Chemistry 2000 Lecture 11: Chemical equilibrium

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1 Chemistry 2000 Lecture 11: Chemical equilibrium Marc R. Roussel February 4, 2019 Marc R. Roussel Chemical equilibrium February 4, / 27

2 Equilibrium and free energy Thermodynamic criterion for equilibrium Recall that under given reaction conditions (concentrations of reactants and products), r G m = r G m + RT ln Q If r G m < 0, the reaction is thermodynamically allowed as written. If r G m > 0, the reverse of the reaction is thermodynamically allowed. What happens if r G m = 0? Neither the forward nor reverse direction of the reaction is thermodynamically allowed, so the reaction is at equilibrium. Marc R. Roussel Chemical equilibrium February 4, / 27

3 Equilibrium and free energy The equilibrium constant At equilibrium, r G m = r G m + RT ln Q = 0 r G m = RT ln Q This last equation implies that there is a fixed value of the reaction quotient Q when we reach equilibrium. We call this value the equilibrium constant K. Thus, r G m = RT ln K For a given reaction, K is a number that depends only on the temperature. At equilibrium, Q = K. Marc R. Roussel Chemical equilibrium February 4, / 27

4 Equilibrium and free energy If we know K, we can calculate the standard free energy change for a reaction by r G m = RT ln K We can also rearrange this equation to calculate K from r G m: ln K = r Gm RT ( K = exp r Gm ) RT [exp(x) = e x ] Marc R. Roussel Chemical equilibrium February 4, / 27

5 Equilibrium and free energy r G m = RT ln K Important: K is related to the standard free energy change r G m (obtained from tables of standard free energies of formation), not to r G m, which is zero at equilibrium. Marc R. Roussel Chemical equilibrium February 4, / 27

6 Conditions for thermodynamically feasibility Thinking in terms of Q and K Roughly, Q = products reactants. If Q < K, then Q needs to grow to reach equilibrium. Q increases if we make more products and consume reactants. Conclusion: If Q < K, a reaction is thermodynamically allowed as written. Converse: If Q > K, the reverse of the reaction is thermodynamically allowed. Marc R. Roussel Chemical equilibrium February 4, / 27

7 Conditions for thermodynamically feasibility Conditions for thermodynamic feasibility r G m < 0 Q < K reaction thermodynamically allowed r G m = 0 Q = K equilibrium r G m > 0 Q > K reverse reaction thermodynamically allowed Marc R. Roussel Chemical equilibrium February 4, / 27

8 Nitrogen dioxide Example: Dimerization of NO 2 2NO 2(g) N 2 O 4(g) r G m = f G (N 2 O 4, g) 2 f G (NO 2, g) = (51.3) kj/mol = 4.8 kj/mol ( K = exp r Gm ) RT ( ) J/mol = exp ( J K 1 mol 1 )( K) = 6.96 (Note the double negative in the exponential.) Marc R. Roussel Chemical equilibrium February 4, / 27

9 Nitrogen dioxide Suppose that, at some particular point in time, p NO2 = 0.4 bar and p N2 O 4 = 1.8 bar. Is the reaction thermodynamically allowed under these conditions? Q = a N 2 O 4 (a NO2 ) 2 = p N 2 O 4 /p (p NO2 /p ) 2 = 1.8 (0.4) 2 = 11 Q > K = 6.96 Conclusion: The reaction will run backwards, i.e. N 2 O 4 will dissociate into NO 2. Marc R. Roussel Chemical equilibrium February 4, / 27

10 Nitrogen dioxide We could have come to the same conclusion by calculating r G m : r G m = r G m + RT ln Q = 4.8 kj/mol + ( kj K 1 mol 1 ) = 1.2 kj/mol > 0, ( K) ln(11) from which we also conclude that the reverse reaction is thermodynamically allowed. Marc R. Roussel Chemical equilibrium February 4, / 27

11 Nitrogen dioxide Calculating an equilibrium mixture 2NO 2(g) N 2 O 4(g) For the composition given above (0.4 bar NO 2, 1.8 bar N 2 O 4, we know that the reaction will run backwards, dissociating N 2 O 4, until equilibrium is reached. What is the equilibrium composition? Marc R. Roussel Chemical equilibrium February 4, / 27

12 Nitrogen dioxide 2NO 2(g) N 2 O 4(g) Use an Initial/Change/Equilibrium (ICE) table to figure out the equilibrium composition. NO 2 N 2 O 4 I C 2x x E x 1.8 x Substitute into the equilibrium relationship: and solve for x. K = 6.96 = a N 2 O 4 a 2 NO 2 = 1.8 x ( x) 2 Find x = , which gives p NO2 = 0.50 bar and p N2 O 4 = 1.75 bar. Marc R. Roussel Chemical equilibrium February 4, / 27

13 Acid ionization Acid ionization constants The acid ionization constant K a of (e.g.) hydrofluoric acid is the equilibrium constant for the reaction HF (aq) H + (aq) + F (aq), K a = Problem: Calculate f G (HF, aq) given f G (F (aq)) = kj/mol. Answer: f G (HF, aq) = kj/mol Marc R. Roussel Chemical equilibrium February 4, / 27

14 Ocean acidification ph Important definition: ph = log 10 (a H +) Marc R. Roussel Chemical equilibrium February 4, / 27

15 Ocean acidification ph of water in equilibrium with atmospheric CO 2 CO 2 in the atmosphere currently has a concentration of approximately ppm. CO 2 reacts with water according to CO 2(g) + H 2 O (l) HCO 3(aq) + H+ (aq) Equilibration with CO 2 therefore acidifies water. What is the ph of water that has been equilibrated with the atmosphere? Marc R. Roussel Chemical equilibrium February 4, / 27

