Lecture 18 - Turn on Camtasia Recording Tami's Review Session Wednesday March 12th 7:00 PM Library Room 1320 1 BioSci Presentation Working in the

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Unformatted text preview: Turn on Camtasia Recording! Tami's Review Session: Wednesday March 12th 7:00 PM Library Room 1320 ... 1 BioSci Presentation Working in the Pharmaceutical Industry. Industry Monday March 10th 5:30 pm Morrill 215 FREE PIZZA & Drinks ... 2 Writing and Manipulating K Expressions Changing direction S(s) + O2(g) ( ) SO2(g) SO2(g) S(s) + O2(g) ... 3 Writing and Manipulating K Expressions Changing coefficients S(s) + 3/2 O2(g) 2 S(s) + 3 O2(g) SO3(g) 2 SO3(g) ... 4 Writing and Manipulating K Expressions Adding equations for reactions S(s) + O2(g) SO2(g) SO3(g) SO2(g) + 1/2 O2(g) Net equation S(s) + 3/2 O2(g) SO3(g) ... 5 Writing and Manipulating K Expressions Concentration Units We h W have been writing K i t b iti in terms of mol/L. f l/L These are designated by Kc But with gases, P = (n/V)RT = conc RT P is proportional to concentration, so we can write K in terms of P. These are designated by Kp. Kc and Kp may or may not be the same. ... 6 Writing and Manipulating K Expressions K using concentration and pressure units Kp = Kc (RT)n For S(s) + O2(g) SO2(g) n = 0 and Kp = Kc For SO2(g) + 1/2 O2(g) SO3(g) n = 1/2 and Kp = Kc(RT)1/2 ... 7 What is K2? K1 = 1 20 1.20 K2 = ? ... 8 What is K2? Ag+ + 2NH3 3Ag+ + 6NH3 Ag(NH3)2+ 3Ag(NH3)2+ K1= 1.6 x 107 K2= ? A. 1.6 x 107 B. 3 2 B 3.2 x 107 C. 4.8 x 107 D. 2.6 x 1014 E. 4.1 x 1021 ... 9 What is K3? A+ B C A+B C D+E D+E A. 1 2 A 1.2 x 102 B. 3.6 x 104 C. 4.0 x 102 D. 4.3 x 106 E. 4.1 x 108 ... 10 K1= 1.2 x 102 12 K2 = 3.6 x 104 K3 = ? Questions to consider: What happens when we alter a reaction at equilibrium? What happens when we raise the temperature? What happens when we remove product or reactant? What h Wh t happens when we change the h h th volume for systems involving gasses? ... 11 EQUILIBRIUM AND EXTERNAL EFFECTS Temperature and concentration changes affect equilibria. The outcome is governed by ___________________________ "...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance. disturbance " Weebles wobble but they don't fall down. ... 12 Le Chatelier's Principle Change T Changes the value of K This causes a change in equilibrium concentrations Add or take away reactant or product: K does not change Reaction adjusts to new equilibrium "position" Use a catalyst: K does not change Reaction comes more quickly to equilibrium ... 13 Le Chatelier's Principle "reactant" "product" Removing a "reactant" from a chemical system. reactants products Blue line= initial state ... Red line = new state Water level= eq. state 14 Le Chatelier's Principle "reactant" "product" Adding a "product" to a chemical system. reactants products ... Blue line= initial state Red line = new state Water level= eq. state 15 Nitrogen Dioxide Equilibrium N2O4(g) (colorless) 2 NO2(g) (brown) ... 16 Nitrogen Dioxide Equilibrium 2 NO2(g) (colorless) (brown) N2O4(g) ... 17 Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) N2O4 + heat (colorless) 2 NO2 (brown) Ho = + 57.2 kJ Kc = [NO2 ]2 [N2O 4 ] Kc (273 K) = 0.00077 Kc (298 K) = 0.0059 ... 18 Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) Ho = + 57.2 kJ Increase T What happens to equilibrium T. position and the value of K? Decrease T. Now what? ... 19 Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) Kc = [NO2 ]2 = 0 0059 at 298 K 0.0059 [N2O4 ] Increase P in the system by reducing the volume (at constant T). ... 20 Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) [NO2 ]2 Kc = = 0.0059 at 298 K [N2O4 ] Increase P in the system by reducing the volume. In gaseous system the equilibrium will shift to the id th side with fewer molecules (in order t ith f l l (i d to reduce the P). Therefore, reaction shifts _________ and [NO2] ________ and [N2O4] ___________. ... 21 Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) + heat ... 22 EQUILIBRIUM AND EXTERNAL EFFECTS Concentration changes: no change in K only the equilibrium composition changes. ... 23 EQUILIBRIUM AND EXTERNAL EFFECTS Catalytic exhaust system Add catalyst ---> ---> no change in K A catalyst only affects the RATE of approach to equilibrium. ... 24 butane ButaneButaneIsobutane Equilibrium K = [isobutane] = 2.5 [butane] isobutane ... 25 Butane Isobutane At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5. Add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? ... 26 Butane Isobutane Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5 Solution Calculate Q immediately after adding more butane and compare with K. Q is LESS THAN K. Therefore, the reaction right will shift to the ____________. (products) ... 27 Butane Isobutane You are at equilibrium with [iso] = 1.25 M and [butane] = [iso] 0.50 M. Now add 1.50 M butane. Solution Q is less than K, so equilibrium shifts right -- away from butane and toward isobutane. isobutane. Set up ICE table [butane] [isobutane] isobutane] Initial Change Equilibrium ... 28 You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution Butane Isobutane K = 2.50 = [isobutane] 1.25 + x = [butane] 2.00 - x x = 1.07 M At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane. ... 29 ...
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This note was uploaded on 04/15/2008 for the course CHEM 112 taught by Professor Hardy during the Spring '08 term at UMass (Amherst).

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