Solution

The equilibrium constant is as follows:

$2{\mathrm{NO}}_2\leftrightarrow {\mathrm{N}}_2{\mathrm{O}}_4{K}_{\mathrm{p}}=\frac{P_{{\mathrm{N}}_2{\mathrm{O}}_4}}{P_{{\mathrm{NO}}_2}^2}$

where:

${\log }_{10}\left(\frac{1}{K_{\mathrm{p}}}\right)=-\frac{2692}{T}+1.75{\log }_{10}T+0.00483\cdot T-7.144\cdot {10}^{-6}{T}^2+3.062$

We use as a base for our computations 1 kmol of N₂O₄.

The mass balance to the equilibrium is presented below. d represents the dissociation degree for the dimer.

	$2{\mathrm{NO}}_2\leftrightarrow {\mathrm{N}}_2{\mathrm{O}}_4$		Inters
Initial	0	1	a
Equilibrium	2d	1-d	a

The molar fraction of each species and the partial pressure are computed as follows:

$\begin{array}{l}{y}_{{\mathrm{N}}_2{\mathrm{O}}_4}^{\mathrm{eq}}=\frac{1-d}{1-d+2d+a}=\frac{1-d}{1+d+a}\Rightarrow P_{{\mathrm{N}}_2{\mathrm{O}}_4}^{\mathrm{eq}}=\frac{1-d}{1+d+a}{P}_{\mathrm{T}}\hfill \\ y_{{\mathrm{NO}}_2}^{\mathrm{eq}}=\frac{2d}{1-d+2d+a}=\frac{2d}{1+d+a}\Rightarrow P_{{\mathrm{NO}}_2}^{\mathrm{eq}}=\frac{2d}{1+d+a}{P}_{\mathrm{T}}\hfill \end{array}$

Substituting both into the equilibrium constant equation it becomes:

$\begin{array}{l}{K}_{\mathrm{p}}=\frac{P_{{\mathrm{N}}_2{\mathrm{O}}_4}}{P_{{\mathrm{NO}}_2}^2}=\frac{\frac{1-d}{1+d+a}{P}_{\mathrm{T}}}{{\left[{\displaystyle \frac{2d}{1+d+a}}{P}_{\mathrm{T}}\right]}^2}=\frac{(1-d)(1+d+a)}{4\cdot d^2\cdot P_{\mathrm{T}}}\hfill \\ d=\frac{-a+\sqrt{a^2+4·(1+a)(1+4·P_{\mathrm{T}}·K_{\mathrm{p}})}}{2·(1+4·P_{\mathrm{T}}·K_{\mathrm{p}})}\hfill \end{array}$

From the definition of equivalent NO₂:

$\begin{array}{l}\%{\mathrm{NO}}_2\mathrm{Eq}=\frac{2}{a+2}100\Rightarrow a=\frac{2\cdot (100-\%{\mathrm{NO}}_2\mathrm{Eq})}{\%{\mathrm{NO}}_2\mathrm{Eq}}\hfill \\ {\mathrm{NO}}_2\mathrm{Eq}={\mathrm{NO}}_2+2{\mathrm{N}}_2{\mathrm{O}}_4=2d+2(1-d)=2\hfill \end{array}$

For a total pressure of 1 atm, and different equivalent NO₂ compositions and temperatures, we compute d. In Fig. 6E5.1 we see that dissociation increases with temperature and with the amount of equivalent NO₂. In Fig. 6E5.2 we present the effect of pressure. N₂O₄ dissociation decreases with pressure, as expected based on Le Chatelier’s principle.

