
(a)
Interpretation:
Whether there is any advantage in processing the reactions at pressure above
Concept introduction:
For a multiple reaction system, the final moles for each of the components present in the products can be estimated by the equation:
Here,
Mole fraction
Here,
Equilibrium constant of this reaction from equation 14.28 can be written as
Where,
According to the Le’ Chatelier’s principle, if a reaction is subject to any change at its equilibrium, then the reaction tends to shift its equilibrium in the direction, so as to undo the effect of that change on its equilibrium.
(a)

Answer to Problem 14.34P
There is no advantage in processing the reactions at pressure above
Explanation of Solution
Given information:
The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii ):
The equilibrium conditions given for these reactions are
For the given reaction to produce synthesis gas, let the extent of the reaction ( i ) be
The overall stoichiometric coefficients for both the reactions are
The individual stoichiometric coefficients for all the components in both the reactions are
The initial and final pressure of the system is taken as
Now, use equation (3) for the equilibrium constant of both the reactions and simplify the expressions as
The expression for the equilibrium constant for reaction ( i ) is dependent on pressure of the system. An increase in the pressure increases the equilibrium constant of the reaction. Thus, according to the Le’ Chatelier’s principle, to undo the effect of the increase in the products, the reaction shifts to the left toward the reactants. Thus, there is more of reactants present at the equilibrium providing no advantage of carrying the reactions at pressure above
(b)
Interpretation:
Whether there is any advantage in processing the reactions at temperatures below
Concept introduction:
According to the Le’ Chatelier’s principle, if a reaction is subject to any change at its equilibrium then the reaction tends to shift its equilibrium in the direction so as to undo the effect of that change on its equilibrium.
(b)

Answer to Problem 14.34P
The decrease in the equilibrium temperature of the reaction ( i ) does not provide any advantage as the equilibrium shifts towards the reactants with decreasing temperature for an endothermic reaction.
Explanation of Solution
Given information:
The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )
The equilibrium conditions given for these reactions are
From Table C.4 the standard heat of reaction and Gibb’s free energy for both the reactions are
Since, the reaction ( i ) is endothermic, the heat required during the reaction acts as one of the reactants of the reaction.
Decreasing the equilibrium temperature of the reaction below
(c)
Interpretation:
In the gas synthesis, the ratio of number of moles of hydrogen to carbon monoxide needs to be determined.
Concept introduction:
For a multiple reaction system, the final moles for each of the components present in the products can be estimated by the equation:
Here,
Mole fraction
Here,
Equilibrium constant of this reaction from equation 14.28 can be written as:
Where,
Gibb’s free energy in terms of equilibrium constant is written as:
Also, Gibbs free energy is calculated using heat of reaction from the equation given as
Here,
Where,
(c)

Answer to Problem 14.34P
The molar ratio of hydrogen to carbon monoxide in the synthesis gas
Explanation of Solution
Given information:
The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )
The equilibrium conditions given for these reactions are
Equimolar mixture of steam and methane is fed to the reactor.
