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16 Chemical Equilibrium

16 Chemical Equilibrium. Reactions seldom go complete such that one of the reactants is exhausted as we have discussed in stoichiometry and limiting reagent. Instead, they are mostly reversible. For example, CO (g) + 3 H 2 (g)  CH 4 (g) + H 2 O (g)

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16 Chemical Equilibrium

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  1. 16 Chemical Equilibrium Reactions seldom go complete such that one of the reactants is exhausted as we have discussed in stoichiometry and limiting reagent. Instead, they are mostly reversible. For example, CO (g) + 3 H2 (g)  CH4 (g) + H2O (g) The two-head arrow (or equal signs =) indicates that the reaction goes both ways. Similarly, N2O4 (colorless gas)  2 NO2 (brown gas) (show a film) When the forward and reverse reaction rates are equal, the system is said to be at equilibrium. 16 Chemical Equilibrium

  2. The NASA Computer program CEA (Chemical Equilibrium with Applications) calculates chemical equilibrium compositions and properties of complex mixtures. Applications include assigned thermodynamic states, theoretical rocket performance, Chapman-Jouguet detonations, and shock-tube parameters for incident and reflected shocks. 16 Chemical Equilibrium

  3. 1 mol CH42 mol S2T = 1000 KP = 1.5 atm Countours show G/RT Equilibrium occur at minimum G/RT 16 Chemical Equilibrium

  4. A System The environment or The surrounding Scientific experiments usually investigate a system, which is isolated from its environment or surrounding. The system can be a nucleus, an atom, a molecule, a plant, an animal, an experimental setup, the Earth, the solar system, the galaxy, or the universe. Matter and energy can be transferred into or out of a system. The open or closed system 16 Chemical Equilibrium

  5. Chemical Equilibrium – a dynamic process Chemical equilibrium is a state of a system when reaction rates in both directions are equal. Changes continue at the molecular level, but the macroscopic properties stay the same. Equilibrium is a dynamic rather than static process. Heat is energy flowing from a high temperature object to a low temperature object. Equilibrium is reached when the temperatures are the same. Molecules still exchange kinetic energies during collision. Water flows from a high potential-energy place to a low potential-energy place. When potential energies are the same, water stops flowing. Water-molecule diffusion continues. 16 Chemical Equilibrium

  6. Some Typical Equilibria Macroscopic properties such as vapor pressure, solubility, distribution of a substance in two immiscible liquids are typical equilibria. Vapor pressure of a liquid is the pressure of the vapor in equilibrium with a liquid. It is a function of temperature. Solubility, amount in solution in equilibrium with solid, depends on the substance and the solvent, often also on temperature. Distribution (or equilibrium) coefficients of a substance in immiscible liquids are useful guides for the separation of mixtures into components. 16 Chemical Equilibrium

  7. The reaction quotients For any chemical reaction in a system, a A + b B  c C + d D we can always write a quotient, [C]c [D]dQ = —————— (Q is reaction quotient) [A]a [B]bfor example, [NO2]2Q = —————— for N2O4 2 NO2 [N2O4] Know the definition of Q. Text: 16-5 defines Q & 16-1 defines K. 16 Chemical Equilibrium

  8. Expressions for reaction quotients For the ionization of an acid, H2SO4 = 2 H+ + SO42- what is the reaction quotient expression? Hint: The reaction quotient is [H+]2 [SO42-] —————— = Q [H2SO4] If M is used for the concentrations, what are the units for Q? 16 Chemical Equilibrium

  9. More expressions for reaction quotients [H2O]2Q = ————— [H2]2 [O2] 2 H2 + O2 2 H2O [H2]2 [O2] 1Q r = ————— = —— [H2O]2Q 2 H2O  2 H2 + O2 [H2O]4Q2 = ————— = Q 2 [H2]4 [O2]2 4 H2 + 2 O2 4 H2O Q =  Q2Note the relationships All relations are due to definitions. See 16-3 16 Chemical Equilibrium

