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(Icn marks) Ure the fLowing information to answer the next question In thc HakT-Bosch process of manufacturing ummonia ntrogen Is ucaled #ith hydrogcn, which rsults...

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(Icn marks) Ure the fLowing information to answer the next question In thc HakT-Bosch process of manufacturing ummonia ntrogen Is ucaled #ith hydrogcn, which rsults in 4n cxoxhcric Iaction Animn catalyst (Fc" uscd in tbe process Jnd aluminum oxide ad potassium oxid 4rc used promolers. The following reactions takes place at 25 "C NlgL:He) = ZNHg) 24Indicale thc dircction in which the equilibrium will shift when lcmpcTalurc lncIcased volume is increascd pressure is decreased catalyst is

(Icn marks) Ure the fLowing information to answer the next question In thc HakT-Bosch process of manufacturing ummonia ntrogen Is ucaled #ith hydrogcn, which rsults in 4n cxoxhcric Iaction Animn catalyst (Fc" uscd in tbe process Jnd aluminum oxide ad potassium oxid 4rc used promolers. The following reactions takes place at 25 "C NlgL:He) = ZNHg) 24 Indicale thc dircction in which the equilibrium will shift when lcmpcTalurc lncIcased volume is increascd pressure is decreased catalyst is addkd the concentration of Ni(e} is incrcascd helium gas is added at constant volume , s0 that thc total prcssur increased B A mixturc of 0.48 mol ofN;(g) and 0.93 mol of H (g) is uscd in thc above proccss. ngd rcuction !csscl of 10.0 L capacity is uscd Calculalc thc valuc of thc cquilibrium constant and thc composition ofthc cquilibrium mixture at 25 *C,if thc cquilibrium conccntntion of H(g) = 049 nVL SCHAU Lesson Assignment Pake 10



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The feed to a distillation column (sketched below) is a 45.0 mole\% $n$ -pentane- 55.0 mole\% n-hexane liquid mixture. The vapor stream leaving the top of the column, which contains 98.0 mole\% pentane and the balance hexane, goes to a total condenser (which means all the vapor is condensed). Half of the liquid condensate is returned to the top of the column as reflux and the rest is withdrawn as overhead product (distillate) at a rate of $85.0 \mathrm{kmol} / \mathrm{h}$. The distillate contains $95.0 \%$ of the pentane fed to the column. The liquid stream leaving the bottom of the column goes to a reboiler. Part of the stream is vaporized; the vapor is returned to the bottom of the column as boilup, and the residual liquid is withdrawn as bottoms product.(a) Calculate the molar flow rate of the feed stream and the molar flow rate and composition of the bottoms product stream. (b) Estimate the temperature of the vapor entering the condenser, assuming that it is saturated (at its dew point) at an absolute pressure of 1 atm and that Raoult's law applies to both pentane and hexane. Then estimate the volumetric flow rates of the vapor stream leaving the column and of the liquid distillate product. State any assumptions you make. (c) Estimate the temperature of the reboiler and the composition of the vapor boilup, again assuming operation at 1 atm.(d) Calculate the minimum diameter of the pipe connecting the column and the condenser if the maximum allowable vapor velocity in the pipe is $10 \mathrm{m} / \mathrm{s}$. Then list all the assumptions underlying the calculation of that number.

Continuing on with our work around physical chemistry, taking a look at gasses, morals, precious faith, precious and things that are like. We can first start off with some contextual learning, so the extraction as a process of separating substances from a mixture based on their soul abilities, are their relative soul abilities. So the salute is extracted from the carrier using the help of some sort of solvent, and the extraction is the minimum of the three components processes. So the extraction is a tertiary system, so from 50 80% relative saturation relation, we have gamma nor P is able to 9.500 p h star Garmin autism or fraction peers. The pressure pH stars of vapor pressure we start substitute in the vapor pressure is one of 68 millimeters of Mercury Pierce. 7 60 millimeters of mercury. Dominant becomes not 0.703 So saturation relation at the outlet stream we have pH. Star. It was another point, not five p get that in millimeters of mercury. So then we multiply No point No. Five by 7. 60 millimeters of mercury. So I got peak Age Star is able to 38 millimeters of mercury. And so from the answer in equations, the temperature at saturation could be calculated as negative 3.26 degrees Celsius. More balance off the condenser is as follows we have. And no, it is ableto end one at 1.5. Kill them all at one take we're not 10.703 multiplied by n dot north people to not 0.95 and one talk. So we have equation to in equation three. So then we generate and don't know is 2.81 killer malls a minute on n dot juan is equal to 9.68 to kill a moles per minute. Eso the most of nitrogen in the outlet stream is no 0.6479 killer models Volume of nitrogen. We have 14.5 us see? And, um so the flurry of nitrogen to the driest 14.5 SCM Uh huh. Moving on to part B now. So we have the relation for the 50% relative saturation, and the Condenser inlet is as follows. So we have not 0.5 zero p h star sequel to govern our P we re arrange and substitute PN for Gap P each star favor. Pressure is equal to turn 685.6 millimeters of mercury. So from equation one that we saw previously, the temperature of saturation t zero is 187 degrees Celsius. So the saturation of the outlet condenser pH star is not point No. Five p. And now we can multiply. No, we're not five by 76 years. There were millimeters of mercury, it's of P H staff comes 3 80 millimeters Are Makary so moving on. So from previous equation, we have the outlet temperature saturation t wan that we can define us 48.2 degrees Celsius So the ratio of the volumetric flow rate probably in moving on entering the condenser. We have fedora want divided by video no secret tow N one r t one of p drive by end to our t to overpay. So we substitute in our moles about n one or not and zero that we defined previously r t one or not zero which is 3 to 1 Kelvin and for 60 Kelvin respectively. So we have the one over v not is not 0.682 kill the malls per minute multiplied by 3 to 1 Kelvin, divided by 2.1 a kill more a minute multiplied by 4 60 kelvin. We got a value off, not point to two. So therefore the final gas temperature is 48.2 degrees Celsius and the volume ratio volumetric flow rate it's now point to to eso for part B cost of compressing and cooling. A 10 atmosphere is required for the first part. The cost of calling to negative 3.24 degrees Celsius as well as the installation costs and the cost of utilities that will need are required to find the cost effectiveness off part A and part B that we looked at. And so therefore we can determine, uh, part a is more cost effective.


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