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Chemical Equilibrium
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  1. Chemical Equilibrium Equilibrium 2009-2010

  2. Chemical Equilibrium • Few chemical reactions proceed in only one direction. Most are reversible, at least to some extent. At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse process begins to take place and reactant molecules are formed from product molecules. Chemical equilibrium is achieved when the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant. Equilibrium 2009-2010

  3. Equilibrium is a dynamic process. • Physical equilibrium H2O(s) H2O(l) At equilibrium, liquid water is reforming ice as fast as ice is melting. Equilibrium 2009-2010

  4. Chemical equilibrium N2O4(g) 2NO2(g) colorless brown Equilibrium is a state in which there are no observable changes as time goes by. When a chemical reaction has reached the equilibrium state, the concentrations of reactants and products remain constant over time, and there are no visible changes in the system. Equilibrium 2009-2010

  5. The Law of Mass Action For a reversible reaction at equilibrium and at a constant temperature, the ratio of reactant and product concentrations has a constant value, k (the equilibrium constant). • Note that although the concentrations may vary, as long as a given reaction is at equilibrium and the temperature does not change, according to the law of mass action, the value of k remains constant. Equilibrium 2009-2010

  6. We can generalize this phenomenon with the following reversible reaction: aA + bB cC + dD where a, b, c and d are the stoichiometric coefficients for the reacting species A, B, C, and D. [C]c[D]d [A]a[B]b k = • kc indicates that the concentrations of the reacting species are expressed in molarity or moles per liter. • kp is used when equilibrium concentrations are expressed in terms of pressure.

  7. For the reaction, N2O4(g) 2NO2(g) k = at 25 oC [NO2]2 [N2O4] • Note that the exponent 2 for [NO2] in this expression is the same as the stoichiometric coefficient for NO2 in the balanced chemical equation. • k is defined as having no units.

  8. Changes in the concentrations of NO2 and N2O4 with time, in three situations. • Initially only NO2 is present • Initially only N2O4 is present • Initially a mixture of NO2 and N2O4 is present • Note that even though equilibrium is reached in all cases, the equilibrium concentrations of NO2 and N2O4 are not the same as in the previous trials. Equilibrium 2009-2010

  9. Solving for k, the equilibrium constant The following table shows some experimental data for the NO2-N2O4 system at 25 oC. The gas concentrations are expressed in molarity, which can be calculated from the number of moles of the gases present initially and at equilibrium and the volume of the flask in liters. Equilibrium 2009-2010

  10. 0.0851 0.102 0.0967 0.0880 0.227 The NO2-N2O4 System at 25 oC Analysis of the equilibrium data shows that the ratio [NO2]/[N2O4] gives scattered values.

  11. 0.0851 0.102 0.0967 0.0880 0.227 The NO2-N2O4 System at 25 oC Analysis of the equilibrium data shows that the ratio [NO2]2/[N2O4] gives a nearly constant value that averages 4.63 x 10-3

  12. The magnitude of the equilibrium constant tells us whether an equilibrium reaction favors the products or reactants. • If k is much greater than 1 (in this context, any number greater than 10 is considered to be much greater than 1) the equilibrium will lie to the right and favors the products. • Conversely, if the equilibrium constant is much smaller than 1 (any number less than 0.1 is much less than 1), the equilibrium will lie to the left and favor the reactants. Equilibrium 2009-2010

  13. The concentrations of reactants and products in gaseous reactions can also be expressed in terms of their partial pressures. For the equilibrium process N2O4(g) 2NO2(g) we can write (PNO2)2 (PN2O4) kp = Where PNO2 and PN2O4 are the equilibrium partial pressures (in atm) of NO2 and N2O4, respectively.

  14. Example – Write the equilibrium constant expression kc and kp if applicable, for the following system: (NH4)2Se(s) 2NH3(g) + H2Se(g) [NH3]2[H2Se] [(NH4)2Se] k = • The “concentration” of a solid or a liquid is a constant, (an intensive property that does not depend on how much of the substance is present) and can be omitted when writing equilibrium expressions. kc = [NH3]2[H2Se] kp = (PNH3)2(PH2Se)

  15. In general, kc is not equal to kp, since the partial pressures of reactants and products are not equal to their concentrations expressed in moles per liter. • A simple relationship exists between kp and kc kp = kc(RT)Dn where Dn = moles of gaseous product – moles of gaseous reactants R = the gas constant 0.0821 L.atm/mol. K and T = temperature in Kelvins Equilibrium 2009-2010

  16. Determine kp for the following reaction at 220 oC. The kc for this reaction at 220 oC is 10.5 CO(g) + 2H2(g) CH3OH(g) Equilibrium 2009-2010

  17. Multiple Equilibria The reactions we have considered so far are all relatively simple. A more complicated situation is one in which the product molecules in one equilibrium system are involved in a second equilibrium process: aA + bB cC + dD k’c cC + dD eE + fF k”c aA + bB eE + fF kc kc = k’c. k”c Equilibrium 2009-2010