  10. Equilibrium constant - mass action law When a chemical reaction in a closed system is at equilibrium, the reaction quotient is defined as the equilibrium constant, K, which depends on T. For a chemical reaction in a closed system, a A + b B  c C + d Dthen [C]c [D]dQ = —————— = K(at equilibrium) [A]a [B]bUnits of K depends on units of [ ]s, M(a+b-c-d) if M is used. If Q < K, the reaction goes forward | to increase [product]if Q = K, the system is at equilibrium |the law of mass actionif Q > K, the reaction goes backward | to decrease [product] 16 Chemical Equilibrium See 16-5 & 16-3

  11. Evaluating K The equilibrium mixture is found to contain 0.07, 0.11, 0.03 and 0.03 moles of CO, H2, CH4 and H2O respectively for the reaction: CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)What is the equilibrium constant, K? Solution CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)start 0.07 0.11 0.03 0.03 (0.03) (0.03)K = —————— = 9.7 (0.07) (0.11)3 What governs the quantitative relationship among equilibrium mixtures? 16 Chemical Equilibrium

  12. Evaluate quantities When 0.10 moles CO and 0.20 moles of H2 are placed in a 1-L vessel at temperature T and allowed to come to equilibrium. The mixture is found to contain 0.03 moles H2O for the reaction: CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)What is the equilibrium constant, K? Solution (note the strategy please) CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)start 0.10 0.20 0 0change - x - 3x +x +x (x=0.03)eqlbm 0.1-x 0.2-3x 0.03 0.03[ ] 0.07 0.11 0.03 0.03 K = (0.03) (0.03) / (0.07) (0.11)3 = 9.7 16 Chemical Equilibrium

  13. Evaluate quantities When 0.08 moles CO, 0.14 moles of H2, and 0.08 moles of CH4are placed in a 1-L vessel at temperature T and allowed to come to equilibrium. The mixture is found to contain 0.01 moles H2O for the reaction: CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)What is the equilibrium constant, K? Solution (note the strategy please) CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)start 0.08 0.14 0.08 0change - x - 3x 0.08+x +x (x=0.01)eqlbm 0.08-0.01 0.14-0.03 0.08+0.010.01[ ] 0.07 0.11 0.09 0.01 K = (0.09) (0.01) / (0.07) (0.11)3 = 9.7 16 Chemical Equilibrium

  14. Entropy We all know that heat flow from a high-T object into a low-T object. This type of change can be described as due to increase of entropy. Entropy S is the amount of energy transferred, E, divided by T.ES = ———— or T S = ET Entropy increases for spontaneous changes (reactions). Entropy is also associated with randomness of ordering. What causes change when no apparent energy is involved? 16 Chemical Equilibrium

  15. Entropy calculation A system consists 10.0 L of water at 300 K and 3.0 g of ice at 273 K. Heat of fusion of ice (molar mass 18) is 6 kJ/mol. What is the change in entropy when the system is at equilibrium. Solution:Since the amount of heat for melting the ice is 6 *(3/ 18) = 1.0 kJ too little to change the temperature of the entire system appreciably from 300 K. Thus the entropy change is – 1/300 + 1/273 = 3.3e – 4 kJ/K = 0.33 J/K (entropy unit) What is the molar entropy for ice melting? +6000 J mol–1 / 273 K = 22 J mol –1 K –1 (entropy unit mol–1) 16 Chemical Equilibrium

  16. Josiah W. Gibbs (1839-1903) New Haven, Connecticut, USA First Doctorate (1863 Yale) in Engineering of U.S.A. Studied in Europe, influenced by Kirchhoff & Helmholtz. 1871 Professor of mathematics and physics at Yale Unassuming in manner, genial and kindly in his intercourse with his fellow-men, never showing impatience or irritation, devoid of personal ambition of the baser sort or of the slightest desire to exalt himself, he went far toward realizing the ideal of the unselfish, Christian gentleman. In the minds of those who knew him, the greatness of his intellectual achievements will never overshadow the beauty and dignity of his life. - Bumstead lunar crater Gibbs 16 Chemical Equilibrium

