# CHM ENG 142 Study Guide - Midterm Guide: Volumetric Flow Rate, Reaction Step, Limiting Reagent

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School
Berkeley
Department
Chemical and Biomolecular Engineering
Course
CHM ENG 142
Professor Chemical Engineering 142
First Midterm Exam (Spring 2020)
February 26, 2020
Problem 1 (35 points)
An inlet gas stream (v0 = 10 liter/min inlet volumetric flow rate and total concentration CT0 = 10
mol/liter) composed of A and B each at a mole fraction of 0.2 with the rest consisting of an inert
I enters the system shown below. A splitter sends 6 liter/min of the inlet stream to the PFR and 4
liter/min to the CSTR. Both reactors are isothermal and isobaric.
1.5 A + B
C + D
r = k CB k = 3 min-1
a. Is this reaction an elementary step? State at least two reasons for your conclusion.
No, this is not an elementary step.
1. Fractional molecules cannot react in a single irreducible step.
2. The reaction stoichiometry is not reflected in the rate law.
b. What is the limiting reactant? What is the inlet molar rate of B entering the CSTR?
The limiting reactant is A. Although the inlet contains equal amounts of A and B, more
moles of A are needed for each reaction event, so A will be depleted first.
FB0 = yB0(FT0) = yB0(v0CT0) = 0.2 (4 L/min * 10 mol/L) = 8 mol/min B entering the CSTR
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Unlock all 6 pages and 3 million more documents. c. Starting from a general mole balance, derive the steady-state design equation for a CSTR.
Calculate the volume required (VCSTR) to convert 40% of the molar rate of B entering the
CSTR (XB,CSTR = 0.4)
General mole balance: FB0 FB + rB dV = dNB/dt
For a CSTR: FB0 FB + rBV = 0 assuming steady state (no accumulation) and perfect
mixing (concentrations, and therefore rate, are the same throughout the reactor).
Species
Initial
Change
Final
A
FA0
- 1.5 XBFB0
FB0(1 1.5 XB)
B
FB0
- XBFB0
FB0(1 XB)
C
0
+ XBFB0
XBFB0
D
0
+ XBFB0
XBFB0
I
FI0
0
FI0
FT = FB0 (2 + FI0/FB0 - 0.5 XB)
FT0 = FB0 (2 + FI0/FB0)
Therefore v/v0 = FT/FT0 = 1 - XB [0.5 / (2 + 0.6/0.2)] = 1 - 0.1 XB
Plug into equation: FB0 (XB) = k FB0/v0 (1 XB) / (1 0.1 XB) V
Move all constants to one side: kV/v0 = XB (1 0.1 XB) / (1 XB)
Plug in values: 3 min-1 / 4 L/min * V = 0.4 (1 0.04) / 0.6
Therefore V = 0.85 L.
d. What is the Damkohler number for the CSTR in part (c)?
Da = kV/v0 = 3 * 0.85 / 4 = 0.64.
e. If the fraction of the molar rate of B entering the PFR that is converted in that reactor
(XB,PFR) is 0.48, what fraction (XB,f) of the total inlet molar rate of B (combined for both
reactors) has been converted at the end of the process (in the combined exit stream)?
Initial amount of B = 0.2 * 10 L/min * 10 mol/L = 20 mol/min B
Of the 8 mol/min B that enter the CSTR, 40% are reacted, so 4.8 mol/min remain.
Of the 12 mol/min B that enter the PFR, 48% are reacted, so 6.2 mol/min remain.
The total remaining amount of B after recombination is 11 mol/min.
Therefore, XB,f = (20 11)/20 = 0.45.
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## Document Summary

An inlet gas stream (v0 = 10 liter/min inlet volumetric flow rate and total concentration ct0 = 10 mol/liter) composed of a and b each at a mole fraction of 0. 2 with the rest consisting of an inert. A splitter sends 6 liter/min of the inlet stream to the pfr and 4 liter/min to the cstr. 1. 5 a + b c + d r = k cb k = 3 min-1. State at least two reasons for your conclusion. No, this is not an elementary step: fractional molecules cannot react in a single irreducible step, the reaction stoichiometry is not reflected in the rate law. Although the inlet contains equal amounts of a and b, more moles of a are needed for each reaction event, so a will be depleted first. Fb0 = yb0(ft0) = yb0(v0ct0) = 0. 2 (4 l/min * 10 mol/l) = 8 mol/min b entering the cstr.