Solution

Assuming an isothermal plug flow reactor between trays we have:

$F_{\mathrm{NO}}\frac{dX}{dV}=-r_{\mathrm{NO}}$

$\begin{array}{l}\mathrm{NO}+\frac{1}{2}{\mathrm{O}}_2+{\mathrm{N}}_2\leftrightarrow {\mathrm{NO}}_2+{\mathrm{N}}_2\hfill \\ F_{A_{\mathrm{o}}}{\mathrm{F}}_{B_{\mathrm{o}}}{\mathrm{F}}_{C_{\mathrm{o}}}\hfill \\ F_{A_{\mathrm{o}}}(1-X)F_{B_{\mathrm{o}}}-\frac{1}{2}{F}_{A_{\mathrm{o}}}\mathrm{X}{F}_{C_{\mathrm{o}}}{F}_{A_{\mathrm{o}}}X\hfill \\ {\mathit{\Theta }}_i=\frac{F_{i_{\mathrm{o}}}}{F_{A_{\mathrm{o}}}}\hfill \\ F_{\mathrm{T}}=F_{A_{\mathrm{o}}}\left[(1-X)+{\mathit{\Theta }}_{{\mathrm{O}}_2}-\frac{1}{2}X+{\mathit{\Theta }}_{{\mathrm{N}}_2}+{\mathit{\Theta }}_{{\mathrm{NO}}_2}+X\right]=F_{A_{\mathrm{o}}}(1+{\mathit{\Theta }}_{{\mathrm{O}}_2}+{\mathit{\Theta }}_{{\mathrm{N}}_2}+{\mathit{\Theta }}_{{\mathrm{NO}}_2})+F_{A_{\mathrm{o}}}\delta X\hfill \\ \delta =-\frac{1}{2}\hfill \\ F_{\mathrm{T}}=F_{T_{\mathrm{o}}}+F_{A_{\mathrm{o}}}\delta X\hfill \end{array}$

Assuming ideal gases:

$\begin{array}{l}{C}_{\mathrm{T}}=\frac{F_{\mathrm{T}}}{v}=\frac{P}{RT}\hfill \\ C_{T_{\mathrm{o}}}=\frac{F_{T_{\mathrm{o}}}}{v_{\mathrm{o}}}=\frac{P_{\mathrm{o}}}{R{T}_{\mathrm{o}}}\hfill \end{array}$

Dividing both:

$v=v_{\mathrm{o}}\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\frac{F_{\mathrm{T}}}{F_{T_{\mathrm{o}}}}=v_{\mathrm{o}}\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\frac{F_{T_{\mathrm{o}}}+F_{A_{\mathrm{o}}}\delta X}{F_{T_{\mathrm{o}}}}=v_{\mathrm{o}}\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\left(1+\frac{F_{A_{\mathrm{o}}}}{F_{T_{\mathrm{o}}}}\delta X\right)$

Then, the concentration of the species i is as follows:

$\begin{array}{l}{C}_i=\frac{F_i}{v}=\frac{F_i}{v_{\mathrm{o}}\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\frac{F_{\mathrm{T}}}{F_{T_{\mathrm{o}}}}}=C_{T_{\mathrm{o}}}\frac{F_i}{F_{\mathrm{T}}}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}\hfill \\ C_i=C_{T_{\mathrm{o}}}\frac{F_{\mathrm{i}}}{F_{T_{\mathrm{o}}}+F_{A_{\mathrm{o}}}\delta X}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}=C_{T_{\mathrm{o}}}\frac{F_{A_{\mathrm{o}}}({\mathit{\Theta }}_i+v_{i}X)}{F_{T_{\mathrm{o}}}+F_{A_{\mathrm{o}}}\delta X}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}\hfill \end{array}$

$\begin{array}{l}{C}_i=C_{T_{\mathrm{o}}}\frac{F_{A_{\mathrm{o}}}}{F_{T_{\mathrm{o}}}}\frac{({\mathit{\Theta }}_i+v_{i}X)}{1+(F_{A_{\mathrm{o}}}/F_{T_{\mathrm{o}}})\delta X}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}=C_{A_{\mathrm{o}}}\frac{({\mathit{\Theta }}_i+v_{i}X)}{1+\epsilon X}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}\hfill \\ \epsilon =y_{A_{\mathrm{o}}}\delta =(F_{A_{\mathrm{o}}}/F_{T_{\mathrm{o}}})\delta \hfill \end{array}$

