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UT CH 302 - Equilibrium Lecture 3
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Equilibrium Lecture Three: A bit more detail and some additional kinds of problems. Lecture Overview: We get even more involved in the details by equilibria by • relating ∆G to K • relating Kc to Kp • comparing Q to K to figure out the direction of a reaction • developing LeChatelier’s principle to explain system stress But first, a reminder of how to calculate Q and K from data. Why do I seem to start every lecture this way? Because it is the essence of how you should view chemical equilbria. What is the ratio of materials in a system and at what point do their rates stop changing so that Q = K. Calculate Q at: Note that Q=K during this interval Time 0 minutes Time 2 minutes Time 4 minutes Time 6 minutes Q=BA=04=0 Q=BA=22=1 Q=BA=31=3 = K Q=BA=31=3 = K 4 3 2 1 Q=K at equilibrium 1 2 3 4 5 6 7 Time(minutes) 0 Q<KTime out for a few interesting thoughts on K Interesting thought 1. Solids and pure liquids are not included in equilibrium expression calculations When you construct an equilibrium expression from the chemical reaction, you put the materials on the right in the numerator and the materials on the left on the numerator. H2O  H+ + OH- K = [H+][OH-]/[H2O] But these values are concentrations, and while the concentration of a gas can change (with pressure) and the concentration of a solute in a solvent can change, what about the concentration of pure liquids and solids? They don’t change—they are constant which means there is no reason to include these constant values in the equilibrium expression. So as a rule, for all solids and pure liquids, assume [ ] = 1. Here with [H2O], which has a constant concentration of 55.4 M (do the calculation yourself), we set [H2O] = 1 and the equilibrium expression for water dissociation is actually K = [H+][OH-] Interesting thought 2. K can have very large and very small values—get used to using the exponent button on your calculator Although I emphasized the point that reactions never go to completion and that there are always both product and reactant in a closed system at equilibrium, this doesn’t mean that there aren’t a lot of reactions that go nearly to completion and produce very large equilibrium constants. Consider the reaction for the dissociation of N2 to make atomic nitrogen: Example: N2  2N ∆G=+456kJ K=8.6 x 10-81 or the solubility product for silver sulfide Example: Ag2S  2Ag+ + S= K=8 x 10-51 What bout reactions that produce really large values of K, are there any of those? Sure, what about the reverse of these reactions: 2N N2 ∆G = -456kJ K=8.6 x 1081Interesting thought 3: In the RICE calculation, how do you know which signs to put in the ∆ row of the array Consider the RICE expression. When you are doing calculations, either the stuff on the left is getting smaller and the stuff on the right is getting larger 3H2 + N2  2NH3 Or the stuff on the right is getting smaller and the stuff on the left is getting larger: 3H2 + N2  2NH3 In other words, the reaction is either shifting to the right (Q<K) or the reaction is shifting left (Q>K) You always have to make this determination when working a problem or your calculation will be incorrect. Examples: Example 1: Start with all stuff on the left and none on the right, so clearly Q<K and the reaction shifts right with the signs as below: 3H2 + N2  2NH3 All on left side  Q = 02/(1)3(1)2 = 0 - - + + + - 1 1 0 - - +Example 2: Start with all stuff on the right and none on the left, so clearly Q>K and the reaction shifts left with the signs as below: 3H2 + N2  2NH3 All on right side  Q = 12/0 = ∞ Example 3: Here you have some of everything to start so you have to do a Q vs K calculation. In this case, Q = 1 which is less than K = 3/8 so the reaction shifts righ with the signs as below: 3H2 + N2  2NH3 Mixture to start, Q = 12/131 = 1 Time in. Calculating K from ∆Go (a problem on quiz). As we have send, equilibria is just another way of expressing thermodynamics concepts about the extent of a reaction and there is a simple mathematical relationship between ∆G and K that requires a simple plug and chug effort. By the way, this means that for every DG in the tables in the back of the text book, you can instantly determine an equilibrium constant values for the reaction. ∆Go = -RT lnK Example: At 298K for the reaction for the formation reaction: 2C + 3H2 + 12 O2  C2H5OH If ∆Go = -168.6 kJ/mole, what is K? 1. Stick the values in, 2. rearrange for K and Step3, fix the units Step 1: -168.6kJ = (-8.3J) (298) lnK 0 0 1 + + - 1 1 1 - - +Step 2: ∆Go = -RT ln K  (∆G0−RT) = ln K exp (−∆G0RT) = K  exp (∆G0−RT) = exp (ln K) (multiple through by ex) Now inserting values and changing kJ to J K = exp (−168.6x103−(8.3)(298)) = 3.5x1029 Note that as expected, a reaction with a high negative free energy will produce a very large positive exponent for K. Another Time Out. Know your exponent math!!! 1) You will see exp, ln, log, and 10x functions a lot in CH302. You must be able to effortlessly move between values inside functions using inverse function concepts. ln (exp x) = x exp (ln x) = x log (10x) = x 10 (logx) = x (note we just used this trick in the ∆G to K problem above when we multiplied through by exp x) 2) You need to remember the simple math involving exponents and be able to do it in your head—remember my joke about how snakes do it by logs because they are adders? Well even if it isn’t funny, you still need to be able to add exponents when working problems. For example, in the problem below, note that we add -5 and -1 to get -6 for the multiplication of the two exponents, and then we perform a square root by dividing by two to get -3 which is the power of the answer to the calculation below. Example x = [(10-5) (.1)] 1//2 = 10-3 This is something you should be something you can do without a calculator, or you have no chance of working tests in the allotted time in this class.Time in.


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UT CH 302 - Equilibrium Lecture 3

Type: Miscellaneous
Pages: 10
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