Documentation:CHBE Exam Wiki/Module 6 - Reactive Energy Balances

CHBE 241 Exam resources wiki
Chemical and Biological Engineering
Welcome to the CHBE Exam Resources Wiki! This wiki is intended to host past exams with fully worked-out hints and solutions
Past Exams
Final Exam 2016W
Midterm Exam 1 2016W
Midterm Exam 2 2016W
Problem Sets
Module 1 - Process Basics
Module 2 - Reactors
Module 3 - Separations 1
Module 4 - Separations 2
Module 5 - Non-reactive Energy Balances
Module 6 - Reactive Energy Balances

Question 1

Trichloroethylene is produced in a two-step reaction sequence as shown in the steps below.

Reaction 1: C₂H₄(g) 2Cl₂ (g) → C₂H₂Cl₄(l) + H₂(g) where, ${\hat {H}}_{r}^{\circ }=-385.76kJ/mol$
Reaction 2: C₂H₂Cl₄(l) → C₂HCl₃(l) + HCl(g)

The standard heat of formation of tricholoroethylene (liquid) is - 276 2 kJ/mol.

Question 1a

Calculate the standard heat of the second reaction.

Solution

We first need to determine the heat of formation of tetrachloroethane

${\begin{aligned}({\hat {H}}_{f}^{\circ })_{C_{2}H_{2}Cl_{4}(l)}&=({\hat {H}}_{r1}^{\circ })+({\hat {H}}_{f}^{\circ })_{C_{2}H_{4}(g)}\\&=-333.48kJ/mol\\({\hat {H}}_{r2}^{\circ })&=({\hat {H}}_{f}^{\circ })_{C_{2}HCl_{3}(l)}+({\hat {H}}_{f}^{\circ })_{HCl(g)}-({\hat {H}}_{f}^{\circ })_{C_{2}H_{2}Cl_{4}(l)}\\&=-35.03kJ/mol\\\end{aligned}}$

Question 1b

Use Hess’s law to calculate the standard heat of the reaction
C₂H₄(g)+ 2Cl₂ (g) → C₂HCl₃(l) + H₂(g) + HCl(g)

Solution

Reaction (1) + Reaction (2)
${\begin{aligned}({\hat {H}}_{r}^{\circ })&=({\hat {H}}_{r1}^{\circ })+({\hat {H}}_{r2}^{\circ })\\&=-385.76+(-35.03)\\&=-420.79kJ/mol\\\end{aligned}}$

Question 1c

If 450 mol/h of C₂HCl₃ (l) is produced in the reaction of part (b) and the reactants and products are all at 25 C and 1 atm, how much heat is evolved or absorbed in the process?

Solution

Assume that
${\begin{aligned}{\dot {Q}}&=\Delta {\dot {H}}\\&=450mol/h\cdot (-420.79kJ/mol)\\&=-1.89\cdot 10^{5}kJ/h\\&=-52.60kW({\text{heat is evolved to the surroundings}})\\\end{aligned}}$

Question 2

Ethylene oxide is produced by the oxidation of ethylene under the following reaction:
C₂H₄(g) + 1/2 O₂ (g) → C₂H₄O(g)

Another competing reaction results in oxidation of ethylene to CO.
In a plant, the feed to the reactor contains 2 mol C₂H₄/ mol O₂ while the yield and the conversion in the reactor are 0.8 mol C₂H₄O produced/ mol C₂H₄ consumed and 40% respectively. The outlet stream from the reactor then separates into the following streams:

• C₂H₄ and O₂ are recycled back to the reactor
• C₂H₄O is sold
• CO₂ and H₂O are disposed
  

Also, the inlet and outlet streams from the reactor are kept at 450°C, while the feed stream and the streams leaving the separator are at 25°C.

Question 2a

Using a basis of 4 moles of ethylene entering the reactor, draw a process flow diagram for the plant.

Solution

* Note that the recycle stream above should be at 25°C according to the question statement.

Question 2b

Determine the flow rates (molar) and compositions of all the streams.

