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.

Volumes and Areas from Group Contributions (3:04)

Group contributions are used widely in property prediction. The volumes and surface areas have been determined by x-ray data and high-temperature collision data. The UNIQUAC and UNIFAC activity coefficient methods use these quantities to calculation volume fractions and surface area fractions. The assignment of functional groups for a molecule must be done carefully to assure agreement with the groups used by the model developers.

Comprehension Questions:

1. Estimate R and Q for 1,4 dihydroxy benzene.

2. Estimate R and Q for n-propyl alcohol and compare them to the values for IPA.

3. Estimate R and Q for methyl-npropyl ketone.

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 P_O2?
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.

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.