Chapter 11 equilibrium lecture notes
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Chapter 11 equilibrium lecture notes

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  • When a reaction is endothermic, an increase in the temperature (as heat is put into the system) causes its equilibrium to shift toward the formation of more products; and a decrease in the temperature causes its equilibrium to shift in the direction of more reactants. When a reaction is exothermic, an increase in the temperature causes its equilibrium to shift in the direction of more reactants; and a decrease in the temperature causes its equilibrium to shift toward the formation of more products.
  • When a reaction is endothermic, an increase in the temperature (as heat is put into the system) causes its equilibrium to shift toward the formation of more products; and a decrease in the temperature causes its equilibrium to shift in the direction of more reactants. When a reaction is exothermic, an increase in the temperature causes its equilibrium to shift in the direction of more reactants; and a decrease in the temperature causes its equilibrium to shift toward the formation of more products.

Chapter 11 equilibrium lecture notes Chapter 11 equilibrium lecture notes Presentation Transcript

  • Chapter 11 Chemical Equilibrium11.1 The Equilibrium Condition11.2 The Equilibrium Constant11.3 Equilibrium Expressions Involving Pressures11.4 The Concept of Activity11.5 Heterogeneous Equilibria11.6 Applications of the Equilibrium Constant11.7 Solving Equilibrium Problems11.8 Le Chateliers Principle11.9 Equilibria Involving Real Gases
  • The Equilibrium Condition (General)Thermal equilibrium indicates two systems in thermal contactwith each do not exchange energy by heat. If two bricks are inthermal equilibrium their temperatures are the same.Chemical equilibrium indicates no unbalanced potentials (ordriving force). A system in equilibrium experiences no changeover time, even infinite time.The opposite of equilibrium systems are non-equilibriumsystems that are off balance and change with time.Example 1 atm O2 + 2 atm H2 at 298K
  • The Equilibrium Condition (Chem Rxn)aA + bB cC + dDThe same equilibrium stateis achieved whetherstarting with pure reactantsor pure products.The equilibrium state canchange with temperature.
  • The Equilibrium State (Chem Rxn)H2O (g) + CO (g) H2 (g) + CO2 (g) Change [CO] to PCO [H2O] to PH2O etc
  • Chemical Reactions and EquilibriumAs the equilibrium state is approached, theforward and backward rates of reactionapproach equality. At equilibrium the ratesare equal, and no further net change occursin the partial pressures of reactants orproducts.Fundamental characteristics of equilibrium states: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.5. No change over time.
  • Arrows: Chemical Symbolism Use this in an equilibrium expression.↔ Use this to indicate resonance.
  • Chemical Reactions and Equilibrium The 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) @ 25oC Temperature (oC) Vapor Pressure (atm) 15.0 0.01683 17.0 0.01912 K = 0.03126 19.0 0.02168H2O(l) H2O(g) @ 30oC 21.0 0.02454 23.0 0.02772 K = 0.04187 25.0 0.03126 30.0 0.04187 50.0 0.1217
  • Law of Mass Action (1)Partial pressures and concentrations of products appear in thenumerator and those of the reactants in the denominator.Each is raised to a power equal to its coefficient in thebalanced chemical equation. aA + bB cC +if gases dD if concentrations c d c d( PC ) ( PD ) [C] [D] =K a b a b =K( PA ) ( PB ) [ A] [B]
  • Law of Mass Action (2)1. Gases enter equilibrium expressions as partial pressures, inatmospheres. 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 equilibriumexpressions by the number 1 (unity); a solvent taking part in achemical reaction is represented by unity, provided that thesolution is dilute. E.g., I2(s) ↔ I2(aq) [I2 (aq) ] = K I 2 ( s ) ↔ I 2 ( aq ) [ I 2 (aq )] [ I 2 ( aq )] K= = = [ I 2 ( aq )] [ I 2 ( s )] 1
  • Activities The concept of Activity (i-th component) = ai = Pi / P referenceH2O (l) H2O (g) Kp = P H2O @ 25oC Kp = 0.03126 atm PH2O =K Pref is numerically equal to 1 Pref K = 0.03126 The 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.