From Table C.4 the standard heat of reaction and Gibb’s free energy for reaction ( i ) is
From Table C.1 the coefficients for the heat capacity of the component gases in reaction ( i ) are given as
Substance | | | | |
| | | | |
| | | | |
| | | | |
| | | | |
Now, use equations set (6) to evaluate the values of
Now, use equation (5) along with the above calculated values to get the value of
Use this calculated value of
From Table C.4 the standard heat of reaction and Gibb’s free energy for reaction ( ii ) is
From Table C.1 the coefficients for the heat capacity of the component gases in reaction ( ii ) are given as
Substance | | | | |
| | | | |
| | | | |
| | | | |
| | | | |
Now, use equations set (6) to evaluate the values of
Now, use equation (5) along with the above calculated values to get the value of
Use this calculated value of
The value of
Since, there is stoichiometric amount of
Accordingt o the stoichiometry of the reaction ( i ),
Therefore, the ratio of moles of
(d)
Interpretation:
In the gas synthesis, the molar ratio of hydrogen to carbon monoxide needs to be determined for the feed ratio of steam-to-methane equals to
Concept introduction:
For a multiple reaction system, the final moles for each of the components present in the products can be estimated by the equation,
Here,
Mole fraction
Here,
Equilibrium constant of this reaction from equation 14.28 can be written as
Where,
Gibb’s free energy in terms of equilibrium constant is written as
Also, Gibbs free energy is calculated using heat of reaction from the equation given as
Here,
Where,
(d)

Answer to Problem 14.34P
The molar ratio of hydrogen to carbon monoxide in the synthesis gas
Explanation of Solution
Given information:
The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )
The equilibrium conditions given for these reactions are
Molar ratio of steam and methane fed to the reactor is
From part (c), the value of
Moles of
For reaction ( ii ), initially
Let the extent of the reaction be
The overall stoichiometric coefficient for this reaction is,
The individual stoichiometric coefficients for all the components in this reaction is
Using equation (1), write the expressions for final moles of all the components present in the products as gases in reaction ( ii )
Total moles of the products will be
Using equation (2) to write the mole fraction of all the species as
Now, use equation (3) and the calculated value of the equilibrium constant and calculate
Now, the ratio of
(e)
Interpretation:
The way to alter the feed composition to yield lower
Concept introduction:
According to the Le’ Chatelier’s principle, if a reaction is subject to any change at its equilibrium, then the reaction tends to shift its equilibrium in the direction so as to undo the effect of that change on its equilibrium.
(e)

Answer to Problem 14.34P
Addition of
Explanation of Solution
Given information:
The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )
The equilibrium conditions given for these reactions are
By adding
This leads to the increase in the amount of
(f)
Interpretation:
Whether there is any danger in the formation of solid carbon by the given side reaction at equilibrium conditions or not needs to be determined for conditions given in part (c) and part (d). Also, a way to alter the feed conditions to prevent this deposition needs to be explained.
Concept introduction:
For a single reaction system, the final moles of each of the components present, can be estimated by the equation:
Here,
Mole fraction
Here,
Equilibrium constant of this reaction from equation 14.28 can be written as
Where,
Gibb’s free energy in terms of equilibrium constant is written as
Also, Gibbs free energy is calculated using heat of reaction from the equation given as
Here,
Where,
(f)

Answer to Problem 14.34P
In part (c), there is danger of formation of solid carbon by the given side reaction at equilibrium conditions.
Addition of
In part (d), there is no danger of formation of solid carbon by the given side reaction at equilibrium conditions.
Explanation of Solution
Given information:
The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )
The equilibrium conditions given for these reactions are
At this equilibrium condition, the side reaction taking place to form carbon is
From Table C.4, the standard heat of reaction and Gibb’s free energy for the formation of carbon is
From Table C.1, the coefficients for the heat capacity of the component gases in carbon formation reaction are given as
Substance | | | | |
| | | | |
| | | | |
| | | | |
Now, use equations set (6) to evaluate the values of
Now, use equation (5) along with the above calculated values to get the value of
Use this calculated value of
This is the equilibrium constant for this reaction and if the actual value of this constant is greater than the equilibrium value, then the reaction tries to shift to the left and reduce the formation of carbon.
Calculate the actual value of this constant as
For the given reaction to produce carbon, let the extent of the reaction be
The overall stoichiometric coefficient for this reaction is (for gases only),
The individual stoichiometric coefficients for all the gaseous components in this reaction is:
Use equation (3) which is applicable only for gaseous species, such that,
From part (c), the actual value of the mole fraction of
Calculate the ratio of the actual constant as:
As the actual value of
Addition of
From part (d), the actual value of the mole fraction of
Calculate the ratio of the actual constant as:
As the actual value of
Want to see more full solutions like this?
