# Shift Reactor Calculations

November 12, 2017 | Author: ankitsamria | Category: Chemical Equilibrium, Mole (Unit), Chemical Reactor, Chemistry, Physical Chemistry

#### Description

ChE 201 Chemical Process Principles and Calculations Fall, 2012

Washington State University Voiland School of Chemical Engineering and Bioengineering Richard L. Zollars

Problem 4.47, Felder and Rousseau At low to moderate pressures, the equilibrium state of the water-gas shift reaction

Is approximately described by the relation ( )

[

( )]

where T is the reactor temperature, Ke the reaction equilibrium constant, and yi is the mole fraction of species i in the reactor contents at equilibrium. The feed to a batch shift reactor contains 20.0 mole% CO, 10.0% CO2, 40.0 % water, and the balance an inert gas. The reactor is maintained at T = 1123 K. (a) Assume a basis of 1 mol feed and draw and label a flowchart. Carry out a degree-of-freedom analysis of the reactor based on extents of reaction and use it to prove that you have enough information to calculate the composition of the reaction mixture at equilibrium. Do no calculations. (b) Calculate the total moles of gas in the reactor at equilibrium (if it takes you more than 5 seconds you’re missing the point) and then the equilibrium mole fraction of hydrogen in the product. (Suggestion: Begin by writing expressions for the moles of each species in the product gas in terms of extent of reaction, and then write expressions for the species mole factions.) (c) Suppose a gas sample is drawn from the reactor and analyzed shortly after startup and the mole fraction of hydrogen is significantly different from the calculated value. Assuming that no calculations mistakes or measurement errors have been made, what is the likely explanation for the discrepancy between the calculated and measured hydrogen yields? SOLUTION (a) The flowchart would look like the one shown below.

Reactor Initial n1,CO = 0.200 mol n1,H2O = 0.400 mol n1,CO2 = 0.100 mol n1,in = 0.300 mol

Final n2,CO n2,H2O n2,CO2 n2,H2 n2,in

There are five unknowns. There is also one independent reaction and four reactive species (CO, H2O, CO2, and H2) plus one non-reactive species. The equilibrium constant gives another equation relating the unknowns. Thus the number of degrees of freedom, using the procedure on p. 130) gives

Therefore, the problem is solvable with the given information. (b) Note that the number of moles of products is the same as the number of moles of reactants. Therefore, if you start with 1 mol you will have 1 mol regardless of the amount of reaction that takes place. Expressing the number of moles of each species in terms of the extent of reaction you get

Since the total number of moles = 1 the number of moles at equilibrium will also equal the mole fraction of that species. Substitute the values into the equitation for the equilibrium constant to get ( (

)( ) )(

( )

)

[

]

Solving this for ξ gives ( √(

) ) (

(

)(

)

)

The only reasonable value is ξ = 0.110 since the other value would give a negative number of moles for CO2 and H2. Thus the mole fraction of H2 is 0.110. (c) If there are no calculation errors or measurement errors the most reasonable explanation is that the mixture has not yet reached equilibrium.