Overall material balance

Entering: Ammonia $+$ EO $+$ water

Leaving: MEA is $99\mathrm{.}5\%(0\mathrm{.}5\%$ other?$)$

$N{H}_3+C_2{H}_{4}O+H_{2}O=C_2{H}_{7}NO+H_{2}O$

$so$

$N{H}_3+C_2{H}_{4}O=C_2{H}_{7}NO.$

Assume complete conversion of $C_2{H}_{4}O.$

Since $N{H}_3$ is $10C_2{H}_{4}O$

Some ammonia will leave the system $($unless a recycle stream is required$)$

$N{H}_3+C_2{H}_{4}O+H_{2}O=C_2{H}_{7}NO+H_{2}O+N{H}_3$

What to consider

$,50$ tons of water leave the system with MEA

But how much water is coming in$?$

If we recycle $NH3$ what is the actual feed of $NH3$

Energy balance

Heat of formation $(Hf)$

$H_{rxn}={\displaystyle \underset{\text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}\text{s}}{\overset{?}{}}}{v}_i{H}_{f,i}-{\displaystyle \underset{reactants}{\overset{?}{}}}{v}_i{H}_{f,i}$

$H_T=H_{T_0}^0+R{\displaystyle \underset{T_0}{\overset{T}{}}}\frac{Cp}{R}dT,\frac{p_p{p}^0}{R}=1+B+C{T}^2+D{T}^{-2}$

$H_{rxn}=n_p{H}_{\text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}\text{s}}^o+n_p{\displaystyle \underset{25}{\overset{T}{}}}{C}_{p}dT-n_r{H}_{reactants}^o$

Reaction equilibrium

$\frac{G_{T_2}^0}{T_2}=\frac{G_{T_1}^0}{T_1}+H_T^0(\frac{1}{T_2}-\frac{1}{T_1}),K(T)=$exp$(\frac{-G_{rxn}(T)}{RT})$

$x=\frac{(F_{A0}-F_A)}{F_{A0}}$