Chapter 14 Solutions
Introduction to Chemical Engineering Thermodynamics
- You are part of a team constructing a pipeline to transfer shale gas produced at the oceanfloor to the coastline. The temperature of the pipeline is nearly constant at 2 oC. The pipelineis made of smooth stainless steel and is 0.3 m in diameter and 100 m long. The averagevelocity of shale gas is 10 m/s and the inlet temperature is 20 oC ** Useful shale gas properties at 20 oC (Table A-12 for propane gas):(use these values for calculations and validate them later)• Density (ρ) = 18.13 kg/m3• Cp = 1974 J/kg-K• Viscosity (μ) = 8.54*10-6 kg/m-s• Pr = 0.918• k = 0.01836 W/m-Ka) Is the flow laminar or turbulent? Is the flow hydrodynamically and thermally fully developed?(circle your answer below and provide justification. • Laminar vs. Turbulent• Hydrodynamically developing vs. developed• Thermally developing vs. fully developedJustification: b) Calculate convective heat transfer coefficient (h). c) Calculate the exit temperature of the shale gas. d) Are the shale gas properties…arrow_forward3) A pilot-plant Podbielniak centrifugal extractor operating at 11,400 x g (this is G₁) is capable of processing 500 mL/min of filtered fermentation broth and 125 mL/min organic solvent, giving a recovery of 95%. The rotating cylinder inside the extractor has a diameter of 20 cm and is 2.5 cm wide. You need to scale up this extraction by using a larger Podbielniak extractor that has a diameter of 91 cm and width of 91 cm and delivers 2,300 x g (G2). What flow rates (in L/min) should be used in the larger extractor to achieve the same recovery efficiency?arrow_forward7) You are tasked with separating two proteins by ion exchange chromatography on a 30 cm long column with an inner diameter of 2 cm. The resin has a diameter of 100 μm and a void fraction of 0.3, and your mobile phase flows through the column at a rate of Q = 5 cm³/min. The Van Deemter coefficients A, B, and C have been determined to be 0.0228 cm, 0.0036 cm²/min, and 0.00053 min, respectively, for both proteins. Protein A elutes from the column with an average retention time of 27 min and standard deviation of 0.8 min. Protein B elutes from the column. with an average retention time of 33.8 min and standard deviation of 1.0. a) How many theoretical plates does the column contain? b) What flow rate (Q) will give you the maximum resolution? c) What is the minimum height of a theoretical plate for the system?arrow_forward
- 1 5) You are asked to design a moving bed adsorption process using two columns (see the figure below). Your feed contains 100 mg/L protein and flows through both columns at 4 m³/h. Fresh resin enters the bottom of both columns (resin does not flow from the bottom column to the top column). The maximum resin flow rate that your pumps can comfortably handle is 120 kg resin/h. Experimental data suggest that the adsorption equilibrium can be modeled as qi=4ci where qi is in g protein/kg resin and c; is in g protein/L broth. (Pay attention with units!) a) What is the lowest concentration of proteins that you could get in the effluent from column 1 (indicated by the *) in mg/L? (Hint: set up a mass balance) 0.25 , * 1 2 b) What should be the flow rate of resin (in kg/h) into the second column (B2) if your overall process needs to remove 99% of the protein?arrow_forward6) Instead of moving bed adsorption, you decide to try fixed bed adsorption with a different resin for removal of your protein. Your column is 25 cm long with an inner diameter of 5 cm. The resin packed in the column has a density of 1.5 g/cm³ and a void fraction of 0.25. Equilibrium data suggests that the protein binding to the column follows a Langmuir isotherm with an Stot = 6.25 g protein/kg resin and Keq = 2.58 L broth/g protein. The feed contains 100 mg/L protein and flows through the column at 500 mL/h. The calculated binding capacity of the column under these conditions is 945 mg protein. a) After 17.7 h, you detect an unacceptable level of protein in the column effluent. What is the length of unused bed? b) After deciding that this process will work well for separation, you need to scale up to a 1 m long column with the same diameter. If all else but length of the column is held constant, how long will you be able to run the column before breakthrough?arrow_forward2) A bioprocess stream contains two components: 60 g/L A (the desired bioproduct) and 40 g/L B (a contaminant). The partition coefficients in an extraction process are KA = 14.0 and KB = 0.5. If the stream is fed into a 5-stage countercurrent extractor at 25 L/min and solvent enters at 5 L/min, calculate the amount of each component in the extract phase at the outlet of the extractor.arrow_forward