16 Ocean acidification ph of water in equilibrium with atmospheric CO 2 Conversion from ppm to bar ppm can refer to mass or mole fraction. In the case of atmospheric CO 2, the concentration is a mole fraction ppm means that for every million molecules in the atmosphere, approximately are CO 2 molecules. Given the proportionality between n and p, this means that p CO2 = ( bar) = bar at average sea level pressure. Marc R. Roussel Chemical equilibrium February 4, / 27

17 Ocean acidification ph of water in equilibrium with atmospheric CO 2 From these data, calculate CO 2(g) + H 2 O (l) HCO 3(aq) + H+ (aq) f G /kj mol 1 CO 2(g) HCO 3(aq) H 2 O (l) r G = 44.7 kj mol 1 K = Marc R. Roussel Chemical equilibrium February 4, / 27

18 Ocean acidification ph of water in equilibrium with atmospheric CO 2 Take a H2 O 1. CO 2(g) + H 2 O (l) HCO 3(aq) + H+ (aq) K = (a HCO )(a H +) 3 (a CO2 )(a H2 O) a CO2 = (from our previous calculation) If we start with pure water, then a HCO = a H +. 3 Solve (a = H +) 2 ( ) to get a H + = ph = log 10 a H + = 5.61 Marc R. Roussel Chemical equilibrium February 4, / 27

19 Ocean acidification Ocean acidification The foregoing calculation shows that surface water acidity is linked to atmospheric CO 2 levels. More CO 2 = increased acidity In the sea (and other waterways) there are other equilibria coupled with this one, notably CO 2(g) + CO 2 3(aq) + H 2O (l) 2HCO 3(aq) which tends to moderate acidification, but doesn t stop it. Marc R. Roussel Chemical equilibrium February 4, / 27

20 Vapor pressure Vapor pressure of a pure substance Imagine taking a pure substance (solid or liquid), putting it in an air-tight container, removing all the air, then letting it come to equilibrium. The equilibrium pressure reached in this experiment is the vapor pressure of the substance and is due to evaporation of the substance. Note: It isn t really necessary to remove the air, but it makes the measurement easier. Also note that if the partial pressure of a substance is lower than the (equilibrium) vapor pressure, it will evaporate. If, on the other hand, the partial pressure is higher than the vapor pressure, it will condense. Marc R. Roussel Chemical equilibrium February 4, / 27

21 Vapor pressure Problem: Calculate the vapor pressure of pure water at 25 C. Species f G /kj mol 1 H 2 O (l) H 2 O (g) Answer: bar Marc R. Roussel Chemical equilibrium February 4, / 27

22 Vapor pressure Vapor pressures of solutions What about the vapor pressure of a solution? This could be due either to the solvent alone if the solute is (relatively) involatile (e.g. ionic compounds, sugar), or to a combination of the solvent and solute. Marc R. Roussel Chemical equilibrium February 4, / 27

23 Vapor pressure Vapor pressures of solutions Raoult s law Let s start by looking at the solvent, again using water as an example. Note that K = p(h 2 O, pure)/p. Notation: p(h 2 O, pure) p H 2 O H 2 O (l) H 2 O (g) K = a g a l = p H 2 O/p X H2 O p H2 O = p H 2 O X H 2 O or, in general, p solvent = p solvent X solvent This equation is known as Raoult s law. Marc R. Roussel Chemical equilibrium February 4, / 27

24 Vapor pressure Let s calculate the vapor pressure of an aqueous solution prepared by dissolving mol of sodium sulfate in kg of water at 60 C. The vapor pressure of pure water at this temperature is 149 Torr. n H2 O = 1000 g = mol g/mol n Na + = 2(1.875 mol) = mol n SO 2 4 = mol n H2 O X H2 O = n H2 O + n Na + + n SO 2 4 = = mol mol p H2 O = p H 2 O X H 2 O = (149 Torr)(0.9080) = 135 Torr Marc R. Roussel Chemical equilibrium February 4, / 27

25 Vapor pressure Vapor pressures of solutions Henry s law We can also consider the vapor pressure of a volatile solute or the solubility of a gas in a solvent using equilibrium theory. For either of these cases, we consider A (g) A (sol) or K = [A] = Kc p p A a(a, sol) a(a, g) = [A]/c p A /p [A] = k H p A This equation is known as Henry s law, and the constant k H is called the Henry s law constant. Marc R. Roussel Chemical equilibrium February 4, / 27

26 Vapor pressure Let us calculate the Henry s law constant for oxygen in water. The standard free energy of formation of an aqueous oxygen molecule is kj/mol. The process is O 2(g) O 2(aq) r G = f G (O 2, aq) f G (O 2, g) = kj/mol ( K = exp r Gm ) RT ( ) J/mol = exp ( J K 1 mol 1 )( K) = k H = Kc p = mol L 1 bar 1 Marc R. Roussel Chemical equilibrium February 4, / 27

27 Vapor pressure Suppose that the atmospheric pressure of oxygen is 0.18 bar (roughly the case in Lethbridge, 920 m above sea level). Then, in water at equilibrium with the atmosphere, we have [O 2 ] = k H p O2 = ( mol L 1 bar 1) (0.18 bar) = mol/l. That s (roughly) how much oxygen the fishies in the Oldman River have to live on. Marc R. Roussel Chemical equilibrium February 4, / 27

Chemistry 2000 Lecture 12: Temperature dependence of the equilibrium constant

Chemistry 2000 Lecture 12: Temperature dependence of the equilibrium constant Marc R. Roussel February 12, 2019 Marc R. Roussel Temperature dependence of equilibrium February 12, 2019 1 / 15 Temperature