Thus the partial pressure of component i is calculated as follows:

$\begin{array}{l}{P}_i=C_{i}RT=C_{A_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{P_{\mathrm{o}}}\frac{({\mathit{\Theta }}_i+v_{i}X)}{1+\epsilon X}R\frac{T}{T}\cdot P=P_{A_{\mathrm{o}}}\frac{({\mathit{\Theta }}_i+v_{i}X)}{1+\epsilon X}\frac{P}{P_{\mathrm{o}}}\hfill \\ P_{A_{\mathrm{o}}}=C_{A_{\mathrm{o}}}R{T}_o\hfill \end{array}$

$\begin{array}{ll}-r_{\mathrm{NO}}\hfill & =-\frac{d{P}_{\mathrm{NO}}}{dV}=0.2166\exp (1399/T)\frac{P_{\mathrm{NO}}^2}{P_{{\mathrm{O}}_2}}\\ F_{\mathrm{NO}}\frac{dX}{dV}\hfill & =0.2166\exp (1399/T)\frac{{\left(P_{\mathrm{NO}}{\displaystyle \frac{(1-X)}{1+(-(F_{\mathrm{NO}}/F_{T_{\mathrm{o}}})0.5)X}}\right)}^2}{P_{\mathrm{NO}}\frac{(P_{{\mathrm{O}}_2}/P_{\mathrm{NO}}-0.5X)}{1+(-(F_{\mathrm{NO}}/F_{T_{\mathrm{o}}})0.5)X}}\\ & =0.2166\exp (1399/T)\frac{P_{\mathrm{N}\mathrm{O}}{(1-X)}^2}{(P_{{\mathrm{O}}_2}/P_{\mathrm{NO}}-0.5X)·(1+(-(F_{\mathrm{NO}}/F_{T_{\mathrm{o}}})0.5)X)}\hfill \end{array}$

The MATLAB code for solving the problem is shown below. For details on how to use MATLAB for reactor modeling we refer the reader to Martín (2014).

[a,b]=ode45(‘TorreOxidaNO’,[0 0.3],[0]);

plot(a,b)

xlabel(‘Height (Z)’)

ylabel(‘x’)

function Reactor = TorreOxidaNO(t,x)

%UNTITLED Summary of this function goes here

%  Detailed explanation goes here

Temp=30+273.15;

Ptotal=15;

k1=0.2166*exp(1399/Temp);

PO2ini=0.06*Ptotal;

PNOini=0.1*Ptotal;

FNO=1; %kmol/s

A=1;%m2

Reactor(1,1)=A*(k1/FNO)*PNOini*(1−x)^2/( (PO2ini/PNOini−0.5*x)*(1-0.5*x*(PNOini/Ptotal)));

End

At 15 bar we get more than 90% conversion after 10 cm (Fig. 6E6.1).

Solution

$\begin{array}{l}1{\mathrm{kmol}}_{\mathrm{air}}=n_{{\mathrm{O}}_2}+3.76n_{{\mathrm{O}}_2}+4.76n_{{\mathrm{O}}_2}\left(\frac{P_{\mathrm{v}}}{P_{\mathrm{T}}-P_{\mathrm{v}}}\right)\hfill \\ \frac{n_{\mathrm{air}}}{12}=n_{{\mathrm{O}}_2,\mathrm{added}}\hfill \\ \frac{n_{{\mathrm{O}}_2,\mathrm{total}}}{2}=n_{{\mathrm{NH}}_3}\hfill \end{array}$

To evaporate the ammonia, the energy provided to the heat exchanger is as follows:

$Q(\mathrm{vaporization})=\mathrm{\lambda }{m}_{{\mathrm{NH}}_3}$

The mixture enters the reactor at the same pressure and temperature, 1.5 bar and 25°C. The summary of the mass and energy balances is presented in Fig. 6E7.2.

