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General Chemistry 202 week 3 test 2

by: Catherine Montz

General Chemistry 202 week 3 test 2 Chem 202

Marketplace > University of Louisville > Chemistry > Chem 202 > General Chemistry 202 week 3 test 2
Catherine Montz
U of L
General Chemistry 2
Dr. Kuta

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About this Document

Again, just what we covered this week and what we will be covering Tuesday before the exam Thursday. Study Guide will be up shortly.
General Chemistry 2
Dr. Kuta
Class Notes
Gen Chem, General Chemistry 2, chem 202
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This 0 page Class Notes was uploaded by Catherine Montz on Saturday February 27, 2016. The Class Notes belongs to Chem 202 at University of Louisville taught by Dr. Kuta in Spring 2016. Since its upload, it has received 26 views. For similar materials see General Chemistry 2 in Chemistry at University of Louisville.


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Date Created: 02/27/16
General Chemistry 202 Professor Kuta Test 2 2 Chapters this week Chapters 52 amp 53 Chapter 52 521 Equilibrium Change Problems of Change I ll start with the problem in the book and break it down At 100 C K in the equilibrium is 13 PC3g C2g e PC5g This process started with PC3 100 atm C2 100 atm at equilibrium We need to nd the pressures of PC3C2 and PCI5 at equilibrium First we need to set up a K expression P Cl 5 13 This means we start with zero PCI5 which means I pcg wcvg K the exergonic direction is on that side so the reaction is going to the right Now were going to talk about the changes so the variable x will be brought in to symbolize changes Follow this chart PPCI3 PCl2 PPCI5 Initial 100 atm 100 atm 0 Change x x x Remember if reagents are decreasing they will have a negative change At equilibrium there is no further net change so we will see them written as initial plus changes Table PPCI3 PCl2 PPCI5 Initial 100 100 0 Changes x x x Equilibrium 100 x 100 x X Iwcg We now put this Info In your K expressmn K PPCZ3PCZZ X 100 x 100 x13 Now this is where your algebra will come in big time Solve for X X 13 100 X2 x 13x2 26x 13 13x2 27x 13 0 2 Now use the quadratic equation which is x um2b 4 a 27z2 41313 z X 27b1 x 13 x 076 a You have two results so what do you do Well plug and chug those values into the equilibrium equations in the table So how do we know which one is right Well notice one gives you negative pressure ls negative pressure possible NO so throw this bad boy out 076 is your answer We can do these problems exactly the same with concentration as well However with these you ll have another equation to deal with but it s super easy t h hd39 39 td dissooated amoun w w moaae x 100 starting amount This is literally a plug and chug equation so for instance you found the concentration to be 00078 and the initial value was 0481 Just plug em in This could pop up in the test so know that percent dissociation is a relative relationship it s the percent of how much acid did dissociate relative to how much acid was originally present 522 Approximation We can approximate for some equilibrium problems based upon the K value to make things simple 0 Simple approximation There s going to be a small relative change between nal equilibrium and the intial amounts so you can approximate for K So if relative change is small you can assume x is small relative to the equilibrium amount You can only approximate where x is being added or subtracted from a substantially bigger number and you cannot approximate an x that stands by itself Approximation can fail miserably I would just suggest solving on the test the old fashioned way described above until you have mastered approximation If a problem is straightforward use the method above not this one But really quick I ll show you x2 18 x 10395 0481 square root the left side to get your approximation value It s simple sometimes but it can get really tricky Sig gs are VERY IMPORTANT Watch them fewer sig gs easier it is to approximate 523 General steps for change Here are the exact general steps Dr Noble lists in his book you need to know for equilibrium calculations which involve amount changes Begin with a balanced equation for the equilibrium process can t do it without one Set up the K expression Start a table with the initial amounts given remember gases and solutes are only K components In the same table set up the changes in amounts using X know the direction of the reaction If no amounts are present on one side the it directs towards that side you may have to solve for Q Also pay attention to coef cients In the same table add the amounts to get the equilibrium amounts these tables are at the top Plug the expressions for the equilibrium amounts into K and solve for x Put the x values you receive into the equilibrium amounts equation to solve for them and see which value makes sense And check your work by plugging the equilibrium amounts into the K expression and solve for K your K value should be close to the starting value doesn t have to be exact


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