  17. Gibbs free energy, G or G Following the definition of entropy S, Gibbs defined a function together with enthalpy of change, H. G = H – T S now called Gibbs free energy. or G = H – T S for differences of these quantities Since H is negative and S is positive for spontaneous change, Gibbs free energy is negative for all spontaneous changes, and it unified enthalpy and entropy for changes. Gibbs free energy is the maximum amount of available energy in any change. Similar to Ho, Gois the Gibbs free energy at the standard T and P. What drives physical and chemical changes? 16 Chemical Equilibrium

  18. Energy and equilibrium constant K A quantity called Gibbs free energy, DG, is defined as the maximum amount of energy from a chemical reaction for doing work.DG is negative for spontaneous reactionDG is positive for non-spontaneous reaction The DG is related to the reaction quotient Q, DG = DGo + R T ln Q When a system is at equilibrium there is no available energy, DG = DGo + R T ln K = 0Thus,DGo = – R T ln K, ln K = – DGo / R T K = exp (–DGo / R T) See 20-2 to 20-6 page 784-808 plse 16 Chemical Equilibrium

  19. More expressions for equilibrium constant Note these relationships in light of ln K = – DGo / R T [H2O]2K = ————— [H2]2 [O2] 2 H2 + O2 2 H2O [H2]2 [O2] 1K r = ———— = —— [H2O] 2K 2 H2O  2 H2 + O2 [H2O]4K2 = ————— = K 2 [H2]4 [O2]2 4 H2 + 2 O2 4 H2O  K2 = K 16 Chemical Equilibrium

  20. An Example Hydrogen iodide, HI, decomposes, 2 HI  H2 + I2 The amount of I2 can be determined from the intensity of the violet color of I2. When 4.00 mol of HI was placed in a 5.00-L vessel at 458oC, the equilibrium mixture was found to contain 0.442 mol I2. What is the value for K for the decomposition of HI at this temperature? See 16-4 and 16-7 for this and the following examples. 16 Chemical Equilibrium

  21. Analysis of Example Hydrogen iodide, HI, decomposes, 2 HI  H2 + I2 The amount of I2 can be determined from the intensity of the violet color of I2. When 4.00 mol of HI was placed in a 5.00-L vessel at 458oC, the equilibrium mixture was found to contain 0.442 mol I2. What is the value for K for the decomposition of HI at this temperature? Analysis of the problem: [HI]0 = 4.0 mol / 5.00 L = 0.800 M [I2]equilibrium = 0.442 mol / 5.0 L = 0.0884 M Know your strategy  16 Chemical Equilibrium

  22. Calculate Kc Example 14.3 The equilibrium concentration of various species are given below their symbols, 2 HI (g) = H2 (g) + I2 (g)start 0.800 0 0change = ? = ? = 0.0884 M from experimentequlbm 0.800-2x x x value 0.6232 0.0884 0.0884 M 0.0884*0.0884Kc = ——————— = 0.0201 (0.6232)2 Skill: re-interpret question. Evaluate other quantities and K from known start conditions and experimental results Evaluate DGo 16 Chemical Equilibrium

  23. Evaluate KcCompare with slide 41 At 300 K and 1 atm, an equilibrium mixture contains 70% NO2 by mass in a tube of NO2 and N2O4. Evaluate Kc for the reaction, N2O4 = 2 NO2 The challenge (hint): assume 100 g total mass (70 g of NO2 and 30 g N2O4) find amount in mole for each (1.522 mol NO2; 0.326 mol N2O4) find volume from pressure for all (45.49 L) find concentration from volume (0.0335 M NO2; 0.00717 M) evaluate Kc(equation dependent) 16 Chemical Equilibrium

