Equilibrium

Dynamic Equilibrium Rate of forward reaction = rate of reverse reaction Changes will cause reaction to re-establish equilibrium. Dynamic b/c the molecules never quit moving DOES NOT mean that concentrations of reactants or products are the same.

Reversible Reactions Reactions are capable of moving forward or backward. “Favored” To the right—products are favored, more product will be made. To the left—reactants are favored, less product will be made. Neutral—neither reactants nor products are favored and there is roughly equal concentrations of both at a given point in time.

Equilibrium Constant—K K = equilibrium concentrations of products and reactants at a given set of conditions. Do not include solids or liquids in K expressions. Only variable that affects K is temperature No units

Significance of K K < 1 Reverse reaction is favored K = 1 Neither direction is favored K > 1 Forward reaction is favored If K < 1 to the 10-3, forward reaction is considered neglible. .

Expressing Equilibrium Constants Express the equilibrium constant for the reactions below: 1. 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g) 2. PCl5 (s)  PCl3 (l) + Cl2 (g) 3. CuSO4 ∙ 5H2O(s)  CuSO4(s) +5H2O(g)

Relationship between K and balanced equations Common modifications: Reverse the equation, invert K. Multiply coefficients, raise K to same factor. Add 2+ equations to obtain overall, multiply K’s by each other to obtain overall K.

Practice Predict the equilibrium constant for the first reaction shown here given the K values for the 2nd and 3rd reactions: CO2(g) + 3H2(g)  CH3OH(g) + H2O(g) K1 = ? CO(g) + H2O(g)  CO2(g) + H2(g) K2 = 1.0×105 CO(g) + 2 H2(g)  CH3OH(g) K3 = 1.4×107

2 N2 (g) + O2(g)  2 N2O(g) K = 1.2 × 10-35 Calculate the value of Kc for the reaction:  2 N2O(g) + 3 O2(g) 2 N2O4(g), using the following information.  2 N2 (g) + O2(g)    2 N2O(g) K = 1.2 × 10-35 N2O4(g)    2 NO2(g) K = 4.6 × 10-3   1 2 N2(g) + O2(g)   NO2(g) K = 4.1 × 10-9 Are products or reactants favored for the overall equation?  

Practice Consider the following reaction: N2(g) + 3H2(g)  2 NH3(g) K = 5.6 × 105 Calculate the equilibrium constant for the following reaction: NH3(g)  ½ N2(g) + 3/2 H2(g) K = ?

Practice Consider the following reaction: CO(g) + 2H2(g)  CH3OH(g) A reaction mixture at 780oC initially contains [CO] = 0.500 M and [H2] = 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of Kc?

More practice Consider the reaction: 2 NO(g) + Br2(g) 2 NOBr(g) Kp = 28.4 @ 298 K In a reaction mixture at equilibrium, the partial pressure of NO is 108 torr and that of Br2 is 126 torr. What is the partial pressure of NOBr in this mixture?

More practice Consider the following reaction: H2(g) + I2(g)  2 HI(g) Complete the table. Assume all concentrations are equilibrium concentrations in M. T (K) [H2] [I2] [HI] Kc 25 0.0355 0.0388 0.922 ---------- 340 ----------- 0.0455 0.387 9.6 445 0.0485 0.0468 50.2

Reaction Quotient (Q) Used to gauge progress of a reaction and predict the direction of change. Same format as equilibrium constant Keq = constant at a given temperature. Q = specific to current state of reaction

Reaction quotient (Q) Reaction containing both reactants and products: Q varies Q > K Q = K Q < K

Practice Consider the reaction and its equilibrium constant: I2(g) + Cl2(g)  2 ICl(g) K = 81.9 A reaction mixture contains PI2 = 0.144 atm, PCl2 = 0.102 atm, and PICl = 0.355 atm. Is the reaction at equilibrium? If not, in which direction will the reaction proceed?

Calculating K from measured concentrations ICE tables I = Initial C = Change E = Equilibrium Can use ICE tables to substitute values into equilibrium expression to calculate K.

A mixture of 0. 10 mol of NO, 0. 050 mol of H2, and 0 A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established: 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g) At equilibrium, [NO] = 0.062 M. a. Calculate the equilibrium concentrations of H2, N2, and H2O. b. Calculate Kc.

Practice with ICE For the reaction H2(g) + I2(g)  2HI(g) A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00L flask at 430oC. The Kc for this reaction is 54.3 at this temperature. Calculate the concentrations of H2, I2, and HI at equilibrium.

Practice Suppose the same reaction at the same temperature had initial concentrations of 0.00623 M, 0.00414 M, and 0.0224 M for H2, I2, and HI respectively. Calculate the concentrations of these species at equilibrium.

Simplifying Approximations Consider the reaction: 2H2S(g)  2 H2(g) + S2(g) Kc = 1.67 ×10-7 @ 800oC A 0.500 L reaction vessel initially contains 0.0125 mol of H2S. Find the equilibrium concentrations of H2 and S2.

K in terms of Pressure Expressed as Kp For gaseous reactants, concentration is proportional to partial pressure. Same as Kc, except we use parentheses and partial pressures. If Δn = 0, Kp = Kc

Le Châtelier’s Principle When a chemical system at equilibrium tends to try to maintain that equilbrium. Factors that effect equilibrium: Concentration Volume Temperature

Concentration Changes Increasing concentration of reactants shifts rxn to right. Increasing concentration of products shifts reaction to the left. Decreasing concentration of reactants shifts rxn to left. Decreasing concentration of products shifts rxn to right.

Concept Check Consider the following reaction: CaCO3(s)  CaO(s) + CO2(g) What is the effect of adding additional CO2 to the reaction mixture? What is the effect of adding additional CaCO3?

Volume Changes Decreasing the volume causes reaction to shift in direction that has fewer moles of gas particles. Increasing volume causes reaction to shift in direction that has more moles of gas particles.

Temperature Changes If reaction is exothermic: A B + heat Adding heat shifts rxn to the left If reaction is endothermic: A + heat  B Adding heat shifts rxn to the right.

Ksp (solubility product) A chemist adds 10.00 g of PbCl2 to 50.0 mL of water at 25oC and finds that only 0.22 g of the solid goes into solution. Find the solubility (in g/L and mol/L) of PbCl2 in water at 25oC. Calculate the Ksp of PbCl2 at 25oC.

From solubility product… Calculate the molar solubility of Cu(OH)2 at 25oC. Ksp = 2.2x10-20

Common Ion Effect Calculate the molar solubility of PbCl2 at 25oC (Ksp = 1.6x10-5) in a 0.100 M solution of KCl.

Precipitate formation Q < Ksp. The solution is unsaturated, and more of the ionic solid, if available, will dissolve. Q = Ksp. The solution is saturated and at equilibrium. Q > Ksp. The solution is supersaturated, and ionic solid will precipitate.

2015 FRQ Answer the following questions about the solubility of Ca(OH)2 (Ksp = 1.3x10-6) (a) Write a balanced chemical equation for the dissolution of Ca(OH)2(s) in pure water. (b) Calculate the molar solubility of Ca(OH)2 in 0.10 M Ca(NO3)2 .

Relating Kp and Kc Kp = Kc(RT)Δn Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxide in the atmosphere according to the equation: NO(g) + O2(g)  NO2(g) Kp = 2.2 × 1012 Find Kc for this reaction.

Relating to ∆G ∆G = -RTlnK R = 8.314 J/mol The overall reaction for the corrosion (rusting) of iron by oxygen is 4 Fe(s) + 3O2(g)2Fe2O3(s) Using the following data, calculate the equilibrium constant for this reaction at 25°C.