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GT CHEM 1310 - Chemical Equilibrium

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Chapter 11 Chemical EquilibriumThe Equilibrium Condition (General)The Equilibrium Condition (Chem Rxn)The Equilibrium State (Chem Rxn)Chemical Reactions and EquilibriumArrows: Chemical SymbolismSlide 7Law of Mass Action (1)Law of Mass Action (2)ActivitiesThe Equilibrium StateSlide 12Slide 13Heterogeneous EquilibriumSlide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Non-Equilibrium Conditions: The Reaction Quotient (1)The Reaction Quotient (2)Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Chapter 11Chemical Equilibrium11.1 The Equilibrium Condition 11.2 The Equilibrium Constant 11.3 Equilibrium Expressions Involving Pressures 11.4 The Concept of Activity 11.5 Heterogeneous Equilibria 11.6 Applications of the Equilibrium Constant 11.7 Solving Equilibrium Problems 11.8 Le Chatelier's Principle 11.9 Equilibria Involving Real GasesThe Equilibrium Condition (General)Thermal equilibrium indicates two systems in thermal contact with each do not exchange energy by heat. If two bricks are in thermal equilibrium their temperatures are the same.Chemical equilibrium indicates no unbalanced potentials (or driving force). A system in equilibrium experiences no change over time, even infinite time.The opposite of equilibrium systems are non-equilibrium systems that are off balance and change with time.Example 1 atm O2 + 2 atm H2 at 298KaA + bB cC + dDThe same equilibrium state is achieved whether starting with pure reactants or pure products.The equilibrium state can change with temperature.The Equilibrium Condition (Chem Rxn)The Equilibrium State (Chem Rxn)H2O (g) + CO (g) H2 (g) + CO2 (g)Change [CO] to PCO[H2O] to PH2OetcAs the equilibrium state is approached, the forward and backward rates of reaction approach equality. At equilibrium the rates are equal, and no further net change occurs in the partial pressures of reactants or products.1. No macroscopic evidence of change.2. Reached through spontaneous processes.3. Show a dynamic balance of forward and backward processes.4. Same regardless of the direction from which they are approached.Fundamental characteristics of equilibrium states:Chemical Reactions and Equilibrium5. No change over time.↔Use this in an equilibrium expression.Use this to indicate resonance.Arrows: Chemical SymbolismChemical Reactions and EquilibriumThe equilibrium condition for every reaction can be described in a single equation in which a number, the equilibrium constant (K) of the reaction, equals an equilibrium expression, a function of properties of the reactants and products.H2O(l) H2O(g) @ 25oCTemperature (oC) Vapor Pressure (atm) 15.0 0.01683 17.0 0.0191219.0 0.0216821.0 0.0245423.0 0.0277225.0 0.0312630.0 0.0418750.0 0.1217H2O(l) H2O(g) @ 30oCK = 0.03126K = 0.04187Partial pressures and concentrations of products appear in the numerator and those of the reactants in the denominator. Each is raised to a power equal to its coefficient in the balanced chemical equation.aA + bB cC + dD€ if gasesPC( )cPD( )dPA( )aPB( )b= K€ if concentrationsC[ ]cD[ ]dA[ ]aB[ ]b= KLaw of Mass Action (1)1. Gases enter equilibrium expressions as partial pressures, in atmospheres. E.g., PCO22. Dissolved species enter as concentrations, in molarity (M) moles per liter. E.g., [Na+]3. Pure solids and pure liquids are represented in equilibrium expressions by the number 1 (unity); a solvent taking part in a chemical reaction is represented by unity, provided that the solution is dilute. E.g., I2(s) ↔ I2(aq) [I2 (aq) ] = K)]([1)]([)]([)]([)()(222222aqIaqIsIaqIKaqIsI===↔Law of Mass Action (2)PH2OPref= K Pref is numerically equal to 1The convention is to express all pressures in atmospheres and to omit factors of Pref because their value is unity. An equilibrium constant K is a pure number.H2O (l) H2O (g) Kp = P H2OThe concept of Activity (i-th component) = ai = Pi / P reference@ 25oCKp = 0.03126 atmK = 0.03126ActivitiesH2O (g) + CO (g) H2 (g) + CO2 (g)COOHCOHpPPPPK222=The Equilibrium StateThe Equilibrium ExpressionsIn a chemical reaction in which a moles of species A and b moles of species B react to form c moles of species C and d moles of species D,The partial pressures at equilibrium are related throughK = PcCPdD/PaAPbBaA + bB cC + dDWrite equilibrium expressions for the following reactions 3 H2(g) + SO2(g) H2S(g) + 2 H2O(g)2 C2F5Cl(g) + 4 O2(g) Cl2(g) + 4 CO2(g) + 5 F2(g)Gases and Solids CaCO3(s) CaO(s) + CO2(g) K=PCO2 K is independent of the amounts of CaCO3(s) or CaO(s)Heterogeneous EquilibriumLiquids SolutionsH2O(l) H2O(g) K=PH2OI2(s) I2(aq) K=[I2]Heterogeneous EquilibriumRelationships Among the K’s of Related Reactions#1: The equilibrium constant for a reverse reaction is always the reciprocal of the equilibrium constant for the corresponding forward reaction.2 H2 (g) + O2 (g) 2 H2O (g)(PH2O)2(PH2)2(PO2)= K12 H2O (g) 2 H2 (g) + O2 (g) (PH2)2(PO2)(PH2O)2= K2K1 = 1/K21KKor K1Krevforrevfor==#1#2aA + bB cC + dD cC + dD aA + bB versus# 2: When the coefficients in a balanced chemical equation are all multiplied by a constant factor, the corresponding equilibrium constant is raised to a power equal to that factor.(PH2O)(PH2)(PO2)½= K3K3 = K1½ Relationships Among the K’s of Related Reactions2 H2 (g) + O2 (g) 2 H2O (g) Rxn 1#1H2 (g) + ½ O2 (g) H2O (g) Rxn 3 = Rxn 1 times 1/2#3(PH2O)2(PH2)2(PO2)= K1# 3: when chemical equations are added to give a new equation, their equilibrium constants are multiplied to give the equilibrium constant associated with the new equation.2 BrCl (g) ↔ Br2 (g) + Cl2 (g)(PBr2)(PCl2)(PBrCl)2Br2 (g) + I2 (g) ↔ 2 IBr (g)(PIBr)2(PBr2) (PI2)2 BrCl (g) + I2 (g) ↔ 2 IBr (g) + Cl2(g)= K1K2= (0.45)(0.051)=0.023 @ 25oCRelationships Among the K’s of Related Reactions(PBr2)(PCl2)(PBrCl)2= K1 = 0.45 @ 25oCX(PIBr)2(PBr2) (PI2)= K2 = 0.051 @ 25oC= K1K2 = K3wrong arrowwrong arrowwrong arrowCalculating Equilibrium ConstantsConsider the equilibrium 4 NO2(g)↔ 2 N2O(g) + 3 O2(g) The three gases are introduced into a container at partial pressures of 3.6 atm (for NO2), 5.1 atm (for N2O), and 8.0 atm (for O2) and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of the NO2 is measured to be 2.4 atm. Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur.4 NO2(g)


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