The reaction taking place in the converter is as follows:

${\mathrm{NH}}_3+\frac{5}{4}{\mathrm{O}}_2\to \mathrm{NO}+\frac{3}{2}{\mathrm{H}}_2\mathrm{O}$

As presented in the description of the process, ammonia is expensive, and so is the platinum. Therefore, a conversion of 100% of ammonia is assumed. Based on the stoichiometry, we compute the flows of NO, H₂O, oxygen, and nitrogen:

$\begin{array}{l}{n_{{\mathrm{NH}}_3}|}_{\mathrm{out}}={n_{{\mathrm{NH}}_3}|}_{\mathrm{in}}(1-X)\hfill \\ {n_{\mathrm{NO}}|}_{\mathrm{out}}={n_{{\mathrm{NH}}_3}|}_{\mathrm{in}}(X)\hfill \\ {n_{{\mathrm{O}}_2}|}_{\mathrm{out}}={n_{{\mathrm{O}}_2}|}_{\mathrm{in}}-\frac{5}{4}{n_{{\mathrm{NH}}_3}|}_{\mathrm{in}}(X)\hfill \\ {n_{{\mathrm{H}}_2\mathrm{O}}|}_{\mathrm{out}}={n_{{\mathrm{H}}_2\mathrm{O}}|}_{\mathrm{in}}+\frac{3}{2}{n_{{\mathrm{NH}}_3}|}_{\mathrm{in}}(X)\hfill \\ {n_{{\mathrm{N}}_2}|}_{\mathrm{out}}={n_{{\mathrm{N}}_2}|}_{\mathrm{in}}\hfill \end{array}$

The temperature of the gases is computed by an adiabatic energy balance to the reactor where:

$\begin{array}{l}{Q}_{\mathrm{in}(\mathrm{flow})}+Q_{\mathrm{reaction}}=Q_{\mathrm{out}(\mathrm{flow})}\hfill \\ \sum m_i\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{in}}}{\int }}{c}_{p,i}dT+Q_{\mathrm{reaction}}=\sum m_i\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p,i}dT\hfill \end{array}$

The summary of the results can be seen in Fig. 6E7.3.

The gases are compressed up to 10 atm using a two-stage compression system with intercooling. The pressure ratio at each compressor is (P₂/P₁)^0.5=2.58. Assuming polytropic behavior of the compressors, the temperature and energy required are computed as follows:

$\begin{array}{l}{T}_2=T_1{\left({\displaystyle \frac{P_2}{P_1}}\right)}^m\hfill \\ m=\frac{1}{2}\frac{1.4-1}{1.4}\hfill \\ W=2T_{1}R\frac{1.4}{1.4-1}\left[{\left({\displaystyle \frac{P_2}{P_1}}\right)}^m-1\right]\hfill \end{array}$

Furthermore, the cooling requirements in between the compression stages so that each compressor receives the gas at the same initial temperature are as follows:

$Q=-2\sum _i{m}_i\underset{T_i}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p}dt$

During the cooling to 30°C, we assume that the conversion of NO to NO₂ is 100% since it is favored at high pressure and low temperature. In this condensation, we assume that nitric acid 3.5% by weight is produced with the water that condenses. Furthermore, there is an equilibrium between the NO₂ and the N₂O₄. The composition of the gas exiting the cooler is computed in three stages:

1. Determine the gas composition after the oxidation:

$\mathrm{NO}+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{NO}}_2$

$\begin{array}{l}{n_{\mathrm{NO}}|}_{\mathrm{out}}={n_{\mathrm{NO}}|}_{\mathrm{in}}(1-X)\hfill \\ {n_{{\mathrm{NO}}_2}|}_{\mathrm{out}}={n_{\mathrm{NO}}|}_{\mathrm{in}}(X)\hfill \\ {n_{{\mathrm{O}}_2}|}_{\mathrm{out}}={n_{{\mathrm{O}}_2}|}_{\mathrm{in}}-\frac{1}{2}{n_{\mathrm{NO}}|}_{\mathrm{in}}(X)\hfill \\ {n_{{\mathrm{H}}_2\mathrm{O}}|}_{\mathrm{out}}={n_{{\mathrm{H}}_2\mathrm{O}}|}_{\mathrm{in}}\hfill \\ {n_{{\mathrm{N}}_2}|}_{\mathrm{out}}={n_{{\mathrm{N}}_2}|}_{\mathrm{in}}\hfill \end{array}$

