1. Keeping your balance

2. Equilibrium Systems at equilibrium are subject to two opposite processes occurring at the same rate Establishment of equilibrium 2NO 2  N 2 O 4 Start with 1.00 mol N 2 O 4 and allow it to react As NO 2 appears, N 2 O 4 disappears The reaction slows and comes to an apparent stop with 1.6 moles NO 2 present and 0.2 moles N 2 O 4 present

3. Equilibrium Equilibrium animation

4. Equilibrium At equilibrium,  G = 0. Neither process is more spontaneous than the other. For the amounts present at equilibrium, considerations of enthalpy and entropy exactly cancel each other The amounts present (the “position of equilibrium”) can be changed by changing the temperature

5. Types of equilibrium systems Phase equilibria Vapor – liquid H 2 O (l)  H 2 O (g) liquid – solid H 2 O (s)  H 2 O (l) Solution equilibrium Solid in liquid – saturated salt water NaCl (s)  NaCl (aq)

6. Types of equilibrium systems Gas in liquid – dissolved oxygen O 2(aq)  O 2(g) Gas phase reaction equilibria H 2 + I 2  2HI Acid-base equilibria HF + H 2 O  H 3 O + + F - NH 3 + H 2 O  NH 4 + + OH -

7. Equilibrium constant expression This is a characteristic of an equilibrium system that governs the position of equilibrium The value is temperature dependent Equilibrium constant is the concentrations of the products divided by the concentrations of the reactants at equilibrium (“Law of Mass Action”). PCl 3(g) + Cl 2(g)  PCl 5(g) K eq = [PCl 5 ]/ [PCl 3 ][Cl 2 ]

8. Equilibrium constant expression In higher order reactions (those with coefficients greater than 1) the concentrations are raised to the power of the coefficients of the reactants and products H 2 + I 2  2HI K eq = [HI] 2 /[ H 2 ][ I 2 ] Solids and pure liquids do not appear in the equilibrium constant expression because their activity is constant.

9. Equilibrium constant expression Example: Nitrogen dioxide dimerizes to form dinitrogen tetroxide as follows: N 2 O 4(g)  2NO 2(g) N 2 O 4 is placed in a flask and allowed to reach equilibrium at a temperature where K p = 0.133. At equilibrium, P N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO 2. Solution: K p = 0.133 = P NO2 2 /P N2O4, and P N2O4 = 2.71atm. Substitute and rearrange: 0.133 = P NO2 2 /2.71, so P NO2 =  [0.133(2.71)] = 0.600atm

10. Solubility equilibrium expressions For solution of ionic materials NaCl (s)  Na + (aq) + Cl - (aq) Concentration of solid does not appear, so K sp = [Na + ][Cl - ] Solubilities can be calculated from K sp, since at equilibrium the solution is saturated Example: Calculate the molar solubility of lead (II) chloride at 25ºC if the solubility product constant is 1.8x10 -10 at that temperature.

11. Solubility product constant Solution: PbCl 2  Pb +2 + 2Cl - Ksp = [Pb +2 ][Cl - ] 2 = 1.8x10 -10 [Pb +2 ] = molar solubility [Cl - ] = 2[Pb +2 ] Set [Pb +2 ] = x 1.8x10 -10 = x(2x) 2 = 4x 3 x = 3.6x10 -4 M

12. Common ion effect The presence of a common ion decreases the solubility of a salt. Example: Find the molar solubility of lead (II) chloride in 0.100M HCl. Assume no changes in volume. Solution: K sp = 1.8x10 -10 = (x)(0.100 + 2x) 2 Ignore 2x since it is small compared to 0.100 K sp = 1.8x10 -10 = (x)(0.100) 2 x = 1.8x10 -8 M Check answer

13. LeChatelier’s Principle Disturbing an equilibrium system results in a shift in the position of equilibrium until equilibrium is reestablished at a new position Change in concentration – when the concentration of one product or reactant is increased, the equilibrium will shift to use up some of the excess material H 2 + I 2  2HI What is the effect of adding additional hydrogen gas? More HI is formed – position of equilibrium is shifted “right”

14. LeChatelier’s Principle Example: You introduce 0.230 mole HI into a 1.00L flask and allow it to come to equilibrium. If K eq for the reaction is 1.75, find the concentrations of H 2, I 2 and HI. Solution: K eq = [HI] 2 /[ H 2 ][ I 2 ] = 1.75 [H 2 ] = [I 2 ] = x [HI] = (0.230 – 2x) 1.75 = (0.230-2x) 2 /x 2 x = [H 2 ] = [I 2 ] = 0.0692M; [HI] = (0.230-2x) = 0.0916M

15. LeChatelier’s Principle Example part 2. Find the new concentrations if 0.100 mole H 2 is added to the reaction vessel from the previous problem. Solution. Make an ICE chart. [H 2 ][I 2 ][HI] Initial0.0692M 0.0916M Change+ 0.100 - x- x+ 2x Equilibrium0.0692 + 0.100 - x0.0692 - x0.0916 + 2x

16. LeChatelier’s Principle Plug the equilibrium values into the equilibrium constant expression and solve for x. 1.75 = [HI] 2 /[H 2 ][I 2 ] = (0.0916+2x) 2 /(0.1692-x)(0.0692-x) 1.75x 2 - 0.417x + 0.0205 = 0.00839 + 0.366x + 4x 2 0 = 2.25x 2 + 0.783x - 0.0121 x = 0.0148 [H 2 ] = 0.1692-x = 0.154M [I 2 ] = 0.0692-x = 0.0544M [HI] = 0.0916+2x = 0.121M

17. LeChatelier’s Principle Check your answer by re-evaluating the equilibrium constant at the new position. 0.121 2 /(0.154)(0.054) = 1.75  Solubility (common-ion effect) example: George adds 0.0500g solid NaC 2 H 3 O 2 to 50.0mL saturated silver acetate (K sp = 2.0x10 -3 ). How much silver acetate precipitates? Assume no volume changes. Solution: Find concentration of saturated silver acetate solution.  (2.0x10 -3 ) = [Ag+] = [silver acetate] = 0.045M

18. LeChatelier’s Principle K sp = 2.0x10 -3 = (0.045–x)(0.057–x) 0 = x 2 – 0.102x + 0.000565 x = 0.005878 (change in molarity = amt of ppt) Represents 2.9x10 -4 mol or 0.049g Check answer [Ag + ][C 2 H 3 O 2 - ] Initial0.045M Change- x+ 0.0122 - x Equilibrium0.045 - x0.057 - x

19. LeChatelier’s Principle Change in temperature will shift the position of equilibrium so as to favor the endothermic direction 2NO 2(g)  N 2 O 4(g) + 57.2 kJ The “+57.2kJ” indicates that heat has left the system, thus the forward direction is exothermic. Raising the temperature means more energy put into the system, and shifting to the left uses the excess energy and relieves the stress.

20. LeChatelier’s Principle Changes in pressure – increasing the pressure in gas systems favors the side with fewer particles. If both products and reactants have the same number of particles, no change results. Which of the following systems would produce more products as a result of increased pressure? Cr 2 O 7 -2 (aq) + H 2 O (l)  2CrO 4 -2 (aq) + H 3 O + (aq) H 2(g) + Br 2(g)  2HBr (g) Ca(OH) 2(s)  CaO (s) + H 2 O (g) N 2(g) + 3H 2(g)  2NH 3(g)