  • The Equilibrium StateH2O (g) + CO (g) H2 (g) + CO2 (g) PH 2 PCO 2 Kp = PH 2 O PCO
  • The Equilibrium Expressions aA + bB cC + dDIn a chemical reaction in which a moles of species A and b moles ofspecies B react to form c moles of species C and d moles of species D, The partial pressures at equilibrium are related through K = PcCPdD/PaAPbB
  • Write equilibrium expressions for thefollowing 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)
  • Heterogeneous EquilibriumGases and CaCO3(s) CaO(s) + CO2(g)Solids K=PCO2 K is independent of the amounts of CaCO3(s) or CaO(s)
  • Heterogeneous EquilibriumLiquids H2O(l) H2O(g) K=PH2O I2(s) I2(aq)Solutions K=[I2]
  • Relationships Among the K’s of Related Reactions#1: The equilibrium constant for a reverse reaction is alwaysthe reciprocal of the equilibrium constant for thecorresponding forward reaction. 1 K for = or K for K rev = 1 K revaA + bB cC + dD versus cC + dD aA + bB (PH2O)2#1 2 H2 (g) + O2 (g) 2 H2O (g) (PH2)2(PO2) = K1 K1 = 1/K2 2 H2 (g) + O2 (g) (PH2) (PO22) 2#2 2 H2O (g) = K2 (PH2O)
  • Relationships Among the K’s of Related Reactions # 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)2#1 2 H2 (g) + O2 (g) 2 H2O (g) Rxn 1 (PH2)2(PO2) = K1#3 H2 (g) + ½ O2 (g) H2O (g) Rxn 3 = Rxn 1 times 1/2 (PH2O) = K3 K3 = K1½ (PH2)(PO2)½
  • Relationships Among the K’s of Related Reactions # 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) = K1 = 0.45 @ 25oC wrong arrow (PBrCl)2 Br2 (g) + I2 (g) ↔ 2 IBr (g) (PIBr)2 wrong arrow = K2 = 0.051 @ 25oC (PBr2) (PI2)2 BrCl (g) + I2 (g) ↔ 2 IBr (g) + Cl2(g) wrong arrow = K1K2 = (0.45)(0.051) (PBr2)(PCl2) (PIBr)2 X (P ) (P ) = K1K2 = K3 (PBrCl)2 Br2 I2 =0.023 @ 25oC
  • Calculating 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 3.6atm (for NO2), 5.1 atm (for N2O), and 8.0 atm (for O2) and react to reachequilibrium at a fixed temperature. The equilibrium partial pressure of theNO2 is measured to be 2.4 atm. Calculate the equilibrium constant of thereaction at this temperature, assuming that no competing reactions occur. 4 NO2(g) ↔ 2 N2O(g) + 3 O2(g) initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm)
  • Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur. 4 NO2(g) ↔ 2 N2O(g) + 3 O2(g) initial partial pressure (atm) 3.6 5.1 8.0 change in partial pressure (atm) – 4x +2x +3x equilibrium partial pressure (atm) 2.4 5.1 + 2x 8.0 + 3x 5.1 + 2(0.3 atm) = 5.7 atm N2O3.6 – 4x = 2.4 atm NO2; x = 0.3 atm 8.0 + 3(0.3 atm) = 8.9 atm O2 (PN2O)2(PO2)3 K= = (PNO2)4
  • The compound GeWO4(g) forms at high temperature in the reaction 2 GeO (g) + W2O6(g) ↔ 2 GeWO4(g)Some GeO (g) and W2O6 (g) are mixed. Before they start to react,their partial pressures both equal 1.000 atm. After their reaction atconstant temperature and volume, the equilibrium partial pressure ofGeWO4(g) is 0.980 atm. Assuming that this is the only reaction thattakes place, (a) determine the equilibrium partial pressures ofGeO and W2O6, and (b) determine the equilibrium constant forthe reaction. 2 GeO (g) + W2O6 (g) ↔ 2 GeWO4(g) initial partial pressure (atm) 1.000 1.000 0 change in partial pressure (atm) – 2x –x +2x equilibrium partial pressure (atm)
  • (a) determine the equilibrium partial pressures of GeO and W2O6, and(b) determine the equilibrium constant for the reaction. 2 GeO(g) + W2O6(g) ↔ 2 GeWO4(g) initial partial pressure (atm) 1.000 1.000 0 change in partial pressure (atm) –2x –x +2x equilibrium partial pressure (atm) 1.000 – 2x 1.000 – x 0.980 0 + 2x = 0.980 atm GeWO4; 1.000 – 2(0.490) = 0.020 atm GeO x = 0.490 atm 1.000 – 0.490 = 0.510 atm W2O6 (PGeWO4)2 K = (P )2(P GeO W2O6 ) =
  • • Skip Solving quadratic equations• Will utilize approximation method – Systems that have small equilibrium constants. – Assume “x” (the change in concentration) is small (less than 5%) of the initial concentration.
