. Initially $($at $t=0),$ the concentrations of the three components are

$C_A(0)=5mo\frac{l}{L}$;$,C_B(0)=12mo\frac{l}{L}$;$,C_C(0)=3mo\frac{l}{L}$

$($a$)$ Calculate $C_A(t),C_B(t)$ and $C_C(t)$ by integrating the mole balance equations over time. Plot all three concentrations on the same graph, with time $t$ on the horizontal axis $($ranging from $0$ to $15s).$

To produce reasonably smooth curves, your time intervals within the first $2$ seconds should be quite small $($e$.$g$.0\mathrm{.}1s).$

$($b$)$ What are the concentrations $C_A,C_B$ and $C_C$ at time $t=1\mathrm{.}5s?$

$($c$)$ What is the value of $C_A$ at $t=15s($which is effectively $t)?$

$($d$)$ Confirm your answer to part $($c$)$ using an approach that does not involve integration of ODEs.