Homework+4+rev1-1 - Chemical Engineering 142 Homework Set...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chemical Engineering 142 Homework Set #4 100 points total Due Wednesday, February 17, 2010 Problem 1 (20 points) Consider the following elementary steps in the gas phase oxidation of NO, which involves the intermediate NO3 Step 1 Step 2 The reverse rate of Step 2 is sufficiently small to consider the step irreversible. (a) (b) Derive the rate expression for the formation of NO2. The observed rate expression for NO decomposition is What assumption(s) can you make to transform the rate expression from part (a) into the observed expression? (c) (d) Repeat part (a) with the assumption that Step 1 is now quasi-equilibrated (QE). How does your assumption in part (b) related to the QE assumption? Use the arrow diagram formalism to compare the relative magnitudes of each reaction rate (Steps 1 and 2, both forward and reverse), as well as the net overall reaction rate when Step 1 is QE. Problem 2 (20 points) Chlorine radicals have been implicated in the mechanism for ozone destruction. Consider the following elementary steps in the decomposition of ozone. Step 1 Step 2 Step 3 Step 4 Step 5 Step 5 can be assumed to be the termination step, as other processes quench the methyl radical quickly. Also assume that the rate of propagation steps are much larger than those of termination or initiation steps. (a) (b) Derive the O3 destruction rate as a function of measurable concentrations and relevant rate constants. Consider another step where chlorine is converted rapidly back to a radical Step 6 HCl is convected slowly (Step 7) out of the system at a rate proportional to its concentration. Derive the O3 destruction rate with the new termination step (Step 7). Include OH• in the expression as other processes determine its concentration. (c) The chain length (ν) describes chemical reactions involving radicals. following definition of the chain length Consider the Compare the chain lengths from parts (a) and (b) Problem 3 (20 points) The following elementary steps describe the gas-phase pyrolysis of acetaldehyde k1 = 150 h-1 k2 = 700 L h-1 mol-1 k3 = 850 L h-1 mol-1 k4 =1000 L h-1 mol-1 (a) (b) Derive an expression for the rate of disappearance of CH3CHO Calculate the volume required to achieve 80% conversion of CH3CHO in an isothermal, isobaric PFR maintained at 750 K and 500 kPa. The inlet flow rate is 10 L h-1 and composed of pure CH3CHO. Consider all species as ideal gases. Problem 4. (15 points) In free radical polymerization, highly reactive radicals propagate to form long monomer chains and in this way produce polymers with the desired physical properties. The general mechanism for this type of polymerization reactions is comprised of three processes: initiation, propagation and termination steps. Initiation: I 2R• (kd) R• + M R + M• (ki) Propagation: M• + M M2 • + M … Mn•+M Termination: Mn • + Mm • Mm+n (kt) M2• (kp) M3• (kp) Mn+1• (kp) where I is a radical initiator that generates radicals R•, which are transferred to monomer units M, which further propagate to generate polymer chains. For the purposes of this problem assume that the rate of propagation is independent of the length of the chain of the polymer (i.e. all radical chains are equivalent M•=M2•= … =Mn•). So that the propagation reaction can be written in general as M• + M M• (kP) (a) Use the PSSH on all radical species to derive an expression for the rate of monomer (M) consumption in terms of all the rate constants and concentrations of initiator [I] and monomer [M] species. (b) The average number of monomer units added to a radical chain (ν) can be calculated as the ratio between the rates of the propagation step (Rp) over the rate at which radical chains are consumed in the termination step (Rt). Given that, derive an expression for ν and for the average molecular weight of the polymer (Mn) given the molecular weight of a monomer unit (m). Problem 5. (25 points) Pure reactant A (CA0 = 50 mmol/L) is fed at a steady state rate of 10 L/h into a PFR (V = 0.15 L) where it dimerizes (2A →B). The rate law expression is second order in A only (k = 1.0 L/mmol/h). The reactor operates at 300K. Assume that A and B are in gas phase and behave ideally unless told otherwise. a) Construct a stoichiometric table for this process. b) Express the outlet concentrations of A as a function of X. c) Write the design equation for VPFR as a function of XA, v0, and CA0. You may leave the expression in its unintegrated form. d) What is the conversion of A in this PFR? e) It is later found that the reaction proceeds as below: 2A(g) → B(g) + 0.5C(l) The reactor conditions are P=2.5 bar, T=300K Would this change your answer for overall conversion of A? Will CA in the outlet of the PFR change? Why and how? f) Someone else discovers that C is not produced, but that B is partially condensable. The vapor pressure, PBvap is 0.3 bar at 300K. The reactor conditions are the same as above. What would be the overall conversion under this scenario? ...
View Full Document

This note was uploaded on 01/23/2011 for the course CHM 142 taught by Professor Hey during the Spring '10 term at Berkeley.

Ask a homework question - tutors are online