Solution

With 40% conversion, 1.6 mol of C₂H₄ is consumed. Therefore,
$n_{3}=2.4mol$
With 80% yield,
${\begin{aligned}n_{5}&=(1.6molC_{2}H_{4}consumed)\cdot 0.8\\&=1.28mol\\\end{aligned}}$

C balance on reactor:
${\begin{aligned}2(4)&=2(2.4)+2(1.28)+\cdot n_{6}\\n_{6}&=0.64mol\\\end{aligned}}$

Since the molar ratio of water to CO₂ produced is 1:1,
$n_{7}=0.64mol$

O balance on the reactor:
${\begin{aligned}2(2)&=2(n_{4})+1.28+2(0.64)+0.64\\n_{4}&=0.4mol\\\end{aligned}}$

Overall C balance:
${\begin{aligned}2n_{1}&=0.64+2(1.28)\\n_{1}&=1.6mol\\\end{aligned}}$

Overall O balance:
${\begin{aligned}2n_{2}&=2(0.64)+0.64+1.28\\n_{2}&=1.6mol\\\end{aligned}}$

Compositions:
Feed stream: 50% C₂H₄, 50% O₂
Recycle stream: 85.7% C₂H₄,14.2% O₂
Reactor inlet: 66.7% C₂H₄, 33.3% O₂
Reactor outlet: 44.8% C₂H₄, 7.5% O₂, 23.9% C₂H₄O, 11.9% CO₂, 11.9% H₂O

Question 2c

Determine the heat required for the whole process and the reactor alone respectively. Use the following information for C₂H₄O,
$({\hat {H}}_{f}^{\circ })=-51kJ/mol$
$C_{p}[J/(mol\cdot K)]=-4.69+0.2061T-9.995\cdot 10^{-5}T^{2}$ where T is in kelvin

Solution

By performing energy balance on the process:
${\begin{aligned}Q_{process}&=\Delta H\\&=H_{out}-H_{in}\\&=-583.7kJ\\Q_{reactor}&=-558.7kJ\\\end{aligned}}$

Question 2d

Determine the flow rate (kg/h) and the energy consumption in kW for a production for 5000kg C₂H₄O/day.

Solution

Mass of C₂H₄O produced with initial basis,
${\begin{aligned}m&=1.28mol\cdot 440.5kg/10^{3}mol\\&=0.0564kg\\{\text{Scale factor}}&={\frac {5000kg/day}{0.0564kg}}\\&=88677/day\\\end{aligned}}$

The feed mass at the initial basis,
${\begin{aligned}m_{feed}'&=1.6mol\cdot (28.05gC_{2}H_{4}/mol)+1.6mol\cdot (32gO_{2}/mol)\\&=96.08g\\\end{aligned}}$

The fresh feed rate for the production basis of 5000kg/day,
${\begin{aligned}{\dot {m}}_{production}&=96.08\cdot 10^{-3}kg\cdot 88677/day\\&=8520kg/day\\\end{aligned}}$

By performing energy balance on the process:
${\begin{aligned}{\dot {Q}}_{process}&=-583.7kJ\cdot 88677/day\cdot {\frac {1day}{24hr}}\cdot {\frac {1hr}{3600s}}\cdot {\frac {1kW}{1kJ/s}}\\&=-573.4kW\\{\dot {Q}}_{reactor}&=-599.1kW\\\end{aligned}}$

Question 3

In a continuous reactor at 25°C, dilute sulphuric acid was used to absorb ammonia which results in the formation of ammonium sulfate,
2 NH₃(g) + H₂SO₄(aq) → (NH₃)₂SO₄ (aq)

Question 3a

If the product solution is at 25°C while the ammonia and sulfuric acid feeds are at 75°C and 25°C respectively, how much heat must be withdrawn from the unit per mol of (NH₃)₂SO₄ produced?

Solution

By setting the reference to be at 25°C and the basis to be 1 mol of (NH₃)₂SO₄ produced,

Ammonia(g,75°C)
${\begin{aligned}{\hat {H}}_{NH_{3}}&=\Delta ({\hat {H}}_{f}^{\circ })_{NH_{3}}+\int _{25}^{75}C_{p}dT\\&=(-46.19+1.83)kJ/mol\\&=-44.36kJ/mol\\\end{aligned}}$

H₂SO₄(aq,25°C)
${\begin{aligned}{\hat {H}}_{H_{2}SO_{4}}&=\Delta ({\hat {H}}_{f}^{\circ })_{H_{2}SO_{4}(aq)}\\&=-907.51kJ/mol\\\end{aligned}}$

