Class Notes Feb 17

Class Notes Feb 17 - 2/17/2010 Announcements: Wednesday,...

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Unformatted text preview: 2/17/2010 Announcements: Wednesday, February 17, 2010 Homework WebAssign #7- Due Wednesday, February 24 at 8:00 AM Recitation Week 6 – Thermodynamics No Lab Titrations lab report due & TA/LA Evaluations. Suggested Textbook Questions Begin working Chapter 20 problems. You are cordially invited to attend an Informational Session to learn more about becoming a Learning Assistant with the STEM-Colorado Program. When: Wednesday, February 24, 2010, at 6 p.m. Where: UMC 235 (hall right of Reception Desk) RSVP: By Feb. 19 to Olivia Holzman (olivia.holzman@colorado.edu) Refreshments will be served, while they last. Applications for Fall 2010 available Feb 24 - Mar 10 at http://stem.colorado.edu/Apps Get more information from faculty and LAs in these departments: CHEM will be hiring for: Applied Math Math • General Chemistry MCDB Astronomy • Physical Chemistry with Physics Chemistry Biochemistry Applications Geological Sciences February 17, 2010 - CT#1 For each of the following pairs, which has a higher entropy? SF6 & SF4 O2 @ 300 K & O2 @ 600K G vs. G G predicts spontaneity at standard conditions ONLY! All gases at 1 atm and all solutions at 1 M. What about at other conditions??? Non-standard. G = G + RT ln Q Q = reaction quotient Q is calculated the same as K but with non-equilibrium concentrations or pressures. (products/reactants) Q, like K, is always greater than 0! A. SF6 & O2 @ 300K B. SF6 & O22@ 600K & O @ 600K C. SF4 & O2 @ 300K D. SF4 & O2 @ 600K #atoms → S° T → S° H2(g) + I2(g) 2 HI(g) Q= PHI2 PH2PI2 Go = -15.94 kJ/mol Kp, 298 K = 6.2 x 102 o 2 H2(g) + I2(g) 2 HI(g) G = -15.94 kJ/mol Q = PHI Kp, 298 K = 620 PH2PI2 What is G if the conditions are such that Q = K? G = Go + RT ln Q G = -15.94 kJ/mol + (0.008314 kJ/molK)(298 K)(ln 620) (0.01)2 (1)(1) G = 0 kJ/mol At equilibrium, G = 0 and Go = -RT ln K Makes sense, at equilibrium the rxn isn’t going one way or the other. G should be 0! What is G if the pressures of HI, I2, and H2 are 0.01 atm, 1.0 atm, and 1.0 atm? G = Go + RT ln Q G = -15.94 kJ/mol + (0.008314 kJ/molK)(298 K) ln G = -38.8 kJ/mol Reaction is spontaneous under the conditions stated. 1 2/17/2010 February 17, 2010 - CT#2 For a reaction with a negative G°, which of the following is true? G and K At equilibrium, Q = K and G = 0 A. K > 1 A. K > 1 B. K = 1 C. K < 1 D. K = 0 E. K < 0 Go = -RT ln K (-) = – (+)(+)(lnK) ln(K) > 0 → K > 1 G = G + RT ln Q Substitute in G = 0 and Q = K to get, G = - RT ln K Rearranging, K = e-G/RT You can get G from K and K from G. G<0 G>0 G=0 K>1 K<1 K=1 reaction is extensive, products favored over reactants reaction isn’t extensive, reactants favored over products reactants and products favored equally! Problem: Calculate K at 25˚C for the equilibrium reaction N2O4(g) 2 NO2(g) Go = -RT ln K K = e-G/RT Go = 2Gof(NO2) - Gof(N2O4) Go = 2(51.31) – (97.89) kJ/mol = +4.73 kJ/mol K = e -(4.73 kJ/mol)/(0.008314 kJ/molK)(298 K) K = 0.15 Q vs. K G = G + RT ln Q G = -RT ln K + RT ln Q G = RT ln Q/K The magnitude of Q vs. K determines spontaneity. Q<K Q>K Q=K G<0 G>0 G=0 reaction is spontaneous to the right () reaction is spontaneous to the left () at equilibrium, equal forward and reverse February 17, 2010 - CT#3 Predict the sign on G for the following reaction when PHI = PH2 = PI2 = 1.0 atm. H2(g) + I2(g) 2 HI(g) Kp, 298 K = 620 G<0 A. G < 0 B. G > 0 C. G = 0 Q<K G = RTln(Q/K) G < 0 Forward rxn is favorable! D. Not enough information. 2 ...
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