2. Compute the amount of HNO₃ produced. Although the moment we have NO₂ the equilibrium to N₂O₄ can be established (as long as nitric acid is produced), the N₂O₄ will decompose to provide for the NO₂ that has been consumed. Thus, we first compute the acid production. The stoichiometry of the reaction is as follows:

$2\mathrm{NO}{}_2+{\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2\to 2{\mathrm{HNO}}_3\mathrm{\Delta }{H}_{\mathrm{r}}=-127.28\mathrm{kJ}$

During the cooling, there is also water condensation. This water is the one that is forming the weak nitric acid solution. Thus, the water remaining in the gas is the humidity that can be handled at 30°C. The rest condenses. Therefore, the acid produced is computed using the stoichiometry of the reaction as follows:

$\begin{array}{c}\hfill {\%}_{\mathrm{acid}}=\frac{n_{{\mathrm{HNO}}_3}\cdot M_{{\mathrm{HNO}}_3}}{n_{{\mathrm{HNO}}_3}·M_{{\mathrm{HNO}}_3}+M_{{\mathrm{H}}_2\mathrm{O}}\left(n_{{\mathrm{H}}_2\mathrm{O}}-y_{\mathrm{sat}}·\mathrm{GAS}-\frac{1}{2}{n}_{{\mathrm{HNO}}_3}\right)}\hfill \\ \mathrm{GAS}=n_{{\mathrm{NO}}_2,\mathrm{fin}}+n_{{\mathrm{N}}_2,\mathrm{fin}}+n_{{\mathrm{O}}_2,\mathrm{fin}}\hfill \end{array}$

where

$\begin{array}{l}{n}_{{\mathrm{NO}}_2,\mathrm{fin}}=n_{{\mathrm{NO}}_{2,\mathrm{ini}}}-n_{{\mathrm{HNO}}_3}\hfill \\ n_{{\mathrm{O}}_2,\mathrm{fin}}=n_{{\mathrm{O}}_2}-\frac{1}{4}{n}_{{\mathrm{HNO}}_3}\hfill \\ n_{{\mathrm{N}}_2,\mathrm{fin}}=n_{{\mathrm{N}}_2,\mathrm{ini}}\hfill \end{array}$

Thus, the amount of water that accompanies the gas is determined as follows:

$m_{{\mathrm{H}}_2\mathrm{O}}=\left(\frac{100-{\%}_{\mathrm{acid}}}{{\%}_{\mathrm{acid}}}\right)m_{{\mathrm{HNO}}_3}$

The energy involved in the water phase change is included in the global energy balance as:

$Q_{\mathrm{cond}}=-m_{{\mathrm{H}}_2\mathrm{O}}\lambda$

3. Finally, with the gas phase computed in the previous stage, we determine the equilibrium concentration as the final gas phase to be sent to the absorption tower,

$2{\mathrm{NO}}_2\leftrightarrow {\mathrm{N}}_2{\mathrm{O}}_4{K}_{\mathrm{p}}=\frac{P_{{\mathrm{N}}_2{\mathrm{O}}_4}}{P_{{\mathrm{NO}}_2}^2}$

where:

${\log }_{10}\left(\frac{1}{K_{\mathrm{p}}}\right)=-\frac{2692}{T}+1.75{\log }_{10}T+0.00483·T-7.144·{10}^{-6}{T}^2+3.062$

For the sake of argument, allow y to be the moles of N₂O₄ and x the moles of NO₂ at equilibrium. Thus the equilibrium constant is given as follows:

