Unit II - Equilibrium - Chapter 13

2.1. Characterize chemical reactions in terms of reversibility and relative concentrations of reactants and products. [Readings 13.1 Problems 18]

Demonstration: Fe+3 + SCN- --> FeSCN+2 Fe+3 SCN- FeSCN+2

Demonstration: Fe+3 + SCN- --> FeSCN+2 Fe+3 SCN- FeSCN+2

Fe+3 + SCN- --> FeSCN+2 Add Fe+3

Fe+3 + SCN- --> FeSCN+2 Add Fe+3

Fe+3 + SCN- --> FeSCN+2 Add Fe+3 Darker More FeSCN+2 Excess SCN- The relative amounts of reactants and products in a chemical reaction are effected by concentration changes.

Fe+3 + SCN- --> FeSCN+2 Add Fe+3 Darker More FeSCN+2 Excess SCN- Add SCN-

Fe+3 + SCN- --> FeSCN+2 Add Fe+3 Darker More FeSCN+2 Excess SCN- Add SCN-

Fe+3 + SCN- --> FeSCN+2 Add Fe+3 Darker More FeSCN+2 Excess SCN- Expt. Shows that Fe3+ and SCN- must be present after the rxn. All three species must exist at the same time. Add SCN- Darker More FeSCN+2 Excess Fe+3 All reactants and products are present during the reaction.

Observations on the chemical reaction: Fe+3 + SCN- --> FeSCN+2 All reactants and products are present during the reaction. Reactions are reversible. The relative amounts of reactants and products in a chemical reaction are effected by concentration changes.

AgNO3(aq) + NaC2H3O2(aq) --> AgC2H3O2(s)+ NaNO3(aq) 25 mL .08 M = .002 moles AgNO3 25 mL .08 M = .002 moles NaC2H3O2 AgNO3(aq) + NaC2H3O2(aq) --> AgC2H3O2(s)+ NaNO3(aq) Ag+ + NO3- + Na+ + C2H3O2- --> AgC2H3O2(s) + Na+ + NO3- Net Ionic Equation Ag+(aq) + C2H3O2-(aq) --> AgC2H3O2(s)

Ag+ + C2H3O2- --> AgC2H3O2(s) Initial .002 moles .002 moles 0 moles Final (expected) 0 moles 0 moles .002 moles Final (experimental) .122 g = .000730 moles Where did the rest go? .00127 moles .00127 moles .000730 moles .002-.00073 = .00127 ==> .00127 M reactants are left.

AgC2H3O2(s) -->Ag+ + C2H3O2- Initial .002 moles 0 moles 0 moles Final .000730 moles .00127 moles .00127 moles ---------------------------------------------------------------------------------- Initial .02 moles 0 moles 0 moles Final .01873 moles .00127 moles .00127 moles Turn the reaction around to show silver acetate dissolving. The .00127 moles Ag+ and Ac- are the max soluble If start with .02 moles of Silver acetate ==> ?? same

AgC2H3O2(s) <==> Ag+(aq) + C2H3O2-(aq) [Ag+]i [C2H3O2-]i [Ag+]e [C2H3O2-]e 0.1 M 0.1 M 0.062 M 0.062 M 0.1 M 0.2 M 0.1 M 0.3 M Initial concentrations | final concentrations after rxn. 0.038 M Ag+ form the silver acetate solid. 0.1 M 0.4 M

AgC2H3O2(s) <==> Ag+(aq) + C2H3O2-(aq) [Ag+]i [C2H3O2-]i [Ag+]e [C2H3O2-]e 0.1 M 0.1 M 0.062 M 0.062 M 0.1 M 0.2 M 0.029 M 0.124 M 0.1 M 0.3 M 0.016 M 0.209 M Final data table 0.1 M 0.4 M 0.012 M 0.302 M

[Ag+] vs. [C2H3O2-] Final concentrations 0.07  0.06 0.05 0.04 [Ag+] 0.03  0.02   0.01 Final concentrations 0.05 0.1 0.15 0.2 0.25 0.3 0.35 [C2H3O2-]