(NH₄)₂SO₄(aq,25°C)
${\begin{aligned}{\hat {H}}_{(NH_{4})_{2}SO_{4}}&=\Delta ({\hat {H}}_{f}^{\circ })_{(NH_{4})_{2}SO_{4}(aq)}\\&=-1173.1kJ/mol\\\end{aligned}}$

Energy balance,
${\begin{aligned}Q_{process}&=\Delta H\\&=\sum \limits _{out}n_{i}{\hat {H}}_{i}-\sum \limits _{in}n_{i}{\hat {H}}_{i}\\&=1mol(-1173.1kJ/mol)-[2mol(-44.36kJ/mol)+1mol(-9.7.51kJ/mol)]\\&=-177kJ\\\end{aligned}}$

177 kJ should be removed from the unit per mol of (NH₃)₂SO₄ produced.

Question 3b

If the product stream contains 1 mole% ammonium sulfate, estimate the final temperature of the reactor.
Assume that there is no heat loss by the reactor and the heat capacity of the solution to be that of pure liquid water [4.184 kJ/(kg C)].

Solution

Determine the total mass of the solution using a basis of 100 moles.
${\begin{aligned}m_{(NH_{4})_{2}SO_{4}}&=1mol\cdot 132g/mol\\&=132g(NH_{4})_{2}SO_{4}\\m_{H_{2}O}&=99mol\cdot 18g/mol\\&=1782gH_{2}O\\m_{total}=1914{\text{g solution}}\\\end{aligned}}$

Since there is not heat lost to the surroundings, the heat transferred from the reactor is the heat absorbed by the solution, resulting in temperature change.
${\begin{aligned}177kJ&=1.914kg\cdot 4.184kJ/(kg^{\circ }C)\cdot (T_{final}-25)^{\circ }C\\T_{final}&=47.1^{\circ }C\\\end{aligned}}$

Question 4

A gas mixture containing n-hexane in nitrogen with a relative saturation of 80% is fed to a condenser where the feed stream is at 75°C and 2 atm (absolute). Two streams leave the condenser where one stream is in the gas phase while the other stream is in liquid phase. The product gas stream leaves the condenser at 0°C and 2 atm (absolute) at a flow rate of 300 m³/h.

Question 4a

Determine the percentage condensation of hexane (moles condensed/mole fed).

Solution

The molar flow rate of the gas stream, ${\begin{aligned}n_{gas}&=300m^{3}/h\cdot {\frac {2atm}{1atm}}\cdot {\frac {1kmol}{22.4m^{3}(STP)}}\\&=26.8kmol/hr\\\end{aligned}}$

Determine the sat. vapor pressure of n-hexane(h) in both the gas streams (inlet and outlet).
${\begin{aligned}p_{v}(0^{\circ }C)&=6.88555-{\frac {1175.817}{224.867+0}}\\&=45.35mmHg\\p_{v}(75^{\circ }C)&=6.88555-{\frac {1175.817}{224.867+75}}\\&=921.34mmHg\\\end{aligned}}$

Determine the vapor fraction of n-hexane(h) in both the gas streams (inlet and outlet).
${\begin{aligned}y_{gas,h}&={\frac {p_{v}(0^{\circ }C)}{P}}\\&={\frac {45.35mmHg}{2\cdot 760mmHg}}\\&=0.0298\\y_{feed,h}&={\frac {0.8\cdot p_{v}(75^{\circ }C)}{P}}\\&={\frac {0.8\cdot 921.34mmHg}{2\cdot 760mmHg}}\\&=0.485\\\end{aligned}}$

Determine the molar flow rates of the feed and liquid streams
N₂ balance:
${\begin{aligned}{\dot {n}}_{feed}\cdot (1-0.485)&=26.8kmol/hr\cdot (1-0.0298)\\{\dot {n}}_{feed}&=50.45kmol/hr\\\end{aligned}}$

n-hexane balance:
${\begin{aligned}50.45kmol/hr\cdot 0.485&=26.8kmol/hr\cdot (0.0298)+{\dot {n}}_{liquid}\\{\dot {n}}_{liquid}&=23.67kmol/hr\\\end{aligned}}$

${\begin{aligned}{\text{Percent condensation}}&={\frac {23.67kmol/hr}{50.45kmol/hr\cdot 0.485}}\cdot 100\%\\&=96.7\%\\\end{aligned}}$

Question 4b

Determine the rate of heat transfer (kW) from the condenser.

