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University of Alberta
Gail Amort- Larson

Last time we had as an example the hydrogenation of unsaturated organic molecules. Cornered by Mike Baldwin Practice A sample exam is posted online. Give it a try under simulated conditions: 45 min, closed book, etc. Solutions are now posted online. 1 2 Unit 2 – Chemical Equilibrium and applications Chapter 17 – Chemical Equilibrium  Principles of chemical equilibrium Chapter 17 Qualitative  What is chemical equilibrium  Acid–baEsquilibria Chapt1 8r, 91 r e t pyathi l i bul oS  Equilibrium constant  Compleixns Chapt1er  Direction of the reaction  Le Chatelier’s principle Quantitative  Determination of K  Calculation of equilibrium positions  Volume and temperature effects 3 4 Chemical equilibrium N 2 –4NO eq2ilibrium system All chemical reactions reach a state of equilibrium. NO dark brown gas poisonous, corrosive 2 That is, they reached a state in which reactants are left, N 2 4 colorless gas products are present but nothing seems to be happening. In a sealed flask at room temperature both exist Wrong !!! together. N 2O 4 (g)  2 NO 2 (g) Reaction is continuing on the microscopic level, but the rate at which the products are formed, equals exactly to The amounts of each gas in the flask are dictated by the the rate at which reactants are formed, so there is No equilibrium constant. NET change. Figure 17.1 stoich coefficients stoich coefficients k1[reactants] = k –1[products] (products = reactant of the reverse reaction) stoich coefficients k1 [products] K = equil. constant =k –1= [reactants]stoich coefficients Note: Equi. Constant is capital K, rate const. is small k. 5 6 Figure 17.3 When the concentrations are constant, the rxn is at equilibrium. 3 different terms we’ll encounter:  The position of equilibrium is defined by the collection of concentrations of the reactants and products at equilibrium.  The direction of reaction  The equilibrium consta.Kt K is constant if the temperature is constant. 7 8 The equilibrium constant In terms of partial pressures  P P m jA (g) kB (g)  C (g) mD
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