1. I.The Reaction Quotient 1.We can write K only for a reaction at equilibrium 2.How can we describe a reaction at other times? N 2 + 3H 2 2NH 3 3.Q = reaction quotient has the same form as K, but is at non-equilibrium times. 4.Q can tell us how a reaction will change to get to equilibrium 1)If Q = K, we are at equilibrium 2)If Q > K, then products > reactants and the reaction will shift left 3)If Q products and the reaction will shift right 5.Example: For the above reaction, K = 6 x 10 -2. Predict change: 1)[N 2 ] 0 = 1 x 10 -5 M, [H 2 ] 0 = 0.002 M, [NH 3 ] 0 = 0.001M 2)[N 2 ] 0 = 1.5 x 10 -5 M, [H 2 ] 0 = 0.354 M, [NH 3 ] 0 = 0.0002M 3)[N 2 ] 0 = 5 M, [H 2 ] 0 = 0.01 M, [NH 3 ] 0 = 0.0001 M Describes the reaction when not at equilibrium QKPredicted Direction of Reaction 0.551.45To the right (towards products) 2.551.45To the left (towards reactants) 1.45 No change (at equilibrium)

2. II.Solving Equilibrium Problems A.Problems where equilibrium concentrations (or pressures) are given 1.Example: N 2 O 4 (g) 2NO 2 (g) a)K P = 0.133 b)P(N 2 O 4 ) at equilibrium = 2.71 atm c)Find P(NO 2 ) at equilibrium 2.Example: PCl 5 (g) PCl 3 (g) + Cl 2 (g) a)[PCl 3 ] 0 = 0.298 M, [PCl 5 ] 0 = 0.0087 M b)At equilibrium, [Cl 2 ] 0 = 0.002 M c)Calculate all concentrations and K B.Procedure when not given equilibrium concentrations 1.Write the balanced equation 2.Write the equilibrium expression 3.List the initial conditions 4.Calculate Q to decide which way the reaction will shift 5.Write equilibrium conditions using x as unknown changes in concentration 6.Substitute the equilibrium conditions into the equilibrium expression and solve for x 7.Check your answer to see if it gives the correct value of K

3. A.Example: 3 mol of hydrogen and 6 mol of fluorine are mixed in a 3 L flask to give hydrogen fluoride gas. K = 115. Find the equilibrium concentrations. 1.Write the balanced equation: H 2 + F 2 2HF 2.Write the equilibrium expression: 3.List the initial conditions: [H 2 ] 0 = 1 M, [F 2 ] 0 = 2 M, [HF] 0 = 0 4.Calculate Q to decide which way the reaction will shift 5.Write equilibrium conditions using x as unknown changes in concentration InitialChangeEquilibrium [H 2 ] = 1-x1-x [F 2 ] = 2-x2-x [HF] = 0+2x2x

4. Shorthand: H 2 + F 2 2HF Initial: 1 2 0 Change -x -x +2x Equilibrium 1-x 2-x 2x Substitute and solve for x Check your answer to see if it gives the correct value of K a = 111, b = -345, c = 230 x = 2.14 M or x = 0.968 M Reject x = 2.14 M since [H 2 ] = 1 – x = 1 – 2.14 = -1.14 M x = 0.968M

5. B.Example: CO(g) + H 2 O(g) CO 2 + H 2 K = 5.1 Find equilibrium conditions if 1 mol of each is mixed in a 1L flask. C.Example: 3 mol of hydrogen and 3 mol of fluorine are mixed in a 1.5 L flask with 3 mol hydrogen fluoride gas. K = 115. Find the equilibrium concentrations. H 2 + F 2 2HF D.Example: H 2 + I 2 2HI K P = 100. If we mix HI (P = 0.5 atm), H 2 (P = 0.01 atm), and I 2 (P = 0.005) in a 5 L flask, what are eq. conc.? E.Approximations when x is small 2NOCl(g) 2NO + Cl 2 (g) K = 1.6 x 10 -5 Initial 0.5 M 0 0 Equil. 0.5 – 2x 2x x Since K is small, x will be very small. x = 1 x 10 -2

6. II.Le Chatelier’s Principle A.We can change the position of an equilibrium by changing conditions 1.N 2 + 3H 2 2NH 3 2.Low temperature and high pressure favor the forward reaction 3.Le Chatelier’s Principle: When a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance. B.Effect of change in concentrations N 2 + 3H 2 2NH 3 K = 0.0596 1)At equilibrium, [N 2 ] = 0.399 M, [H 2 ] = 1.197 M, [NH 3 ] = 0.202 M 2)Let’s add 1 M N 2. How does the equilibrium shift? a)Too much N 2 for equilibrium b)Reaction must adjust to get back to equilibrium c)Reaction must shift to the right Temp. ( o C)300 atm400 atm500 atm 40048% NH 3 55% NH 3 61% NH 3 50026% NH 3 32% NH 3 38% NH 3 60013% NH 3 17% NH 3 21% NH 3 Eq. shifts to the right

7. 3.We could do the x calculation and find the new concentrations if we wanted, but sometimes you just want to know which way the reaction will shift. Le Chatelier’s principle lets us do that simply. 4.Another Example: What would happen if we add N 2 O 4 ?

9. 5.Example: As 4 O 6 (s) + 6C(s) As 4 (g) + 6CO(g) a)Add CO b)Add or remove As 4 O 6 (s) c)Remove As 4 (g) C.Effect of Change in Pressure/Volume 1.If we add/remove reactant/product, we change concentration (see above) 2.Adding an inert gas, doesn’t change concentrations, so no effect on equilibrium 3.Changing the size of the flask: V↓ = P↑ V↑ = P↓

10. a)The volume changes, so the concentrations must change b)To re-establish equilibrium, the concentrations will adjust i.Fewer gas particles are favored at higher pressures ii.This reduces the pressure (Le Chat’s principle) c)N 2 (g) + 3H 2 (g) 2NH 3 (g) Increase the pressure i.There are 4 particles on the left and 2 on the right ii.To reduce the pressure, the equilibrium will shift right d)If we increased the volume and decreased the pressure, the reaction will shift to increase the number of particles (shift left) 4.Example: Reduce the volume on the following reactions: a)P 4 (s) + 6Cl 2 (g) 4PCl 3 (l) b)PCl 3 (g) + Cl 2 (g) PCl 5 (g) c)PCl 3 (g) + 3NH 3 (g) P(NH 2 ) 3 (g) + 3HCl(g)

11. D.Effect of change in Temperature 1.The above changes (P, Concentration) effect the equilibrium position, but not the equilibrium constant K 2.Changing Temperature changes the equilibrium constant K 3.We can predict the changes by treating heat as a product or reactant of every reaction 4.Exothermic reactions produce heat as a product a)Adding heat (increasing temp.) shifts away from heat, to left, K decreases b)Removing heat (decreasing temp.) shifts towards heat, to right, K increases c)N 2 + 3H 2 2NH 3 + 93 kJ/mol (  H = - 93 kJ/mol) d)Temperature ( o C): 500600700800 K: 90 3 0.30.04 5.Endothermic reactions require heat as a reactant a)Add heat = shift to right, K increases b)Remove heat = shift to left, K decreases c)556 kJ/mol + CaCO 3 (s) CaO(s) + CO 2 (g) (  H = + 556 kJ/mol) 6.Examples: N 2 (g) + O 2 (g) 2NO(g)  H = 181 kJ/mol 2SO 2 (g) + O 2 (g) 2SO 3  H = - 198 kJ/mol)