Chemical Equilibrium General Concepts Ch 14 Dynamic Equilibrium

Chemical Equilibrium - General Concepts (Ch. 14) Dynamic Equilibrium a. A+b. B ⇌ c. C+d. D This is called a reversible reaction. At Equilibrium: • Reaction is proceeding in both directions at the same rate. • There is no net change in concentrations of reactants and products.

Equilibrium Constant Expression (a) Mass action expression: Q= Q = “reaction quotient” [C]c[D]d [A]a[B]b where [A], [B], etc are “non-equilibrium” concentrations (b) At Equilibrium: Kc = [C]c[D]d [A]a[B]b “Law of Mass Action” where Kc = “equilibrium constant” [A], [B], etc. are the “equilibrium” concentrations! • If Q > K, then reaction will go backward to attain equilibrium. • If Q < K, then reaction will go forward to attain equilibrium.

Equilibrium Constant Expressions (c) Magnitude of K K ≈ [products]/[reactants] – if K is very large: mostly products at equilibrium – if K is very small: mostly reactants at equilibrium – if K ≈ 1: roughly equal conc’s of reactants and products (d) Manipulating equilibrium constant expressions – Reverse equation --> invert Kc – Add two equations --> multiply their Kc’s – Multiply equation by coefficient “n” --> (Kc)n

Equilibrium Constants for Gas Phase Reactions (a) since P ∝ molar conc. , the equilibrium constant can be expressed in terms of partial pressures (in atm): Kp = (Pc)c(Pd)d (Pa)a(Pb)b (b) relationship of Kc and Kp – based on ideal gas law… Kp = Kc(RT)Dn Dn = moles gaseous products - moles gaseous reactants R = 0. 0821 L atm / mol K

Heterogeneous Equilibria Concentrations of pure liquids and solids are effectively constants, ∴ they do not appear in K expressions For example: NH 3(aq) + H 2 O(l) ⇌ NH 4+(aq) + OH–(aq) Kc = [NH 4+][OH–]/[NH 3] Ca. O(s) + SO 2(g) ⇌ Ca. SO 3(s) Kp = 1/PSO 2

Equilibrium Calculations Two basic types of problems: (a) Given the equilibrium concentrations (or partial pressures), determine Kc (or Kp) General Method: Put conc values into K expression! Or, if necessary, set up a “concentration table” based reaction stoichiometry, e. g. on PCl 3(g) + Cl 2(g) ⇌ PCl 5(g) In an experiment, 0. 20 mol of PCl 3 and 0. 10 mol Cl 2 were placed in a 1. 00 L container. After the reaction had come to equilibrium the concentration of PCl 3 was found to be 0. 12 mole/L. Determine the Kc value.

Equilibrium Calculations, cont. (b) Given K and one or more initial concentrations, determine the equilibrium concentrations • General Method: – Set up a “concentration table” based on an unknown “x” -- usually one of the equilibrium concentrations or a change in a conc. – Express all equilibrium concentrations in terms of “x” and initial values – Insert these equilibrium values (in terms of “x”) into the proper equation for K, and – If possible, solve directly for “x” using simple math !No need for quadratic formula on Exams! Or, make a chemically reasonable approximation!

Sample Problem Initially, 0. 20 mole PCl 3 and 0. 10 mol Cl 2 are placed in a 1. 00 -L container. The following reaction occurs. PCl 3(g) + Cl 2(g) ⇌ PCl 5(g) Consider the two different types of equilibrium problems: (a) if the equilibrium concentration of PCl 3 is found to be 0. 12 M, calculate Kc. (b) Alternatively, if Kc is known to be 33. 3, calculate the concentration of PCl 5 at equilibrium.

Another Sample Problem For the following reaction, Kc = 64. Initially, 0. 10 mol H 2 and 0. 10 mol I 2 are placed in a 500 -m. L (3 sig fig) container. Calculate the equilibrium concentrations of all three species. H 2(g) + I 2(g) ⇌ 2 HI(g)

Yet More Sample Problems (1) For the following reaction, which occurs in air, Kc = 4. 8 x 10– 31. If the initial concentrations are [N 2] = 0. 033 M and [O 2] = 0. 0081 M, calculate the concentration of NO at equilibrium. N 2(g) + O 2(g) ⇌ 2 NO(g) (2) For the following reaction, Kc = 4. 3 x 104. Initially, 0. 200 mole of H 2 CO 2 is placed in a 1. 0 -L container. Calculate the equilibrium concentrations of all three species. H 2 CO 2(g) ⇌ CO(g) + H 2 O(g)

Le Chatelier’s Principle “When a system at equilibrium is subjected to an external stress, the system will shift in a direction to counteract the stress and achieve a new equilibrium state. ” Common factors (“external stresses”) to consider: – – Addition or removal of some reactant or product Change in volume (important for gaseous reactions) Change in temperature Addition of a catalyst

Example Consider a series of possible changes to the following equilibrium reaction: PCl 3(g) + Cl 2(g) ⇌ PCl 5(g) + 88 k. J (the reaction is exothermic : DHd = -88 k. J) What happens to [Cl 2] if (a) some PCl 3 is added? the “stress” is more PCl 3 system needs to remove PCl 3 ∴ reaction shifts forward and [Cl 2] decreases (but no change in Kc)

Example, cont. PCl 3(g) + Cl 2(g) ⇌ PCl 5(g) + 88 k. J (the reaction is exothermic : DHd = -88 k. J) (b) some PCl 5 is added? the “stress” is more PCl 5 system needs to remove PCl 5 ∴ reaction shifts in reverse and [Cl 2] increases (no change in Kc) (c) the temp is increased? (this requires a DHº value) the “stress” is added heat the system needs to remove heat since the reaction is exothermic, heat is a “product” adding a “product” causes reaction to shift in reverse ∴ [Cl 2] increases (Note: Kc decreases) Only temperature affects the actual value of Kc!

Example, cont. PCl 3(g) + Cl 2(g) ⇌ PCl 5(g) + 88 k. J (d) the volume is increased (or pressure is decreased) the “stress” is more space for gas molecules system tries to produce more gas to fill the space reaction shifts in direction of more moles of gas in this case, it shifts in reverse (2 mol gaseous reactants vs 1 mol gaseous product) ∴ [Cl 2] increases (but no change in Kc) (e) a catalyst is added? Addition of a catalyst does not affect the position of equilibrium. ∴ there is no change in any of the equilibrium concentrations. The catalyst merely lowers the time needed to attain equilibrium. (no change in K c)