$\begin{array}{l}{K}_{\mathrm{p}}=\frac{\frac{y}{n_{\mathrm{T}}}}{P_{\mathrm{T}}{\left(\frac{x}{n_{\mathrm{T}}}\right)}^2}=\frac{y·n_{\mathrm{T}}}{P_{\mathrm{T}}{x}^2}\hfill \\ 2y+x={\mathrm{NO}}_{2,\mathrm{eq}}=cte\hfill \\ K_{\mathrm{p}}\left(P_{\mathrm{T},\mathrm{atm}}\right)x^2=\left(\frac{{\mathrm{NO}}_{2,\mathrm{eq}}-x}{2}\right)\left(n_{{\mathrm{H}}_2}+n_{{\mathrm{O}}_2}+n_{{\mathrm{N}}_2}+n_{{\mathrm{H}}_{2}O}+x+\left(\frac{{\mathrm{NO}}_{2,\mathrm{eq}}-x}{2}\right)\right)\hfill \\ n_{{\mathrm{H}}_2}+n_{{\mathrm{O}}_2}+n_{{\mathrm{N}}_2}+n_{{\mathrm{H}}_{2}O}=N_{\mathrm{ini}}\hfill \\ x=\frac{-0.5N_{\mathrm{ini}}+\sqrt{{(0.5N_{\mathrm{ini}})}^2-4K_{\mathrm{p}}(P_{T,\text{atm}}+0.25)(0.25{\mathrm{NO}}_{2,\mathrm{eq}}^2+0.5{\mathrm{NO}}_{2,\mathrm{eq}}{N}_{\mathrm{ini}})}}{2K_{\mathrm{p}}(P_{T,\text{atm}}+0.25)}\hfill \end{array}$

The energy involved is computed as (Houghen et al., 1959):

$\begin{array}{l}{Q}_{\mathrm{ref}}=\sum _i{m}_i\left(\mathrm{\Delta }{H}_{\mathrm{f}}^{25}+\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p,i}dT\right)-\sum _i{m}_i\left(\mathrm{\Delta }{H}_{\mathrm{f}}^{25}+\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{in}}}{\int }}{c}_{p,i}dT\right)\hfill \end{array}$

where

$\begin{array}{l}\mathrm{\Delta }{H}_{{\mathrm{f},\text{HNO}}_3}^{25}={\mathrm{\Delta }{H}_{\mathrm{f}}^{25}|}_{\mathrm{pure}}+\mathrm{\Delta }{H}_{\mathrm{solution}}\hfill \\ \sum \mathrm{\Delta }{H}_{\mathrm{f},\mathrm{products}}(25°\mathrm{C})=-15,941.257\mathrm{kcal}\end{array}$

For the products, remember that nitric acid is produced and diluted and that energy is also involved in that stage. Water leaves partially as liquid and partially by saturating the gas, thus we have both formation energies for the different aggregation stages:

$\begin{array}{l}\sum \mathrm{\Delta }{H}_{\mathrm{f},\mathrm{reactants}}(25°\mathrm{C})=-9914.745\mathrm{kcal}\hfill \\ {\mathrm{Q}}_{\mathrm{in}}\mathrm{flow}=11,891\mathrm{kcal}\hfill \\ {\mathrm{Q}}_{\mathrm{out}}\mathrm{flow}=52.5\mathrm{kcal}.\hfill \end{array}$

The summary of the compression and cooling stages is presented in Fig. 6E7.4.

The exit gases are fed to the absorption column. The tower operates at 30°C and 10 atm. The liquid-to-gas mass ratio of operation is 0.25 and the nitric acid produced has a concentration of 60% by weight. To compute the mass of liquid added, the gas phase fed to the tower is used as follows (the liquid fed to the column is nitric acid 2.5%):

$\begin{array}{l}\mathrm{Balance}{\mathrm{HNO}}_3\hfill \\ n_{{\mathrm{HNO}}_3,\mathrm{total}}·M_{{\mathrm{HNO}}_3}=0.60·{\mathrm{Liq}}_{\mathrm{out}}=(n_{{\mathrm{HNO}}_3,\mathrm{ini}}+n_{{\mathrm{HNO}}_3,\mathrm{prod}})M_{{\mathrm{HNO}}_3}\hfill \\ n_{{\mathrm{HNO}}_3,\mathrm{prod}}=\frac{0.60·\mathrm{Liq}}{M_{{\mathrm{HNO}}_3}}-n_{{\mathrm{HNO}}_3,\mathrm{ini}}\hfill \end{array}$

