PSU CHEM 110 - Chemical Reactions Equilibrium Part 1

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Chemical Reactions: Equilibrium Part 1Chemical Reactions: Equilibrium Part 1Chemical EquilibriumEquilibrium – GraphicallyEquilibrium ConstantHow are Kc and Kp related?Practice ProblemPractice ProblemEquilibrium DemoPractice ProblemPractice ProblemKeq for Related ReactionsKeq for Related ReactionsKeq for Related ReactionsPractice ProblemWhat You Should KnowChemical Reactions:Equilibrium Part 1CH110 FA11 SAS 1Prepare for Recitation December 8thALEKS Objective 14 due SUNDAY December 4thANGEL Quiz 14, December 8thALEKS FINAL ASSESSMENT:Deadline December 9thLecture 40: November 30thLecture 41: December 2ndLecture 42: December 5thRead: Ch. 15.1 – 15.5HW Preparation: BLB 15: 21,28,29,32,35,46; Packet 15: 1-10CH110 FA11 SAS 2By the end of Today’s Lecture you should know:Chemical EquilibriumEquilibrium Constant KeqEquilibrium constant expressionRelationship between Kp and KcKEY QUESTIONS: THE BIG PICTUREChemical Reactions:Equilibrium Part 1Do real reactions always proceed to completion?What forces drive the reaction process?How do the properties of heterogeneous reactions differ from homogeneous?Remember: macroscopic properties come from microscopic originsWhat is the meaning of chemical equilibrium?How does the equilibrium constant help explain the extent of a reaction?What is the relationship between various equilibrium constant expressions?Chemical EquilibriumCH110 FA11 SAS 3At Equilibrium:1.2.Equilibrium is not always achieved quickly!Examples we’ve already seen:– vapor pressure above liquid– saturated solution– phase equilibriumChemical Equilibrium:Reactants ProductsEquilibrium – GraphicallyCH110 FA11 SAS 4CO + 3 H2CH4+ H2O1200 Kforward & reverse reactions ____ proceed–––321molestimeH2COCH4or H2Ostart: 1 mol CO, 3 mol H2, in 10.0 L volumeAt equilibrium:1.839 mol H20.613 mol CO0.387 mol H2O & CH4• reaching equilibrium _________________• What affect do starting concentrations have on equilibrium concentrations?Equilibrium ConstantCH110 FA11 SAS 5aA + bB cC + dDKeq= Kc= Keq= Kp=KEQis a _______ for a given temperature. It is a function of the temperature (and _____for gases), not the ___________________If units are given in ______________ :If units are given in ____________ :Keqprovides the ______ products to reactants at equilibrium. It is a ________How are Kcand Kprelated?CH110 FA11 SAS 6CO + 3 H2CH4+ H2OKc= =Kp=P = nVRTKp=Kp= = Kp= T= 1200K= ∆n = change in number of moles going from reactants to products=Practice ProblemCH110 FA11 SAS 7Acetic acid is a weak acid that dissociates as follows in aqueous solution:At equilibrium at 25oC a 0.1000 M solution of acetic acid has the following concentrations: [CH3COOH] = 0.0990 M, [CH3COO-] = 1.33x10-3M and [H+] = 1.33x10-3M. What is the equilibrium constant Kcfor the ionization of acetic acid at 25oC?CH3COOH (aq)A) 5.71x104B) 1.00x10-1C) 1.75x10-7D) 1.79x10-5E) 5.71x106CH3COO-(aq) + H+(aq)Practice ProblemCH110 FA11 SAS 8A 2.00 L contains A (g) and B (g), which follow the equilibrium shown below. At 0oC, the equilibrium partial pressures are A (g) = 0.360 atm, B (g) = 0.780 atm. What is the value of Kp?A (g)2B (g)A) 1.14B) 2.17C) 0.845D) 1.69E) 7.65Equilibrium DemoCH110 FA11 SAS 92NO2N2O4Keq is a function of temperature only (the volume of the tube is constant).As I change the temperature of the sealed tube, I shift the equilibrium and you see a color change.Practice ProblemCH110 FA11 SAS 10Which of the following is the correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxideand nitrogen dioxide?N2O4(g) 2NO2(g)A)B)C)D)E)[NO2]2[N2O4][NO2][N2O4][NO2]2[N2O4][NO2][N2O4]2[NO2] [N2O4]Practice ProblemCH110 FA11 SAS 11In the reaction above, the following equilibrium concentrations of NO, N2, and O2were determined at 800oC. What are Kcand Kpfor these conditions?[NO] = 0.0120 M[N2] = 0.508 M[O2] = 0.410 M2NO(g)N2(g) + O2(g)Keqfor Related ReactionsCH110 FA11 SAS 122 NO(g)N2(g) + O2(g)N2(g) + O2(g) 2 NO(g)Keq=Keq = • Keqfor a reaction in the reverse direction is the _______ of Keqin the forward directionRule 1: The same equilibrium is reached regardless of the direction it is approached fromKeqfor Related ReactionsCH110 FA11 SAS 132 NO(g)N2(g) + O2(g)Keq=Keq´=• Sometimes, it is useful to manipulate the overall stoichiometry of a reaction for thermochemical calculationsRule 2 (Stoichiometry): If the coefficients in a reaction are multiplied or divided by an integer, their Keqare related by the exponential of that integer NO(g)½ N2(g) + ½ O2(g)Keqfor Related ReactionsCH110 FA11 SAS 14N2(g) + O2(g) 2 NO(g)Keq = • It is often necessary to break a reaction into two or more steps for Hess’s Law calculations.Rule 3: The Keqfor a net reaction is the product of the Keqfor the individual reactions2NO(g) + O2(g) 2 NO2(g)Keq = N2(g) + 2O2(g) 2 NO2(g)Keq =Practice ProblemCH110 FA11 SAS 15The following reaction reaches equilibrium at a certain temperature and Keqis determined:H2+ I22 HIAt the same Temperature, what is Keqfor the reactionHIH21212+ I2Keq= 16A) 16B) 4C) 1/4D) 1/16E) There is not enough information to answer this problemWhat You Should Know CH110 FA11 SAS 16Chemical Equilibrium and Keq• What is happening in a system that has reached chemical equilibrium?• What is the meaning of the equilibrium constant (Keq) and how does it help predict the outcome of a reaction?• What is the relationship between Kcand Kpand what does each tell you?• If you know Keqfor a given reaction, how does it help you understand what will happen in a related


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PSU CHEM 110 - Chemical Reactions Equilibrium Part 1

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