  24. Application of Kc For N2O4 (g) = 2 NO2 (g) Kc = 4.20e-3 at 300 K,What are the equilibrium concentrations if 0.100 mol of N2O4 is placed in a 5.00-L flask? Strategy:write down the equation figure out the initial condition assume the change to be x figure out the equilibrium concentrationuse the concentration-Kc relationshipsolve for x and then check the results. N2O4 (g) = 2 NO2 (g)0.0200-x 2 x (2x)2 ———— = 4e-3 (K) 0.0200-x 4x2 + 0.0042 x – 0.0000926 = 0 -0.00463 + {(4.63e-3)2 – 4*4*(-9.265e-5)}x = ———————————————————— = 4.27e-3 2*4 N2O4 (g) = 2 NO2 (g)0.0157 8.54e-3 M 16 Chemical Equilibrium

  25. Approximations Evaluate equilibrium concentrations of all for the reaction condition as given in the start: (at T1) N2 (g) + 3 H2 (g)  2 NH3 (g) K = 1e-5start 1.0 2.0 0change - x - 3x +2x (= ?) eqlbm 1.0-x 2.0-3x2x (2x)2—————— = 1e-5 (1-x)(2.0-3x)3 4x2—————— = 1e-5 (1.0-x)(2.0-3x)3 x2 = 1e-5*1.0*2.03/4 = 2e-5 x = 4.5e-3 (justified) Results:[N2] = 0.995 M[H2] = 1.99[NH3] = 9e-3 Checking (2*4.5e-3)2————— = 1e-50.995*1.993 Discuss 16 Chemical Equilibrium

  26. Trial and error or Newtons method When 0.10 moles CO and 0.20 moles of H2 are placed in a 1-L vessel at temperature T and allowed to come to equilibrium. CO (g) + 3 H2 (g)  CH4 (g) + H2O (g) Kc = 9.7What are the concentrations of various species? Solution(note the strategy please) CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)start 0.10 0.20 0 0change - x - 3x +x +x eqlbm 0.1-x 0.2-3x x x (= ?) Kc = (x)2 / (0.1-x) (0.2-3x)3 = 9.7Too difficult to solve.You need not worry.Trial-and-error (Newton’s method) results shown>> x = Kc =0.05* 4000.06 15174 0.04 520.03 11.30.029 8.210.0299 9.50*x < 0.1 16 Chemical Equilibrium Why cannot use successive approximation? The first x=0.44 is unreasonable.

  27. Apply math skill Exactly 0.0100, 0.0100 and 0.0200 moles of H2, I2, and HI are mixed in a 1-L vessel, and then raised to 277 K, at which Kc = 50.3. What concentration will increase?Determine from Qc What are the equilibrium concentrations of H2, I2 and HI? Solution(summary) H2 (g) + I2 (g) = 2 HIstart 0.0100 0.0100 0.0200change - x - x + 2 x assumeeqlbm [ ] 0.01-x 0.01-x 0.02+2 xvalue 0.0044 0.0044 0.0312 M x = 0.00560 (0.02+2x)2 (0.02+2x) ————— = 50.3 > > note > ————— = 7.09 (0.01-x)2 (0.01-x) 16 Chemical Equilibrium

  28. Successive approximation For 2 H2S (g) = 2 H2 (g) + S2 (g) Kc = 4.2e-6 at 831oCWhat is [S2] when 0.0700 mol H2S is allowed to reach equilibrium in a 1.00-L vessel. Solution: 2 H2S (g) = 2 H2 (g) + S2 (g) 0.0700-2x 2 x x (2x)2 x ————— = 4.2e-6 x = 3(4.2e-6*0.07002/4) = 1.73e-3 (0.0700-2x)2 assume 0 (2x)2 x ————— = 4.2e-6 x = 3(4.2e-6*0.06652/4) = 1.67e-3 (0.0700-2*0.00173)2 successive inclusion Carry out the calculation plse 4x3(0.07-2*0.00167) 16 Chemical Equilibrium = 4.2e-6 x = 3(4.2e-6*0.06672/4) = 1.67e-3