  • A vessel holds pure CO (g) at a pressure of 1.282 atm and a temperature of354K. A quantity of nickel is added, and the partial pressure of CO (g)drops to an equilibrium value of 0.709 atm because of the reaction Ni (s) + 4CO (g) ↔ Ni(CO)4 (g)Compute the equilibrium constant for this reaction at 354K. PNi(CO)4 PNi(CO)4K= 4 = (PCO ) [Ni(s)] (PCO ) 4 (1) Ni (s) + 4CO (g) ↔ Ni(CO)4 (g)Construct an “ICE” table P CO (atm) P Ni(CO)4 (atm) initial partial pressure (atm) 1.282 0 change in partial pressure (atm) -4x +1x equilibrium partial pressure (atm) 0.709 x At equil. Pco= x = PNi(CO)4 =
  • Equilibrium CalculationsAt a particular temperature, K = 2.0 x 10-6 mol/L for thereaction 2CO2 (g) 2CO (g) + O2 (g)If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculatethe equilibrium concentrations of all species. 2CO2 (g) 2CO (g) + O2 (g) initial partial pressure (mol/L) 0.4 0 0 change in partial pressure (mol/L) – 2x +2x +1x equilibrium partial pressure (mol/L) 0.4 -2x 2x 1x
  • At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction 2CO2 (g) 2CO (g) + O2 (g) If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. 2CO2 (g) 2CO (g) + O2 (g) initial partial pressure (mol/L) 0.4 0 0 change in partial pressure (mol/L) – 2x +2x +1x equilibrium partial pressure (mol/L) 0.4 -2x 2x 1x [CO]2 [O 2 ] (2 x) 2 ( x)Kp = = 2.0x10-6 = [CO 2 ]2 (0.40 − 2 x) 2 check assumptionassume 2x << 0.40 2x 2(4.3x10-3) = x 100 = 2.2% 0.40 0.40 (2x) 2 (x) assumpiton valid, less than 5%2.0x10-6 = ⇒ x = 4.3x10-3 M (0.40) 2
  • At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction 2CO2 (g) 2CO (g) + O2 (g) If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. 2CO2 (g) 2CO (g) + O2 (g) initial partial pressure (mol/L) 0.4 0 0 change in partial pressure (mol/L) – 2x +2x +1x equilibrium partial pressure (mol/L) 0.4 -2x 2x 1x [CO]2 [O 2 ]Kp = = 2.0x10-6 [CO 2 ]2 [CO 2 ] = 0.40 - 2x = (2 x) 2 ( x)= (0.40 − 2 x) 2 = 0.40 - 2(4.3x10-3 ) = 0.39Mx = 4.3x10-3M -3 [CO] = 2x = 2(4.3 x 10 ) -3[O 2 ] = x = 4.3 x 10 M = 8.6 x 10-3 M
  • Non-Equilibrium Conditions: The Reaction Quotient (1) φορωαρδ  → aA + bB ρεϖερσε cC + dD wrong arrow ←  (PC ) (PD ) c d (PC ) (PD ) c d =Q ? =K Q= K(PA ) (PB ) a b (PA ) (PB ) a b K (the Equilibrium Constant) uses equilibrium partial pressures Q (the reaction quotient) uses prevailing partial pressures, not necessarily at equilibrium
  • The Reaction Quotient (2) φορωαρδ arrow→  aA + bB ρεϖερσε cC + dD wrong ←  (PC )c (PD )d =Q ? Q= K (PC )c (PD )d =K(PA )a (PB )b (PA )a (PB )b If Q < K, reaction proceeds in a forward direction (toward products); If Q > K, reaction proceeds in a backward direction (toward reactants); If Q = K, the reaction is in equilibrium.
