Documentation:CHBE Exam Wiki/6.6 - Practice Problem 1

6.6 – Practice Problem 1

Nitric acid is a crucial chemical that is used in the production of ammonium nitrate which is used for fertilizers, making plastics, and manufacturing of dyes. To produce nitic acid, you need nitric oxide and to produce nitric oxide you must oxidize ammonia. The two reactions that occur during the oxidation of ammonia are:

$4{\text{ }}NH_{3}+5{\text{ }}O_{2}\longrightarrow 4{\text{ }}NO+6{\text{ }}H_{2}O$

$2{\text{ }}NH_{3}+{\frac {3}{2}}{\text{ }}O_{2}\longrightarrow N_{2}+3{\text{ }}H_{2}O$

You are a chemical engineer in charge of this step in the process. A flow chart shown bellow details the reactor you are working on.

• 

  Attribution: Said Zaid-Alkailani & UBC

You must:

a) Take the elemental species ${\big [}N_{2}(g),{\text{ }}H_{2}(g),{\text{ }}O_{2}(g){\big ]}$ at $25^{\circ }C$ as references and prepare and fill an inlet-outlet enthalpy table.

b) Calculate the heat transfer to or from the reactor in MW.

Answer

a) Take the elemental species ${\big [}N_{2}(g),{\text{ }}H_{2}(g),{\text{ }}O_{2}(g){\big ]}$ at $25^{\circ }C$ as references and prepare and fill an inlet-outlet enthalpy table.

First, we must prepare the enthalpy table:

• 

  Attribution: Said Zaid-Alkailani & UBC

Using heat capacity and enthalpy tables found online such as on the [NIST](https://www.nist.gov/) website, we can find the enthalpies of the inputs and outputs.

${\hat {H}}_{i}=\Delta {\hat {H}}_{fi}^{o}+\int _{25}^{T}{C_{pi}dt}$

$NH_{3}(g,{\text{ }}25^{\circ }C)$

${\hat {H}}_{1}=\Delta {\hat {H}}_{f{\text{ }}NH_{3}}^{o}+\int _{25}^{T}{C_{p{\text{ }}NH_{3}}dt}=-46.19{\text{ }}{\frac {kJ}{mol}}+\int _{25}^{25}{C_{p{\text{ }}NH_{3}}dt}={\big (}-46.19+0{\big )}{\text{ }}{\frac {kJ}{mol}}$

${\hat {H}}_{1}=-46.19{\text{ }}{\frac {kJ}{mol}}$

${\text{Air}}(g,{\text{ }}25^{\circ }C)$

${\hat {H}}_{2}=\Delta {\hat {H}}_{f{\text{ }}{\text{Air}}}^{o}+\int _{25}^{T}{C_{p{\text{ }}{\text{Air}}}dt}=3.67{\text{ }}{\frac {kJ}{mol}}+\int _{25}^{25}{C_{p{\text{ }}{\text{Air}}}dt}={\big (}3.67+0{\big )}{\text{ }}{\frac {kJ}{mol}}$

${\hat {H}}_{2}=3.67{\text{ }}{\frac {kJ}{mol}}$

$NO(g,{\text{ }}700^{\circ }C)$

${\hat {H}}_{3}=\Delta {\hat {H}}_{f{\text{ }}NO}^{o}+\int _{25}^{T}{C_{p{\text{ }}NO}dt}$

${\hat {H}}_{3}=90.37{\text{ }}{\frac {kJ}{mol}}+\int _{25}^{700}{{\bigg [}{\big (}2.950\times 10^{-2}{\big )}+{\big (}8.188\times 10^{-6}{\big )}T-{\big (}2.925\times 10^{-9}{\big )}T^{2}+{\big (}3.652\times 10^{-13}{\big )}T^{3}dt{\bigg ]}}{\text{ }}{\frac {kJ}{mol}}$

${\hat {H}}_{3}={\big (}90.37+21.60{\big )}{\text{ }}{\frac {kJ}{mol}}=111.97{\text{ }}{\frac {kJ}{mol}}$

$H_{2}O(g,{\text{ }}700^{\circ }C)$

${\hat {H}}_{4}=\Delta {\hat {H}}_{f{\text{ }}H_{2}O}^{o}+\int _{25}^{T}{C_{p{\text{ }}H_{2}O}dt}={\big (}-241.83+23.92{\big )}{\text{ }}{\frac {kJ}{mol}}$

${\hat {H}}_{4}=-216.91{\text{ }}{\frac {kJ}{mol}}$

$N_{2}(g,{\text{ }}700^{\circ }C)$

${\hat {H}}_{5}=\Delta {\hat {H}}_{f{\text{ }}N_{2}}^{o}+\int _{25}^{T}{C_{p{\text{ }}N_{2}}dt}={\big (}0+20.59{\big )}{\text{ }}{\frac {kJ}{mol}}$===

${\hat {H}}_{5}=20.59{\text{ }}{\frac {kJ}{mol}}$

$O_{2}(g,{\text{ }}700^{\circ }C)$

${\hat {H}}_{6}=\Delta {\hat {H}}_{f{\text{ }}O_{2}}^{o}+\int _{25}^{T}{C_{p{\text{ }}O_{2}}dt}={\big (}0+21.86{\big )}{\text{ }}{\frac {kJ}{mol}}$

${\hat {H}}_{6}=21.86{\text{ }}{\frac {kJ}{mol}}$

Now we can complete the table.

• 

  Attribution: Said Zaid-Alkailani & UBC

b) Calculate the heat transfer to or from the reactor in MW.

The heat transferred to or from the reactor will just be the change in enthalpy of the inlets vs. the outlets. Units are excluded from calculations.

${\dot {Q}}=\Delta {\dot {H}}=\sum _{out}{\dot {n}}_{i}{\hat {H}}i-\sum _{in}{\dot {n}}_{i}{\hat {H}}i$

${\dot {Q}}={\bigg [}{\big (}90\cdot 111.97{\big )}+{\big (}150\cdot -216.91{\big )}+{\big (}716\cdot 20.59{\big )}+{\big (}69\cdot 21.86{\big )}{\bigg ]}-{\bigg [}{\big (}100\cdot -46.19{\big )}+{\big (}900\cdot 3.67{\big )}{\bigg ]}$

${\dot {Q}}=-4890{\text{ }}{\frac {kJ}{min}}$

Now we must convert to the proper units.

${\dot {Q}}=-4890{\text{ }}{\frac {kJ}{min}}\times {\frac {1{\text{ }}min}{60{\text{ }}s}}\times {\frac {1{\text{ }}MJ}{1000{\text{ }}kJ}}=-0.0815{\text{ }}MW$