[Ag+] vs. [C2H3O2-] ---------- [Ag+] vs. 1/[ C2H3O2 -] 0.07 0.07   0.06 0.06 0.05 0.05 0.04 0.04 [Ag+] [Ag+] 0.03  0.03  0.02 0.02     0.01 0.01 Silver vs. 1/Ac gives a linear relationship [Ag+] = K 1/[Ac-] or K = [Ag+][Ac-] 0.05 0.1 0.15 0.2 0.25 0.3 0.35 2 4 6 8 10 12 14 16 18 [C2H3O2-] 1/[C2H3O2-] [Ag+] = 4.0 x 10-3 / [C2H3O2-] + 0 K = [Ag+] [C2H3O2-] [Ag+] [C2H3O2-] = 4.0 x 10-3

2.2. Determine equilibrium expressions for homogeneous and heterogeneous chemical reactions from stoichiometry. [Readings 13.2, 13.3, & 13.4 Problems 1, 2, 26, 28, 34, 36, 38, & 46]

aA + bB <==> cC + dD [ C ] [ D ] K = a b [ A ] [ B ] In heterogeneous equilibria, pure liquids or solids are not included. In aqueous equilibria, pure water is not included.

Superscripts? A + B <==> 2C A + B <==> C + C K = [C] [C] / [A] [B] K = [C]2 / [A] [B]

Example Reaction N2 (g) + 3 H2 (g) –––> 2 NH3 (g)

Example Reaction N2 (g) + 3 H2 (g) –––> 2 NH3 (g) For rxn: 2 NH3 (g) –––> N2 (g) + 3 H2 (g) K’ = ?

Example Reaction N2 (g) + 3 H2 (g) –––> 2 NH3 (g) For rxn: 2 NH3 (g) –––> N2 (g) + 3 H2 (g) K’ = 1/K

Pure solids or liquids? What is the [AgC2H3O2] ? = moles / Liter {moles = weight / MW} = weight / Lx MW = density / MW = constant for pure substance

Water not included? What is the concentration of water in pure water? 1000g / L (1 mole / 18 g) = 55.6 M What is the concentration of water in 0.1 M NaCl solution? (5.9 g) 994g / L (1 mole / 18 g) = 55.2 M Therefore [H2O] in solutions is a constant.

aA + bB <==> cC + dD [ C ] [ D ] K = a b [ A ] [ B ] In heterogeneous equilibria, pure liquids or solids are not included. In aqueous equilibria, pure water is not included.

Example Write equilibruim expressions for the following reactions: (in a 1 liter container) A. 1/8 S8 (s) + O2 (g) –––> SO2 (g) B. NH3 (aq) + H2O (l) –––> NH4+ (aq) + OH- (aq) A. K = [SO2]/[O2] B. H2O is not included

Gas Phase Reactions Often partial pressures are used instead of concentrations for gas phase reactions. Rxn: N2O4 (g) –––2 NO2 (g) What is the eauilibrium constant? Solve for Kp

Gas Phase Reactions Often partial pressures are used instead of concentrations for gas phase reactions. Rxn: N2O4 (g) –––2 NO2 (g) What is the eauilibrium constant? Solve for Kp

K is temperature dependent: Units: Kc ----- M Kp ----- atm Kp = Kc (RT) ∆n K > 1 K < 1 K = 1 N2 + 3H2 ––> 2 NH3 {>,<, OR = 1} M=mol/L = n/v Thus: P = N/V * RT = M * RT

K Values for Related Chemical Equations (1) A <=> B K1= x (2) B <=> A K2= K1-1 = x-1 (3) 2A <=> 2B K3= K12 = x2

2.3. Determine the stoichiometric relationship between initial and equilibrium concentrations of reactants and products. [Readings 13.5 Problems] 2.4. Determine values for K from equilibrium concentrations of reactants and products in a chemical reaction. [Readings 13.5 Problems 80, ]

H2O + HC7H5O2 <=> H3O+ + C7H5O2- Initial 0.10 M 0 M 0 M Equilibrium ? 2.57 x 10-3 ? Change -2.57 x 10-3 +2.57 x 10-3 +2.57 x 10-3 Equilibrium 9.74 x 10-2 2.57 x 10-3 2.57 x 10-3 Initial ----- amount based on experiment Change ----- amount based on stoichiometry Equilibrium ----- amount based on equilibrium