  18. Among the many known examples of multiple equilibra is the ionization of diprotic acids in aqueous solutions. The following equilibrium constants have been determined for carbonic acid at 25 oC. H2CO3(aq) H+(aq) + HCO3 -(aq) k’c= 4.2 x 10-7 HCO3-(aq) H+(aq) + CO3-2(aq) k’’c= 4.8 x 10-11 H2CO3(aq) 2H+(aq) + CO3 -2(aq) kc= ? Equilibrium 2009-2010

  19. The balanced chemical equation must be written when quoting the numerical value of an equilibrium constant. • If a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. • The value of k also depends on how the equilibrium equation is balanced. (If you double the chemical equation throughout, the corresponding equilibrium constant will be the square of the original value.) Equilibrium 2009-2010

  20. The equilibrium constant, kc, for the reaction 2HCl(g) H2(g) + Cl2(g) Is 4.17 x 10-34 at 25 oC. What is the equilibrium constant for the reaction ½H2(g) + ½Cl2(g) HCl(g) at the same temperature? Equilibrium 2009-2010

  21. Q, the Reaction Quotient The equilibrium constant kc for the formation of hydrogen iodide from molecular hydrogen and molecular iodine in the gas phase H2(g) + I2(g) 2HI(g) is 54.3 at 430 oC. Suppose that in a certain experiment we place 0.243 mole of H2, 0.146 mole of I2, and 1.98 moles of HI all in a 1.00 L container at 430 oC. Will there be a net reaction to form more H2 and I2 or more HI? Q= Equilibrium 2009-2010

  22. Because the quotient [HI]2/[H2][I2] is greater than kc, this system is not at equilibrium. Consequently, some of the HI will react to form more H2 and I2 (decreasing the value of the quotient). Thus the net reaction proceeds from right to left to reach equilibrium. • For reactions that have not reached equilibrium, such as the formation of HI considered above, we obtain the reaction quotient (Qc), instead of the equilibrium constant by substituting the initial concentrations into the equilibrium constant expression. Equilibrium 2009-2010

  23. To determine the direction in which the net reaction will proceed to achieve equilibrium, we compare the values of Qc and kc. • If Qc > kc, the reaction proceeds from right to left (consuming products, forming reactants) to reach equilibrium • If Qc < kc, the reaction proceeds from left to right (consuming reactants, forming products) to reach equilibrium. • If Qc = kc, the initial concentrations are equilibrium concentrations. The system is at equilibrium. Equilibrium 2009-2010

  24. Calculating concentrations in an equilibrium system If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the equilibrium mixture from the initial concentrations. Let’s consider the following system involving two organic compounds, cis-stilbene and trans-stilbene cis-stilbene trans-stilbene The equilibrium constant for this system is 24.0 at 200 oC. Equilibrium 2009-2010

  25. cis- trans- stilbene stilbene Suppose that initially only cis-stilbene is present at a concentration of 0.850 mol/L. How do we calculate the concentrations of cis and trans-stilbene at equilibrium? Equilibrium 2009-2010

  26. cis- trans- stilbene stilbene When the system proceeds towards equilibrium, cis-stilbene will decrease by x, and trans-stilbene will increase by the same value, x, (because of the 1:1 ratio in the BCE). The equilibrium concentration of cis-stilbene is initial concentration minus x; trans-stilbene will be initial concentration plus x. Equilibrium 2009-2010

  27. [trans-stilbene] [cis-stilbene] kc = Next, we set up the equilibrium constant expression

  28. Factors that Affect Equilibrium Chemical equilibrium represents a balance between forward and reverse reactions. In most cases, this balance is quite delicate. Changes in experimental conditions (concentration, pressure, volume and temperature) may disturb the balance and shift the equilibrium position so that more or less of the desired product is formed. When we say that an equilibrium position shifts to the right, for example, we mean that the net reaction proceeds from left to right. Equilibrium 2009-2010

  29. Le Chatelier’s Principle There is a general rule that helps us to predict the direction in which an equilibrium reaction will move when a change in concentration, pressure, volume, or temperature occurs. The rule, known as Le Chatelier’s principle, states that if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset. The word “stress” here means a change in concentration, pressure, volume, or temperature that removes a system from the equilibrium state. Equilibrium 2009-2010

  30. Changes in concentration FeSCN+2(aq) Fe+3(aq) + SCN-(aq) red pale yellow colorless An aqueous solution of Fe(SCN)3 Equilibrium 2009-2010

  31. FeSCN+2(aq) Fe+3(aq) + SCN-(aq) red pale yellow colorless What happens if we add some sodium thiocyanate, NaSCN, to the solution? In this case, the stress applied to the equilibrium system is an increase in the concentration of SCN- (from the dissociation of NaSCN). To offset this stress, some Fe+3 ions react with the added SCN- ions, and the equilibrium shifts from right to left, and the red color of the solution deepens. Equilibrium 2009-2010

  32. FeSCN+2(aq) Fe+3(aq) + SCN-(aq) red pale yellow colorless Similarly, if we added iron (III) nitrate to the original solution, the red color would also deepen because the additional Fe+3 ions (from the Fe(NO3)3 solution) would shift the equilibrium from right to left. Equilibrium 2009-2010