  29. Equilibria involving solvents Since [solvent] is essentially constant, it does not appear in Kc expression. You just have to know and apply. CH3COOC2H5 (aq) + H2O (l) = CH3COOH (aq) + C2H5OH (aq) [CH3COOH] [C2H5OH] K’c = ———————————— [CH3COOC2H5] [ H2O] [CH3COOH] [C2H5OH] Usually, Kc = K’c [ H2O] = ———————————— [CH3COOC2H5] 16 Chemical Equilibrium

  30. Equilibria involving solvents - application For the reaction,CH3COOC2H5 (aq) + H2O (l) = CH3COOH (aq) + C2H5OH (aq)Kc = 0.25 at 25oC. Calculate [C2H5OH] at equilibrium starting with 1.00 M of ethyl acetate. Solution:CH3COOC2H5 (aq) + H2O (l) = CH3COOH (aq) + C2H5OH (aq)1.00 – x x x= [C2H5OH ] x2———— = 0.25(1.00 – x) x2 + 0.25 x – 0.25 = 0 – 0.25 + 0.252– 4(-0.25)x = ———————————— = 0.39 M 2 16 Chemical Equilibrium

  31. Equilibrium Calculations – a summary Several methods have been introduced in evaluating x Simple methodApproximationTrial and error (Newtons method)Recognize math relationship (taking square root on both sides)Successive approximation 16 Chemical Equilibrium

  32. Kpand Kc;Kp= Kc (R T) {c+d-(a+b)} The equilibrium constants depends on the units used for quantities. In general, the equilibrium constant is represented by K When concentrations are used, Kc is the notation;when partial pressures are used, Kp is used Since PV = n R T, P = (n/V) R T = [ ] R T [ ]= (n / V) is the concentration For the reaction a A + b B  c C + d D PCc PDd [C]c [D]d (R T)c + dKp = ———— = ———————— = Kc (R T){c+d-(a+b)} PAa PBb [A]a [B]b (R T)a + b See page 635 16 Chemical Equilibrium

  33. Kpmore appropriate for H2O (l) = H2O (g) Kp = vapor pressure of water H2O (s) = H2O (g) Kp = vapor pressure of ice At the triple point, it’s obvious that vapor pressure of ice = vapor pressure of waterif these are equal, the three phases are present. CaCO3 (s) = CaO + CO2Kp = partial v. p. of CO2 16 Chemical Equilibrium

  34. Kc and Kp conversion For 2 H2S (g) = 2 H2 (g) + S2 (g) Kc = 4.2e-6 at 831oCWhat is Kp? Solution:Kp =Kc (R T){c+d – (a+b)} (recall or derive) = Kc (R T)(2+1 – 2) = 4.2e-6 *8.314 kJ mol – 1 K – 1 * (831+273) K = 3.9e-2 2 H2S (g) = 2 H2 (g) + S2 (g) Kpconfirmation0.0667 3.34e-3 1.67e-3 Kc = 4.2e-6 at 831oC612 kPa 30.7 15.3 Kp = 3.9e-2 J mol 1 K 1 at 831oC6.04 atm 0.30 0.15 Kp = 3.7e-4 L atm mol 1 K 1 16 Chemical Equilibrium

  35. Le Chatelier’s Principle The system will restore its equilibrium after changing conditions. An open bottle of water will dry eventually, because the partial pressure never reach equilibrium (removing vapor pressure). Increase pressure will increase the formation of NH3 in N2 (g) + 3 H2 (g) = 2 NH3 (g) Hof =  91.8 kJIncrease T will decrease the formation of NH3 in this exothermic reaction. How does [CH4] change when pressure increase for an equilibrium of CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)?Increasing pressure tends to reduce the number of molecules. Study sections: 16.6 16 Chemical Equilibrium

  36. Le Chatelier’s Principle – cont. Possible changes: Temperature, T Increase T favors forward reaction in an endothermic reaction but favors reverse reaction in an exothermic reaction Pressure, P and volume V Pressure and volume are related. Decreases of volume increase the pressure Amounts of product and reactants Increase amounts of reactants, products or both increases P and V 16 Chemical Equilibrium