$\begin{array}{l}\mathrm{Balance}\mathrm{water}\hfill \\ {(n_{{\mathrm{H}}_2\mathrm{O},\text{vap}}+n_{{\mathrm{H}}_2\mathrm{O},\mathrm{liq}})}_{\mathrm{ini}}=0.40·{\mathrm{Liq}}_{\mathrm{out}}+n_{{\mathrm{H}}_2\mathrm{O},\mathrm{vap}}+0.5·n_{{\mathrm{HNO}}_3,\mathrm{prod}}\hfill \\ n_{{\mathrm{H}}_2\mathrm{O},\mathrm{vap}}=(n_{{\mathrm{N}}_2}+n_{{\mathrm{O}}_2}+n_{{\mathrm{NO}}_2,\mathrm{eq}})\frac{P_{\mathrm{v}}}{P_{\mathrm{T}}-P_{\mathrm{v}}}\hfill \\ n_{{\mathrm{NO}}_2,\mathrm{out}}=n_{{\mathrm{NO}}_2,\mathrm{ini}\mathrm{t}}-n_{{\mathrm{HNO}}_3,\mathrm{prod}}\hfill \\ n_{{\mathrm{O}}_2,\mathrm{out}}=n_{{\mathrm{O}}_2,\mathrm{ini}}-\frac{1}{4}{n}_{{\mathrm{HNO}}_3,\mathrm{prod}}\hfill \\ n_{{\mathrm{N}}_2,\mathrm{out}}=n_{{\mathrm{N}}_2,\mathrm{ini}}\hfill \\ {\mathrm{Liq}}_{\mathrm{ini}}=0.25{\mathrm{V}}_{\mathrm{ini}}\hfill \end{array}$

The gas phase is determined from the coupled balances. Again, the equilibrium between the NO₂ and the N₂O₄ is to be recomputed after the production of nitric acid at the pressure and temperature of operation. The reaction is also highly exothermic, and therefore requires cooling.

$2\mathrm{NO}{}_2+{\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2\to 2{\mathrm{HNO}}_3$

An energy balance to the tower is performed as follows (Sinnot, 1999):

$\begin{array}{l}{Q}_{\mathrm{ref}}=\sum _i{m}_i\left(\mathrm{\Delta }{H}_{\mathrm{f}}^{25}+\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p,i}dT\right)-\sum _i{m}_i\left(\mathrm{\Delta }{H}_{\mathrm{f}}^{25}+\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{in}}}{\int }}{c}_{p,i}dT\right)\hfill \\ \mathrm{\Delta }{H}_{\mathrm{f},{\mathrm{HNO}}_3}^{25}={\mathrm{\Delta }{H}_{\mathrm{f}}^{25}|}_{\mathrm{pure}}+\mathrm{\Delta }{H}_{\mathrm{solution}}\hfill \end{array}$

When computing the energy balance, beware of the different aggregation stages of water, fact that the solution of the concentrated acid has a specific heat capacity while the liquid in flow is mostly water and dilution energies as a function of the concentration (Houghen et al., 1959, Sinnot, 1999):

$\begin{array}{l}\sum \mathrm{\Delta }{H}_{\mathrm{f},\mathrm{products}}(25°\mathrm{C})=-30097\mathrm{kcal}\hfill \\ \sum \mathrm{\Delta }{H}_{\mathrm{f},\mathrm{reactants}}(25°\mathrm{C})=-28513\mathrm{kcal}\hfill \\ Q_{\mathrm{in}},\mathrm{flow}=65\mathrm{kcal}\hfill \\ Q_{\mathrm{out}},\mathrm{flow}=29\mathrm{kcal}.\hfill \end{array}$

The summary of the mass and energy balances is given in Fig. 6E7.5.