- What flow rate (Q, in L/min) will give the maximum resolution if you scale up the process in problem 7 to a column with a diameter of 50 cm? (Use the conditions in the problem statement, not the optimal flow conditions calculated in the problem).arrow_forwardYou are asked to separate two proteins via size exclusion chromatography. Your column is 5 cm in diameter and 25 cm in length. You add 50 g (dry weight) of Sephacryl gel (density = 1.25 g/cm³) to the column. The volume outside of the gel and pores is 77 cm³. The column operates at a flow rate of 1 mL/min to not compress the resin. The average retention times of proteins A and B are 244.5 and 273 min, respectively. Under the given conditions, you calculate that there are 1600 theoretical plates in the column. (This problem continues onto the next page). a) What is the volume inside of the pores? b) What are the elution limits of this system (i.e., what re the minimum and maximum volumes that particles will elute in)? c) What are the gel partition coefficients of proteins A and B for this system? d) What are the peak widths for both proteins? e) Assuming the peak widths are 22 mL and 28 mL for proteins A and B, respectively (these values are not necessarily correct), what is the…arrow_forward1) Indicate whether each statement below is True or False: a) In extractions, small proteins usually end up in the heavy phase. b) In extractions, large proteins usually end up in the heavy phase. c) Tie line length on a phase diagram has no effect on protein partitioning. d) For extraction, we assume that each stage reaches equilibrium. e) Isotherms that are concave down lead to self-sharpening solute fronts in adsorption. f) When scaling up an adsorption unit, the LUB scales proportionally with adsorber length. g) For chromatography, separation efficiency increases when HETP increases. h) In size exclusion chromatography, smaller molecules elute from the column first. i) In hydrophobic interaction chromatography, very hydrophobic proteins elute in water. j) Molecules at very high concentrations elute from the column as a perfect Gaussian peak. k) In chromatography scale-up, resolution will increase as column diameter increases. 1) A good adsorbent typically has no porosity. m)…arrow_forward
- Chemical Engineering Use the psychrometric chart and demonstrate the linear interpolation method to obtain -0.52 KJ/KgDA. This is the enthalpy deviation. The exercise is uploaded below.arrow_forwardChemical Engineering Use the psychrometric chart. The remaining curves on the psychrometric chart are almost vertical and convex to the left, with labeled values (on Figure 8.4-1) of 0.05, 0.1, 0.2, and so on. (The units of these numbers arekJ/kg DA). Thesecurves are usedto determine theenthalpyof humid air that is not saturated. The procedure is as follows: (a) locate the point on the chart corresponding to air at its specified condition; (b) interpolate to estimate the enthalpy deviation at this point; (c) follow the constant wet-bulb temperature line to the enthalpy scale above the saturation curve, read the value on that scale, and add the enthalpy deviation to it. Also, you will see the exercise on the piece of paper.arrow_forwardCalculate the permeability of the bed of ion-exchange particles in Example 11.1.arrow_forward
- Introduction to Chemical Engineering Thermodynami...Chemical EngineeringISBN:9781259696527Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark SwihartPublisher:McGraw-Hill EducationElementary Principles of Chemical Processes, Bind...Chemical EngineeringISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEYElements of Chemical Reaction Engineering (5th Ed...Chemical EngineeringISBN:9780133887518Author:H. Scott FoglerPublisher:Prentice Hall
- Industrial Plastics: Theory and ApplicationsChemical EngineeringISBN:9781285061238Author:Lokensgard, ErikPublisher:Delmar Cengage LearningUnit Operations of Chemical EngineeringChemical EngineeringISBN:9780072848236Author:Warren McCabe, Julian C. Smith, Peter HarriottPublisher:McGraw-Hill Companies, The





