Payment for Second product Solution manual for 5th edition. Payment for Third product three solution manuals for 4th and 3rd editions. We try to make prices affordable. Contact us to negotiate about price. If you have any questions, contact us here. Assume that the reactions are irreversible and first order. Case 1: gal v0 X 0. Also, the reverse reaction begins to overtake the forward reaction near the exit of the reactor. See Polymath program Pc. The cost of this storage could prove to be the more expensive alternative.
A cost analysis needs to be done to determine which situation would be optimal. What is the maximum number of moles of ethylene glycol CH2OH 2 you can make in one 24 hour period? The feed rate of ethylene cholorhydrin will be adjusted so that the volume of fluid at the end of the reaction time will be dm3.
Now suppose CO2 leaves the reactor as fast as it is formed. First try equal number of moles of A and B added to react. See Polymath program Pb. This results in a conversion of. CDGA d We must reexamine the mole balance used in parts a-c.
The flow rates have changed and so the mole balance on species A will change slightly. Because species B is added to two different reactors we will also need a mole balance for species B. Since for every mole of CH3 2O consumed there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure.
This would result in the pressure increasing faster and less time would be need to reach the end of the reaction. The opposite is true for colder temperatures. See Polymath program Pi. P j For equal molar feed in hydrogen and mesitylene. As the two components are created, the reactant concentration drops and equilibrium forces the production to slow. As those reactions reach equilibrium, the reactions that are still producing the two components are still going and the concentration rises again.
Finally the reactions that consume the two components lower the concentration as the products of those reactions are used up in other reactions.
P m Individualized solution P Solution is in the decoding algorithm given with the modules ICM problem P a Assume that all the bites will deliver the standard volume of venom. This means that the initial concentration increases by 5e-9 M for every bite.
After 11 bites, no amount of antivenom can keep the number of free sites above This means that the initial concentration of venom would be 5. The best result occurs when a dose of antivenom such that the initial concentration of antivenom in the body is 5.
This means that the initial concentration of venom is 0. From the program below, we see that if an amount of antivenom such that the initial concentration in the blood is 7e-9 M, the patient will die.
The concentration of D only increases with time P b Conc. However, the tradeoff is that the reaction rate of species B, and therefore production of B, decrease as temperature drops. So we have to compromise between high selectivity and production. To do this we need expressions for k1, k2, and k3 in terms of temperature. From the given data we know: Ei ki Ai exp 1.
At this temperature the selectivity is only 5. This may result in too much of X and Y, but we know that the optimal temperature is not above K. The optimal temperature will depend on the price of B and the cost of removing X and Y, but without actual data, we can only state for certain that the optimal temperature will be equal to or less than K.
See Polymath program Pf. A moderate pressure would probably be best. Then run the Program for 0. This will give us the concentration of A and B at the time the second martini is ingested. The Polymath code for after the second drink is shown below. See Polymath program Pd. For the first hour the differential equation for CA becomes: dC A k1C A 2t after that it reverts back to the original equations.
So the person has about 4 minutes and 40 seconds to get to their destination. P g A heavy person will have more body fluid and so the initial concentration of CA would be lower.
This means a heavier person will reach the legal limit quicker. The opposite is true for a slimmer person. They will take longer to reach the legal limit, as their initial concentration will be higher. Doing so will result in build up of the drug in the bloodstream that can cause harmful effects. P d If the drug is taken on a full stomach most of it will not reach the wall at all.
The processed food can also drag the drug to the intestines and may limit its effectiveness. This effect can be seen in the adsorption constant k1 and elimination constant k2 values. If k1 decreases this means that the adsorption process is slow and if k2 increases means that the rate of elimination of Tarzlon increases.
Note that the maximum amount of the drug in the bloodstream is reduced by two. The maximum amount of Tarzlon in the bloodstream is 0. Consequently, the reaction system should be operated at highest possible temperature to maximize SDU.
The reaction should also take place in high concentration of A and the concentration of D should be limited by removing through a membrane or reactive distillation. Membrane reactor in which D is diffusing out can be used. Also low temperatures will help keep the selectivity high.
Try to remove D with a membrane reactor or reactive distillation. The selectivity is not dependant on temperature. Also removing D will help keep selectivity high. This is because the first reaction is very fast and the second reaction is slower with no reverse reactions. This is because after B is formed it will not further get converted to C because the reverse reaction is fast. See Polymath program Pa. K K K Similarly, k2' 3. See Polymath program Pb1.
K K K k2' 0. P c Use the Polymath program from part a and change the limits of integration to 0 to P e When the appropriate changes to the Polymath code from part a are made we get the following. P The reactions are 1. The concentration of species are affected by two factors: reaction and pressure drop. Further down the reactor, the concentration of B drops because the effect of pressure drop outweighs the production of B by reaction.
Species E: The reasoning for species E is similar to species B. However, species C and D are also reactants in reaction 3; this contributes to their decline at later times in addition to the pressure drop. But the rate constants for the reactions have been reversed.
If concentration of M increases the slope of line will decrease. P b Example The inhibitor shows competitive inhibition. P e Example See Polymath program Pe. Part 1 Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 24 24 Cc 2. Consequently, the concentration of product is very low compared to the case without uncompetitive inhibition. Part 3 Change the observed reaction rate constant: 0. Thus CP the plots for this part are approximately the same as the plots in part 1.
And at low temperature PSSH results show greatest disparity. Eventually everyone is ill and people start dying. This explains the shape of the figure. Case2: With drug inhibition Reactions: 1. Thus the stoichiometry equation will be changed. This is not realistic as at some point there will be too many cells to fit into a finite sized reactor. Either a cell death rate must be included or the cells cannot be recycled.
PROD max 1 0. Then, Rate of decrease of conc. Now, since the concentration is very less assuming there is no constraint of sunlight. Since the number of days is coming less than 4. Hence, the initial assumption is verified. P The following errors are present in this solution- 1. It should be S for Hanes-Woolf form. The expression for intercept is correct but the slope is given wrong. The correct expressions are - 5. As concentration of inhibitor I increases, slope increases.
In the given plot, slope for line 1 is more as compared to line 2 in spite of having lower concentrations. This implies that the concentration values are switched. Also the numerically calculated slope values are wrong. At 40 atm, we had While, at 80 atm we have This is not possible and the model should be discarded. Model f is the worst model of all.
Assume H2 in the gas phase reacts with C2H6 adsorbed on the surface and ethane goes directly into the gas phase. P c Individualized solution.
S expression, the constant should be KA2 instead of KDC 2 The overall site balance should include the product, as it too is getting adsorbed. S rC kc CC. Therefore, the inlet temperature we would recommend is K. P d Aspen Problem P e 1 2 At high To, the graph becomes asymptotic to the X-axis, that is the conversion approaches 0.
At low To, the conversion approaches 1. Report message as abuse. Show original message. Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message.
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