Equilibrium Keeping your balance

Equilibrium Systems at equilibrium are subject to two opposite processes occurring at the same rate Establishment of equilibrium 2NO2  N2O4 Start with 1.00 mol N2O4 and allow it to react As NO2 appears, N2O4disappears The reaction slows and comes to an apparent stop with 0.2 moles NO2 present and 1.6 moles N2O4 present

Equilibrium Equilibrium animation

Equilibrium At equilibrium, G = 0. Neither process is more spontaneous that 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

Types of equilibrium systems Phase equilibria Vapor – liquid H2O(l)  H2O(g) liquid – solid H2O(s)  H2O(l) Solution equilibrium Solid in liquid – saturated salt water NaCl(s)  NaCl(aq)

Types of equilibrium systems Gas in liquid – dissolved oxygen O2(aq)  O2(g) Gas phase reaction equilibria H2 + I2  2HI Acid-base equilibria HF + H2O  H3O+ + F- NH3 + H2O  NH4+ + OH-

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”). PCl3(g) + Cl2(g)  PCl5(g) Keq = [PCl5]/ [PCl3][Cl2]

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 H2 + I2  2HI Keq = [HI]2/[ H2][ I2] Solids and pure liquids do not appear in the equilibrium constant expression because their activity is constant.

Equilibrium constant expression Example: Nitrogen dioxide dimerizes to form dinitrogen tetroxide as follows: N2O4(g)  2NO2(g) N2O4 is placed in a flask and allowed to reach equilibrium at a temperature where Kp = 0.133. At equilibrium, PN2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2. Solution: Kp = 0.133 = PNO22/PN2O4, and PN2O4 = 2.71atm. Substitute and rearrange: 0.133 = PNO22/2.71, so PNO2 = [0.133(2.71)] = 0.600atm

Solubility equilibrium expressions For solution of ionic materials NaCl(s)  Na+(aq) + Cl- (aq) Concentration of solid does not appear, so Ksp = [Na+][Cl-] Solubilities can be calculated from Ksp, 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.

Solubility product constant Solution: PbCl2  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 = 4x3 x = 3.6x10-4M

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: Ksp = 1.8x10-10 = (x)(0.100 + 2x)2 Ignore 2x since it is small compared to 0.100 Ksp = 1.8x10-10 = (x)(0.100)2 x = 1.8x10-8M Check answer

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 H2 + I2  2HI What is the effect of adding additional hydrogen gas? More HI is formed – position of equilibrium is shifted “right”

LeChatelier’s Principle Example: You introduce 0.230 mole HI into a 1.00L flask and allow it to come to equilibrium. If Keq for the reaction is 1.75, find the concentrations of H2, I2 and HI. Solution: Keq = [HI]2/[ H2][ I2] = 1.75 [H2] = [I2] = x [HI] = (0.230 – 2x) 1.75 = (0.230-2x)/x2 x = [H2] = [I2] = 0.105M; [HI] = (0.230-x) = 0.125M

LeChatelier’s Principle Example part 2. Find the new concentrations if 0.100 mole H2 is added to the reaction vessel from the previous problem. Solution. Make an ICE chart. [H2] [I2] [HI] Initial 0.105M 0.125M Change + 0.100 - x - x + 2x Equilibrium 0.105 + 0.100 - x 0.105 - x 0.125 + 2x

LeChatelier’s Principle Plug the equilibrium values into the equilibrium constant expression and solve for x. 1.75 = [HI]2/[H2][I2] = (0.125+2x)2/(0.205-x)(0.105-x) 1.75x2 - 0.543x + 0.0377 = 0.0156 + 0.500x + 4x2 0 = 2.25x2 + 1.04x - 0.0221 x = 0.0204 [H2] = 0.205-x = 0.185M [I2] = 0.105-x = 0.085M [HI] = 0.125+2x = 0.166M

LeChatelier’s Principle Check your answer by re-evaluating the equilibrium constant at the new position. 0.1662/(0.185)(0.085) = 1.75  Solubility (common-ion effect) example: George adds 0.0500g solid NaC2H3O2 to 50.0mL saturated silver acetate (Ksp = 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

LeChatelier’s Principle [Ag+] [C2H3O2-] Initial 0.045M Change - x + 0.0122 - x Equilibrium 0.045 - x 0.057 - x Ksp = 2.0x10-3 = (0.045–x)(0.057–x) 0 = x2 – 0.102x + 0.000565 x = 0.005878 (change in molarity = amt of ppt) Represents 2.9x10-4mol or 0.049g Check answer

LeChatelier’s Principle Change in temperature will shift the position of equilibrium so as to favor the endothermic direction 2NO2(g)  N2O4(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.

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? Cr2O7-2(aq) + H2O(l)  2CrO4-2(aq) + H3O+(aq) H2(g) + Br2(g)  2HBr(g) Ca(OH)2(s)  CaO(s) + H2O(g) N2(g) + 3H2(g)  2NH3(g)