  37. Henri Louis Le Chatelier (1850-1936) A mining engineer investigated structure of metals and alloys, translated work of Gibbs into French stated Henricus Van’t Hoff’s (Nobel Prize 1901) theory inword andwrote on industrial efficiency and labor management relations Scientist would be wise to refrain from using the mathematical equation unless he understands the theory that it represents, and can make a statement about the theory that does not consist just in reading the equation. L. Pauling 16 Chemical Equilibrium

  38. Effect of a catalyst E There are two factors regarding the speed of reactions: kinetics and thermodynamics. Kc and Kp are thermodynamic parameters, related to DG o. Thus, the equilibrium constants are not affected by a catalyst. Catalyst only affect reaction constant k, which is related to the activation energy. kinetic thermodynamic Reaction See page 643-646 16 Chemical Equilibrium

  39. Equilibria - Summary Apply stoichiometry to equilibrium mixture in a system Write reaction quotient and equilibrium constant expression. Rewrite Qs and Ks when the equations change.Predict direction of reaction by comparing Q and K. Evaluate Ks from some know conditions. Know how energies H and G are related to equilibrium constants. Use various methods to evaluate concentrations of an equilibrium system. Apply Le Chatelier’s principle to predict changes in P, [ ], and T. Separate kinetic and thermodynamic factors of reactions. 16 Chemical Equilibrium

  40. Equilibrium review -1 Recall that DG = DGo + R T ln K = 0and DGo = – R T ln K, orln K = – DGo / R T Since DGo refers to the standard condition (1 atm and 273K), units for pressures to calculate Kp should theoretically be in atm. However, the standard condition is redefined from time to time, and one has to be aware of the relationship here for engineering calculations. In Chem123, we take a simple approach to specify the units and adhere to the unit atm for pressures. 16 Chemical Equilibrium

  41. Equilibrium review – 2Compare with slide 23 At 600K, the partial pressure of NO2 in a equilibrium mixture is 0.10 atm, N2O4 = 2 NO2Kp = 3.06 L atm mol 1 K 1 What is the partial pressure of N2O4? What is the percent of dissociation of N2O4?What percentage of molecules are NO2? Solution: N2O4 = 2 NO2Kp = 3.06 L atm mol 1 K 1x 0.10 let x be the unknown 0.102 / x = 3.06; x = 0.12/3.06 = 0.0033 atm P(N2O4)percent dissociated = 0.05 / (0.0033+0.05) = 0.94 = 94%% of NO2 molecule = 0.10/(0.10+0.0033) = 0.97 = 97% 16 Chemical Equilibrium

  42. At 1123K, Kc = 0.153 for the reaction: C(s) + CO2 (g) = 2 CO (g) Ho = 172.5 kJWhat is Kp?If initially the partial pressure of CO2 was 1.50 atm, what is the total pressure at equilibrium?How will the equilibrium shift if temperature is decreased to 900K? Solution: Equilibrium review - 3 16 Chemical Equilibrium

  43. At 1123K, Kc = 0.153 for the reaction: C(s) + CO2 (g) = 2 CO (g) Ho = 172.5 kJWhat is Kp?If initially the partial pressure of CO2 was 1.50 atm, what is the total pressure at equilibrium?How will the equilibrium shift if temperature is decreased to 900K? Solution: R in L-atm mol-1 K-1Kp = 0.153 (R T)1 = 0.153(0.08206*1123) = 14.1C(s) + CO2 (g) = 2 CO (g) 1.50-x 2xequilibrium pressure(2x)2 / (1.50 – x) = 14.1; 4x2 + 14.1x –21.2 = 0x = {- 14.1 + (14.12 + 4*4*21.1)} / (2*4) = 1.13 atmtotal pressure = 1.50-x+2x = 1.50+x = 1.50 + 1.13 = 2.63 atmP(CO2) = 1.50 – 1.13 = 0.37 atm; P(CO) = 2*1.13 = 2.26 P(CO) decreases at 900K (<1123K) in the endothermic reaction (+ Ho). shift left Equilibrium review - 3 16 Chemical Equilibrium

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