  • The equilibrium constant for the reaction P4(g) ↔ 2 P2(g) is1.39 at 400oC. Suppose that 2.75 mol of P4(g) and 1.08 mol ofP2(g) are mixed in a closed 25.0 L container at 400oC. ComputeQ(init) (the Q at the moment of mixing) and state the direction inwhich the reaction proceeds. ? (PP 2 )2 = K = 1.39 Q= K PV = nRT (PP 4 )1 K = 1.39 @ 400oC; nP4(init) = 2.75 mol; nP2(init) = 1.08 mol PP4(init) = nP4(init)RT/V = [(2.75mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L) = 6.08 atm PP2(init) = nP2(init)RT/V = [(1.08mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L) = 2.39 atm Q=
  • Henri Louis Le Châtelier (1850-1936)Highlights – 1884 Le Chateliers Principle: A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress – If a chemical system at equilibrium experiences a change in concentration, temperature or total pressure the equilibrium will shift in order to minimize that change. – Industrial chemist involved with industrial efficiency and labor-management relationsMoments in a Life – Le Chatelier was named "chevalier" (knight) of the Légion dhonneur in 1887, decoration established by Napoléon Bonaparte in 1802.
  • Effects of External Stresses on Equilibria: Le Châtelier’s PrincipleA system in equilibrium that is subjected to astress reacts in a way that counteracts the stress.Le Châtelier’s Principle provides a way to predict theresponse of an equilibrium system to an externalperturbation, such as…1. Effects of Adding or Removing Reactants or Products2. Effects of Changing the Volume (or Pressure) of the System3. Effects of Changing the Temperature
  • Effects of Adding or Removing Reactants or Products PCl5(g) PCl3(g) + Cl2(g) K = 11.5 @ 300oC = Q add extra PCl5(g) add extra PCl3(g) remove some PCl5(g) remove some PCl3(g)A system in equilibrium that is subjected to a stress reacts in away that counteracts the stress. In this case adding orremoving reactants or products
  • Effects of Changing the Volume of the System PCl5(g) PCl3(g) + Cl2(g) 1 mole 1+1 = 2 moles Let’s decrease the volume of the reaction container Less room :: less amount (fewer moles) Shifts reaction to restore equilibrium Let’s increase the volume of the reaction container More room :: more amount (greater moles) Shifts reaction to restore equilibriumA system in equilibrium that is subjected to astress reacts in a way that counteracts the stress.In this case a change in volume
  • Volume Decreased Volume Increased (Pressure Increased) (Pressure Decreased) Equilibrium shift right Equilibrium Shifts leftV reactants > V products (toward products) (toward reactants) Equilibrium Shifts left Equilibrium shift rightV reactants < V products (toward reactants) (toward products)V reactants = V products Equilibrium not affected Equilibrium not affected 2 P2(g) P4 (g) PCl5(g) PCl3(g) + Cl2(g) CO (g) +H2O (g) CO2 (g) + H2 (g)Boyles Law:PV = Constant A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in volume (or pressure)
  • Effects of Changing the Temperature Endothermic: heat is aborbed by a reaction Reactants + heat gives Products Exothermic: heat is liberated by a reaction Reactants gives Products + heatA system in equilibrium that is subjected to a stress reactsin a way that counteracts the stress. In this case a changein temperature
  • Effects of Changing the TemperatureEndothermic: absorption of heat by a reaction Reactants + heat gives ProductsLet’s increase the temperature of the reaction, whatdirection does the equilibrium reaction shiftLet’s decrease the temperature of the reaction
  • Effects of Changing the TemperatureExothermic: heat liberated by a reaction Reactants ↔ Products + heat Let’s increase the temperature of the reaction Let’s decrease the temperature of the reactionA system in equilibrium that is subjected to a stress reactsin a way that counteracts the stress. In this case a changein temperature
  • Temperature Raised Temperature LoweredEndothermic Equilibrium shift right Equilibrium Shifts left Reaction (toward products) (toward reactants)(absorb heat) Exothermic Equilibrium Shifts left Equilibrium shift right Reaction (toward reactants) (toward products)(liberate heat) If a forward reaction is exothermic, forward aA + bB → cC + dD + Heat Then the reverse reaction must be endothermic reverse cC + dD + Heat  aA + bB → φορωαρδ  → aA + bB ρεϖερσε cC + dD ←  
  • Driving Reactions to Completion/ Increasing YieldIndustrial Synthesis of Ammonia (Haber)N2 (g) + 3H2 (g) ↔ 2NH3 (g)Forward reaction is exothermicWhat conditions do we need to increase the yield, i.e.,produce more ammonia? Volume Decreased Volume Increased (Pressure Increased) (Pressure Decreased)V reactants > Equilibrium shift Equilibrium Shifts left right (towardV products (toward products) reactants) Temperature Raised Temperature Lowered Exothermic Equilibrium Shifts left Equilibrium shift right Reaction (toward (toward (liberate heat) reactants) products)