Solution

Determine the enthalpy by using the following relationships:

Nitrogen
${\hat {H}}={C_{p}}(T-25)$

n-hexane
${\hat {H}}=\int _{0}^{T_{vap}}C_{p,l}dT+\Delta H_{v}+\int _{T_{vap}}^{T}C_{p,v}dT$

The values obtained for the streams are as follows:

${\begin{aligned}{\dot {Q}}&=\Delta {\dot {H}}\\&=\sum \limits _{out}n_{i}{\hat {H}}_{i}-\sum \limits _{in}n_{i}{\hat {H}}_{i}\\&=-312.6kW\end{aligned}}$

Question 5

In an adiabatic dryer, a sheet of film which contains 4 wt% liquid acetone enters where 80% of the acetone evaporates into air. The film enters at 35°C (T_f1) and exits at T_f2 while air enters at T_a1 and leaves at 49°C with relative saturation of 50%. The air enters at 1.01 atm and leaves the dryer at 1 atm Also, use these following assumptions:

• C_p for dry film is 1.33 kJ/(kg °C)
• C_p for liquid acetone is 0.129 kJ/(mol °C)
• The heat of vaporization of acetone is independent of temperature

Question 5a

Calculate the feed ratio [liters dry air (STP)/kg dry film] by assuming a basis of 100kg of film being fed into the dryer.

Solution

Determine the sat. vapor pressure of acetone
${\begin{aligned}p_{v}(49^{\circ }C)&=7.11714-{\frac {1210.595}{229.664+0}}\\&=592.73mmHg\\\end{aligned}}$

Determine the molar flow rate of acetone in exit gas,
${\begin{aligned}n_{g,acetone}&=3.2kg\cdot {\frac {1kmol}{58.08kg}}\cdot {\frac {10^{3}mol}{1kmol}}\\&=55.1mol\\\end{aligned}}$

Determine the vapor fraction of n acetone in exit gas,
${\begin{aligned}y_{gas,acetone}&={\frac {0.5\cdot p_{v}(49^{\circ }C)}{P}}\\&={\frac {0.5\cdot 592.73mmHg}{760mmHg}}\\&=0.39\\\end{aligned}}$

Determine the molar flow rates of the feed air
${\begin{aligned}y_{gas,acetone}&={\frac {n_{g,acetone}}{n_{g,acetone}+n_{g,air}}}\\n_{g,air}&=86.2mol\\\end{aligned}}$

${\begin{aligned}{\text{Feed ratio}}&=86.2mol\cdot {\frac {22.4L(STP)}{mol}}\cdot {1}{96kgDF}\\&=20.1L(STP)/kgDF\\\end{aligned}}$

Question 5b

Derive an expression for T_a1 in terms of the film temperature change, (T_f2 - 35)

Solution

Reference temperatures: Air (25°C), Acetone,l (35°C), Dry film (35°C)

Determine the enthalpy by using the following relationships:

Dry film
${\hat {H}}={C_{p}}(T-35)$

Acetone
${\hat {H}}=\int _{35}^{T_{vap}}C_{p,l}dT+\Delta H_{v}+\int _{T_{vap}}^{49}C_{p,v}dT$

Air
${\hat {H}}={C_{p}}(T-25)$

Using energy balance,
${\begin{aligned}\Delta {\dot {H}}&=\sum \limits _{out}n_{i}{\hat {H}}_{i}-\sum \limits _{in}n_{i}{\hat {H}}_{i}\\0&=96\cdot 1.33(T_{f2}-35)+13.7\cdot 0.129(T_{f2}-35)+55.1\cdot 32.3+86.2\cdot 0.70-86.2\cdot \int _{25}^{T_{a1}}(C_{p})_{air}dT\\\int _{25}^{T_{a1}}(C_{p})_{air}dT&={\frac {129.5(T_{f2}-35)+1840.3}{86.2}}\\\end{aligned}}$

Question 5c

Calculate the film temperature change if the inlet air temperature is 120°C.

Solution

If T_a1 is 120°C,
${\begin{aligned}\int _{25}^{120}(C_{p})_{air}dT&=2.78kJ/mol\\&={\frac {129.5(T_{f2}-35)+1840.3}{86.2}}\\T_{f2}&=22.6^{\circ }C\\\end{aligned}}$