  33. FeSCN+2(aq) Fe+3(aq) + SCN-(aq) red pale yellow colorless Now suppose we add some oxalic acid, H2C2O4, to the original solution. Oxalic acid ionizes in water to form the oxalate ion, C2O4-2, which binds strongly to the Fe+3 ions. This formation of Fe(C2O4)3-3 removes free Fe+3 ions in solution. Consequently, more FeSCN+2 units dissociate and the equilibrium shifts from left to right, and the solution will turn yellow. Equilibrium 2009-2010

  34. Changes in Volume and Pressure Changes in pressure do not affect the concentrations of solids, liquids or aqueous solutions because they are virtually incompressible. On the other hand, concentrations of gases are greatly affected by changes in pressure. Equilibrium 2009-2010

  35. Suppose that the equilibrium system N2O4(g) 2NO2(g) is in a cylinder fitted with a movable piston. What happens if we increase the pressure on the gases by pushing down on the piston at constant temperature? Brown NO2(g) and colorless N2O4(g) in equilibrium. Equilibrium 2009-2010

  36. Since the volume decreases, the concentration of both NO2 and N2O4 increases. Because the concentration of NO2 is squared in the equilibrium constant expression, the increase in pressure increases the numerator more than the denominator. The system is no longer at equilibrium, so we write [NO2]2 [N2O4] Qc = Equilibrium 2009-2010

  37. Brown NO2(g) and colorless N2O4(g) in equilibrium. • When the volume in the syringe is suddenly decreased, the concentration of NO2 and N2O4 are both increased, (indicated by the darker brown color). • A few seconds after the sudden volume decrease, the color is much lighter brown as the equilibrium shifts from the brown NO2(g) to colorless N2O4(g). Qc > Kc, and the net reaction will shift to the left until Qc = kc. (a) (b) (c)

  38. In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases (the reverse reaction, in this case), and a decrease in pressure (increase in volume) favors the reaction that increases the total number of moles of gases. For the reactions in which there is no change in the number of moles of gases, a pressure (or volume) change has no effect on the position of equilibrium. Equilibrium 2009-2010

  39. We can also increase the pressure of a system without changing its volume by adding an inert gas (helium, for example) to the equilibrium system. Adding helium to the equilibrium mixture at constant volume increases the total gas pressure and decreases the mole fractions of both NO2 and N2O4; but the partial pressure of each gas, given by the product of its mole fraction and total pressure, does not change. Thus the presence of an inert gas does not affect the equilibrium. Equilibrium 2009-2010

  40. Changes in Temperature A change in concentration, pressure, or volume may alter the equilibrium position, but it DOES NOT change the value of the equilibrium constant (k). Only a change in temperature can alter the equilibrium constant. Equilibrium 2009-2010

  41. 2NO2(g) N2O4(g) DH = -58.0 kJ 2NO2(g) N2O4(g) + heat The formation of NO2 from N2O4 is an endothermic process. Another way to write this reaction would be N2O4(g) 2NO2(g) DH = 58.0 kJ N2O4(g) + heat 2NO2(g) The formation of N2O4 from NO2 is an exothermic process. This process could be written or Equilibrium 2009-2010

  42. What will happen if the following system N2O4(g) 2NO2(g) is heated at constant volume? Equilibrium 2009-2010

  43. Since endothermic processes absorb heat from the surroundings, heating favors the dissociation of colorless N2O4 into brown NO2 molecules, (an endothermic process). N2O4(g) 2NO2(g) Equilibrium 2009-2010

  44. Another way of looking at this is that the stress is too much heat, so the reaction will shift to the right, away from the heat. N2O4(g) + heat 2NO2(g) Equilibrium 2009-2010

  45. Cooling favors the exothermic process (the formation of colorless N2O4). N2O4(g) 2NO2(g) Equilibrium 2009-2010

  46. In other words, cooling the system results in too little heat, so the shift would be towards the heat. N2O4(g) + heat 2NO2(g) Equilibrium 2009-2010

  47. The Effect of a Catalyst A catalyst enhances the rate of a reaction by lowering the reaction’s activation energy of the forward and reverse reaction to the same extent. We can therefore conclude that the presence of a catalyst does not alter the equilibrium constant, nor does it shift the position of an equilibrium system. Equilibrium 2009-2010

  48. Adding a catalyst to a reaction mixture that is not at equilibrium will simply cause the mixture to reach equilibrium sooner. The same equilibrium mixture could be obtained without the catalyst, but we might have to wait much longer for it to happen. Equilibrium 2009-2010

  49. ksp, Solubility Equilibria Precipitation reactions are important in industry, medicine, and everyday life. Although useful, the solubility rules do not allow us to make quantitative predictions about how much of a given ionic compound will dissolve in water. Equilibrium 2009-2010

  50. Consider a saturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s) Ag+(aq) + Cl-(aq) Because salts are considered strong electrolytes, all the AgCl that dissolves in water is assumed to dissociate completely into Ag+ and Cl- ions. Equilibrium 2009-2010