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For the A following reaction, 2 Alts) +3/2 Ozte) AlzO3s} calculate the valucs of 4H"and AS" at 900 K (10%)B For the following reaction; CaOts) TiOz(s) CaT...

Question

For the A following reaction, 2 Alts) +3/2 Ozte) AlzO3s} calculate the valucs of 4H"and AS" at 900 K (10%)B For the following reaction; CaOts) TiOz(s) CaTiOxis) calculale the values of AH" and AS" at H100 K (10%)C Aluminum mclts at 937 K Estimate thc molar transition cnthalpy and entropy of Alby rapid frcczing at 900 K (10%6) HablatA Eneiny Chengod Seterl Aractions Wan TE_R Hncelo Go- -J0 540 40 ~4J.uu on AD_ Leed 790-203o -944,100 7a7o0 Cado Loldi cee 220 4770 T cO2al Camhat

For the A following reaction, 2 Alts) +3/2 Ozte) AlzO3s} calculate the valucs of 4H"and AS" at 900 K (10%) B For the following reaction; CaOts) TiOz(s) CaTiOxis) calculale the values of AH" and AS" at H100 K (10%) C Aluminum mclts at 937 K Estimate thc molar transition cnthalpy and entropy of Alby rapid frcczing at 900 K (10%6) HablatA Eneiny Chengod Seterl Aractions Wan TE_R Hncelo Go- -J0 540 40 ~4J.uu on AD_ Leed 790-203o -944,100 7a7o0 Cado Loldi cee 220 4770 T cO2al Camhataon 4-I-O : Jetano Aljj T T-240C Hnninicn 220-12 0 LCRD 4or; 007 SeD ot ~Oar Fmrd 034 21j cud J04 %0 + 8B4U 1355-'50] Cu or Ea7amm LeSes= Cn LAMEM #0 03 [ Tuh_L 1 ICco_I oonti nn Aa_464a tor' ~ocot Ien8-2c00 -1102200 308-H; JFDE 42_473 ry4a Cln 2r 469211 12 80 T 200-2000 00#0*0 Ju d 55B5( Alo H0 , Cananul Euat lOaa Ho HgoeJi Hen 1209-20C0 Uacint lea JIAT E4t_I7 "723,Uco + 204T UJLz*20J Io, 30. Mes Oa CaTE1 28-2171 CO. enle MI76C0 + 170 Nlo . Mono -35,360 subatance 4o , Lo. LegiATn Xi_IS7 ihe FiO 8J6 Hdr 3177 296-2080 L7so7 244,, 471200 298-1726 Bao -506 a0 4022T 11726-2a1 Hamo 548 %o 5,87 T -14 Mo _ o 0_ 208 70n Moz9i 601-II52 Ce Mo , 1158-1808 -163.200- ino 501 able AlE Hcat 2137 Udty subtliee renalurueni n Urans GieEon Ciio ~Iesdeco Moe SCaD AL,O, 380, -Rcs AO, 10,.700 CaO Al,O, Si0: -3773 ?00 WuteOo Cao Al,O, 250_ 4217C0 ZCao-AL,O; 50. Seatnn 1933 ~AN7i Cc,0, -241,800 z29 097n HO. 21 Caf, 31,200 _0oeoo So.~ -74bo1 SiN; 1377o 4J0OJ Ho 21,000 Kob O 52 AoO To; 7245 MgF; To: -914,00) Na,o B,03 erulu TiO, ~od6o0 AOt 27-0 En 20 1,885 22,BA0 2,193 490 1,136 5,007