2.5. Determine the equilibrium concentrations of reactants and products of a chemical reaction from initial concentrations and values of K. [Readings 13.5 Problems 10, 11, 25, 48, 52, 54, 56, & 58]

H2O + HC7H5O2 <=> H3O+ + C7H5O2- If 61 g of HC7H5O2 is dissolved in 1 L of solution, what is [H3O+] at equilibrium? Kc = 6.3x10-5 61 g = 0.50 mol (122g/mol) ICE Table!! H2O + HC7H5O2 <=> H3O+ + C7H5O2- I 0.5M 0 0 C -x +x +x E 0.5-x x x Because magnitude is -4, very little of the reactant breaks down, thus we can assume that 0.5-x ≈.5 (If the magnitude is -1 or -2, then use the quadratic eqn.

2.6. Determine if equilibrium has been reached in a chemical reaction; determine the direction the reaction will shift if equilibrium has not been reached. [Readings 13.5 Problems ex 13.6 & prob. 13.7]

rR <==> pP K = [P]ep / [R]er Q = [P]ip / [R]ir

2.7. Use Le Châtelier’s Principle to predict the direction a reaction at equilibrium will shift as a result of changes in concentration, pressure/volume, and temperature as it approaches a new equilibrium. [Readings 13.6, 13.7, 13.8, & 13.9 Problems 66, 68, 70, & 72]

Le Châtelier’s Principle If a stress is put on a system in equilibrium, it will shift (adjust) to minimize (offset, reverse) the effect of the stress.

Le Châtelier’s Principle If a stress is put on a system in equilibrium, it will shift (adjust) to minimize (offset, reverse) the effect of the stress.

Equilibrium: r a t e f o r w a r d R P r a t e r e v e r s e A system is at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction. This does not mean: The amounts are equal The reaction has stopped

. . A <==> B

Stress -- changes that effect rates of forward and reverse reactions differently concentration temperature catalyst? pressure?

Shift -- An adjustment in the concentrations of reactants and products Shift to right -- increase [P], decrease [R] Shift to left -- increase [R], decrease [P]

Minimize A(g) + B(g) <=> 2C(g) K = [C]2 / [A] [B] I 2 5 0 E 1 4 2 K = 1 add 19 A 20 4 2 Q ≠ 1 C -2 -2 +4 new E 18 2 6 K = 1

N2 + 3 H2 <==> 2 NH3 .

N2 + 3 H2 <==> 2 NH3 Concentration H 2 NH 3 N 2 Time -------> . . N2 + 3 H2 <==> 2 NH3 Concentration H 2 NH 3 N 2 Time ------->

Demo: 2 NO(g) + O2(g) --> 2 NO2(g) Initial 20 mL 10 mL Predicted 20 mL Actual 15 mL 2 NO2(g) <==> N2O4(g) + ∆ brown colorless

Change Temperature 2 NO2(g) <==> N2O4(g) + ∆ Increase T -- More Heat -- P constant Equilibrium Shifts to Reduce Heat Color Deepens K = [N2O4] / [NO2]2 , K decreases Therefore K is T dependent

Change Pressure (by decreasing V) 2 NO2(g) <==> N2O4(g) + ∆ PT Kp Equilibrium 20 atm 20 atm 40 atm .05 P x 2 (1/2 V) 40 atm 40 atm 80 atm .025 Change -10 atm +5 atm New 30 atm 45 atm 75 atm .05 Equilibrium Increase pressure ==> a decrease in the number of moles of gas decrease in pressure ==> an increase in the number of moles of gas, thus achieving equilibrium.

Self Test X(g) + H2O(l) <==> Q (s) + Y(g) ∆H = +150 kJ Change Shift Change in K increase Px increase [Y] remove 1/2 Q increase PT increase T catalyst Change Shift ∆K? increase Px right no increase [Y] left no remove 1/2 Q right no increase PT no chng no increase T right yes catalyst no chng no