Top-rated ScreenCasts

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09.05 - Fugacity and Fugacity Coefficient Click here. 100 1

What is fugacity? (10min) (learncheme.com) Defines fugacity in terms of Gibbs Energy and describes the need for defining this new property as a generalization of how pressure affects ideal gases.
Comprehension Questions
1. The phases in this video start with concentrations 0.0007kg/L and 1.0 kg/L, when not at equilibrium. What are the equilibrium concentrations?
2. Why is concentration an unreliable indicator for the direction of mass transfer?
3. Name two indicators for the direction of mass transfer that are superior to concentration.  

09.04 - Changes in Gibbs Energy with Pressure Click here. 100 1

Gibbs Energy - Nuts to Soup. (learncheme.com, 8min) It is straightforward to start from the definition of Gibbs Energy and derive all the changes in Gibbs energy. These can be graphed for H2O to see how familiar quantities from the steam tables relate to changes in this unfamiliar property.

08.02 - The Internal Energy Departure Function Click here. 100 1

Departure Function Derivation Principles (8:03) (msu.edu)
This screencast covers sections 8.2 - 8.8. Concepts of using the equation of state to evaluate departure functions. The screencasts also discusses the choice of density integrals or pressure integrals. The use of a reference state is discussed.

10.07 - Nonideal Systems Click here. 100 1

This screencast shows how to quickly visualize Pxy phase diagrams for nonideal systems using Excel (5min, uakron.edu). These sample calculations for methanol+benzene apply the simplest nonideal solution model: ΔHmix = A12*x1*x2. Rigors of this model are discussed in Chapter 11. Nevertheless, its basic elements are simple enough that they can be understood in Chapter 10. When x1=0 or x2=0, a pure fluid is indicated, corresponding to no mixing and zero heat of mixing. When A12=0, the ideal solution approximation is recovered. When A12>0, the model indicates an endothermic interaction (like 2-propanol+water, Fig. 10.8c), giving rise to "positive deviations from Raoult's Law." When A12<0, the model indicates an exothermic interaction (like acetone+chloroform, Fig. 10.9c), giving rise to "negative deviations from Raoult's Law." With this spreadsheet, you can quickly change your components and A12 values to see how the phase diagram changes and gain "hands-on" familiarity with the principles discussed in Section 10.7. 

Note: This is a companion file in a series. You may wish to choose your own order for viewing them. For example, you should implement the first three videos before implementing this one. Also, you might like to see how to quickly visualize the Txy analog of the Pxy phase diagram. If you see a phase diagram like the ones in section 11.8, you might want to learn about LLE phase diagrams. The links on the software tutorial present a summary of the techniques to be implemented throughout Unit3 in a quick access format that is more compact than what is presented elsewhere. Some students may find it helpful to refer to this compact list when they find themselves "not being able to find the forest because of all the trees."

Comprehension Questions:
1. Make a Pxy diagram for cyclohexane+toluene at 80C and A12=200. What kind of system is this?
2. Make a Pxy diagram for cyclohexane+benzene at 80C and A12=200. What kind of system is this?
3. Why does the system's qualitative behavior change so much when the components and model parameters are changed so little?

17.06 Determining the Spontaneity of Reactions Click here. 100 1

Which way will a reaction go? (3:40) (msu.edu)

When both reactants and products are present in a reactng mixture, the direction the reaction will proceed is not necessarily indicated by the sign of ΔGo or Ka. Rather, it is determined by ΔG. This screencasts provides guidance for understanding this concept.

Comprehension Questions: (Hint: review Example 17.1 before answering.)

1. CO and H2 are fed in a 2:1 ratio to a reactor at 500K and 20 bars with a catalyst that favors only CH3OH as its product. When the conversion of H2 is 32%, will the reaction go forwards towards product or back to reactants?
2. CO and H2 are fed in a 2:1 ratio to a reactor at 500K and 20 bars with a catalyst that favors only CH3OH as its product. When the conversion of CO is 52%, will the reaction go forwards towards product or back to reactants?
3. CO and H2 are fed in a 1:1 ratio to a reactor at 500K and 20 bars with a catalyst that favors only CH3OH as its product. When the conversion of H2 is 42%, will the reaction go forwards towards product or back to reactants?
4. CO and H2 are fed in a 1:1 ratio to a reactor at 500K and 20 bars with a catalyst that favors only CH3OH as its product. When the conversion of H2 is 42%, will the reaction go forwards towards product or back to reactants?

17.05 - Effect of Pressure, Inerts, Feed Ratios Click here. 100 1

Partial pressures and reactor sizing are among the keys to chemical engineering calculations (uakron.edu, 7 min, review from Section 1.6). Partial pressures (uakron.edu, 7 min) also play an essential role in reaction equilibrium calculations. Partial pressure calculations basically involve straightforward mass balances, but specific vocabulary and a need for systematic precision can cause difficulty. The calculations involve six elements that must be carefully computed:

(1) Stoichiometry - the reaction equation must be stoichiometrically balanced such that the number of atoms of each element are the same on both sides of the equation. This balance is achieved by adjusting the stoichiometric coefficients. The change in the number of moles of each component must be in correct stoichiometric proportions relative to the "key component." Inert compounds (see below) are NOT included in the stoichiometric equation. For the example in this presentation, the objective of the reactor is to oxidize carbon monoxide (CO) in a catalytic converter by reacting it with oxygen (O2). So, CO + 0.5 O2 = CO2.
(2) Limiting reactant (aka. "key component") - It is common to feed an excess of one of the components in order to promote complete conversion of the other components. The limiting reactant is the component that is NOT in excess. For this example, O2 is fed in excess so that CO conversion can be promoted. CO becomes the limiting reactant in that case and conversion must be computed relative to CO, NOT O2. If you think about it, expressing the conversion with respect to the excess component would mean that 100% conversion could result in a negative mole number for the limiting reactant. Such an implication is obviously physically impossible (and potentially embarrassing if you appear not to know that).
(3) %Excess - The number of moles of an excess component in the feed is (1+Xs) times the stoichiometric amount relative to the key component, where the stoichiometric amount is the number of moles necessary to perfectly balance the key component, and Xs is the fractional form of the %excess. For this example,  the stoichiometric ratio of CO:O2 would be 1:0.5 and for 50% excess, Xs = 0.50, and the actual ratio would be 1:0.75.
(4) %Conversion - the %conversion is the fraction of the entering amount of the limiting reactant that is transformed into product(s). Note that this might be different from the "extent of reaction," ξ. For example, if 50 moles/h of CO enter the reactor and the conversion is 90%, then 5 moles of CO exit the reactor. If you express the number of moles of CO as 50-ξ, you might conclude that the moles of CO exiting the reactor is 49.1. Take a minute to think about what the words mean before you start to calculate, then make a mental estimate of what the results should be, then get out your calculator. Another common mistake is to apply the % conversion to all the components, wrongly including the excess component. For example, if 45 moles of CO react, then 22.5 moles of O2 react. With 50% excess O2 in the feed, the O2 exiting should be 37.5-22.5=15, NOT 3.75. This is what it means to be careful and systematic. You must compute the conversion of limiting reactant first, then compute the conversion of other components relative to the limiting reactant.
(5) Inerts - These are components that may enter the reactor by coincidence or convenience but do not participate in the reaction. Therefore, their number of moles exiting the reactor is simply equal to their number of moles entering the reactor. A common mistake is to apply the %conversion to all components entering the reactor, including the inerts. In this example, the source of O2 is air, with roughly 4:1 ratio of nitrogen (N2) to O2. The N2 is inert.
(6) Total Pressure - Once the mole numbers and mole fractions have been computed, don't forget to multiply the mole fractions by the total pressure to get the partial pressure. The partial pressure is equal to the mole fraction only in the case that the reactor operates at 1.00 bar.

Comprehension Questions:

1. What is the value of the total pressure (bar) applied in the presentation of this example?
2. What equation is used to compute the mole number of O2? What is the final overall equation used to compute PO2?
3. Suppose 100 moles/h of ammonia (NH3) at 100bars is to be produced from N2 and hydrogen (H2) with 10% excess N2. Methane (CH4) is included with the N2+H2 as a result of the synthesis process with a ratio of 1:10 CH4:H2. (a) Write a stoichiometrically balanced equation (b) Identify the limiting reactant (c) Calculate the number of moles and partial pressures of each component entering the reactor. (d) Calculate the number of moles and partial pressures of each component exiting the reactor assuming 25% conversion.

17.07 - Temperature Dependence of Ka Click here. 100 2

You can customize Kcalc.xlsx (uakron.edu, 17min) to facilitate whatever calculations you may need to perform. This presentation shows how to implement VLOOKUP to automatically load the relevant Hf, Gf, and Cp values. It also shows how to automatically use the Cp/R value when a,b,c,d values for Cp are not available. Finally, it shows how a fairly general table of inlet flows, temperatures, and pressures can be used to set up the equilibrium conversion calculation. The initial set up is demonstrated for the dimethyl ether process, then revised to initiate solution of Example 17.9 for ammonia synthesis.

Comprehension Questions:

1. The video shows how the shortcut Van't Hof equation can be written as lnKa=A+B/T. What are the values of A and B for the dimethyl ether process when a reference temperature of 633K is used?
2. The video shows how the shortcut Van't Hof equation can be written as lnKa=A+B/T. What are the values of A and B for the ammonia synthesis process when a reference temperature of 600K is used?

11.02 - Calculations with Activity Coefficients Click here. 100 3

Bubble Temperature (2:43) (msu.edu)

The culmination of the activity coefficient method is application of the fitted activity coefficients to extrapolate from limited experiments in a Stage III calculation. The bubble temperature is the easiest after bubble pressure. The recommended order of study is 1) Bubble Pressure; 2) Bubble Temperature; 3) Dew Pressure; 4) Dew Temperature. Note that an entire Txy diagram can be generated with bubble temperature calculations; no dew calculations are required.

10.08 - Concepts for Generalized Phase Equilibria Click here. 100 1

Concepts for General Phase Equilibria (12:33) (msu.edu)

The calculus used in Chapter 6 needs to be generalized to add composition dependence. Also, we introduce partial molar properties and composition derivatives that are not partial molar properties. We introduce chemical potential These concepts are used to show that the chemical potentials and component fugacities are used as criteria for phase equilibria.

10.07 - Nonideal Systems Click here. 100 1

Nonideal Mixtures (4:58) (msu.edu)

Raoult's law is an easy way to calculate VLE, but it is inaccurate for most detailed VLE calculations. This screencast provides an overview of the problems, and introduces the concept of an azeotrope. The VLE K-ratio is shown to be less than one or greater than one dependenting on the overall system concentration relative to the azeotrope composition where K=1. The concept of positive and negative deviations is introduced.

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