Answers

In an enzyme-catalyzed reaction with stoichiometry $\mathrm{A} \rightarrow \mathrm{B}, \mathrm{A}$ is consumed at a rate given by an expression of the Michaelis-Menten form: $$r_{\mathrm{A}}[\operatorname{mol} /(\mathrm{L} \cdot \mathrm{s})]=\frac{k_{1} C_{\mathrm{A}}}{1+k_{2} C_{\mathrm{A}}}$$ where $C_{\mathrm{A}}(\operatorname{mol} / \mathrm{L})$ is the reactant concentration, and $k_{1}$ and $k_{2}$ depend only on temperature. (a) The reaction is carried out in an isothermal batch reactor with constant reaction mixture volume $V$ (liters), beginning with pure $A$ at a concentration $C_{\mathrm{A} 0}$. Derive an expression for $d C_{\mathrm{A}} / d t$, and provide an initial condition. Sketch a plot of $C_{\mathrm{A}}$ versus $t,$ labeling the value of $C_{\mathrm{A}}$ at $t=0$ and the asymptotic value as $t \rightarrow \infty$ (b) Solve the differential equation of Part (a) to obtain an expression for the time required to achieve a specified concentration $C_{\mathrm{A}}$ (c) Use the expression of Part (b) to devise a graphical method of determining $k_{1}$ and $k_{2}$ from data for In versus the pour plot should involve fitting a straight line and determining the two parameters $C_{\mathrm{A}}$ (int the parting of the partating and and the conting are a contation a conting from the slope and intercept of the line. (There are several possible solutions.) Then apply your method to determine $k_{1}$ and $k_{2}$ for the following data taken in a 2.00 -liter reactor, beginning with A at a concentration $C_{\mathrm{A} 0}=5.00 \mathrm{mol} / \mathrm{L}$ $$\begin{array}{|l|l|l|l|l|l|} \hline t(\mathrm{s}) & 60.0 & 120.0 & 180.0 & 240.0 & 480.0 \\ \hline C_{\mathrm{A}}(\mathrm{mol} / \mathrm{L}) & 4.484 & 4.005 & 3.561 & 3.154 & 1.866 \\ \hline \end{array}$$

So, in order to figure out the order with respect to each of these reactions were looking for, uh, basically using two experiments where two of the reactors are held constant one that has changed. And we look at the painting, the rate. So put some values on that idea. And so I have the age to SC Um 03 If we want to figure out the order of this, we're looking for two experiments where the contradiction of H plus the concentration of I minus our constant between them but the concentration of H two s, c 03 changes and that one of the experiments that size by that are the 1st and 2nd 1 And so the concentration of H two at C 03 goes from one time center Negative fourth to being too time sensitive before and so obviously that's a two times increase. Now we want to look at how the rate is gonna change. And so it goes from 1.6 600 cents the negative seven to 3.33 time sensitive. Your seventh in this to you can see pretty obviously or you plug into a calculator is a two times increase. Instantly compare these numbers since they're the same. In other words, we're taking you to to the first power. This is going to be a first order reaction with respect to the H two s, c 03 We're gonna do this process for all of them. Now we look at the age plus looking for two experiments where the concentration of the other two reactions are constant and so one that would satisfy that is Experiment one experiment four The concentration of H plus goes from being too some sense in a second being four times tense negus second. And so this is obviously a two times increase. And the rate simultaneously I went from being 1.66 I'm sentinel your seventh being 6.66 times since the negus seventh. And so you can listen Jo Cox later, or just know and you'll realize just a four times increase in the rate. So he compared these two numbers. If we take to square, we get four. And so that tells us that this number this is the order of the reaction with respect to H plus. And so this is a second order reaction with respect to age plus. So we're on our way to, um over already locked. It's tedious, certainly. And finally we have I minus the experiments we could use. We could use Experian number one and then experiment number six. So the concentration of I minus between those two goes from two times 10 today, a second and then 24 I'm sensing a second, and the rate simultaneously goes from 1.66 times Centenario. Seven, 2, 13 0.2. Time center Negative. Seven. You'll see that this is a two times increase. This is an eight times increasing soap, just like before. We have to to the third power equals eight now. And so third power tells us the order with respect to I minus. And so it's first started with respect to age two at 03 second with H plus and third with I'm ice and we can use. That's right. The rate locks the rail always equals a rate constant first, and then we take H two F C 03 to the first power. So just like that, an H plus, but it's squared and then I minus. But it's cute. And so this is our a lot right here. We want to figure out OK, we can plug it any of the experience, find a rate plug in all of these, and we confined. Okay, if you that for multiple of them and find an average, you'll find that the cake now you should get is right around 5.2 times 10 to the fifth. And the units are gonna be definitely unusual. Leaders to the fifth Moles to the fifth per second. This is necessary in order to cancel out all these concentration use that your rate ends up with the commandments. So this is our final answer.


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