1. Chem. 1B – 9/6 Lecture

2. Announcements 2 nd Quiz (and all future quizzes) will be on Monday and Tuesday in lab First Mastering Homework due Tuesday Today’s Lecture – cont. –Equilibrium Problems: STARTING AT INITIAL CONDITIONS –Reaction Quotient and Direction –Le Châtelier’s Principle (Stresses resulting in equilibrium shifts)

3. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions Example Problem 1 –At a certain temperature, K C = 0.38 for N 2 O 4 (g) ↔ 2NO 2 (g) –If a 10.0 L container initially has 0.100 mol of N 2 O 4, what is the equilibrium concentration of NO 2 ?

4. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions Problem 1 required the quadratic – Is this needed always? No. Depends on K value and stoichiometry Example Problem 2: A 10.0 L flask is filled with 0.0020 mol NO 2 (g) and it is expected to decompose as (ignoring the N 2 O 4 formation reaction previously mentioned): 2NO 2 (g) ↔ 2NO (g) + O 2 (g) With K C = 4.5 x 10 -16 Calculate the equilibrium concentration of each gas

5. Chem 1B - Equilibrium Equilibrium Problems – Overview Does problem ask to calculate K or an unknown concentration at equilibrium? KUnknown conc. Are concentrations of all species given at equilibrium? Yes No ICE table needed ICE table needed along with given equil. conc. No Are concentrations of all but 1 species given at equilibrium? Yes No ICE table needed ICE table needed No

6. Chem 1B - Equilibrium Equilibrium Problems – Example to Select Method Example Problem (somewhat tricky): CaCO 3 (s) is placed in a sealed vessel and heated to 800 K and establishes the following equilibrium: CaCO 3 (s) ↔ CaO(s) + CO 2 (g) If K C at this temperature is 6.4 x 10 -4, determine the equilibrium concentration of CO 2. Go back to flow diagram – what is being asked for?, is an ICE table needed? Is it possible to solve this problem – if yes, what is the answer?

7. Chem 1B – Equilibrium The Reaction Quotient and Reaction Direction For a given “system” (e.g. closed flask containing chemicals), the system can either be AT EQUILIBRIUM or under some other conditions (e.g. initial conditions) The equilibrium equation and constant only applies to equilibrium conditions A second quantity, the REACTION QUOTIENT = Q, can be calculated under any conditions (also Q C and Q P ) For generic reaction: aA + bB ↔ cC + dD Q = equilibrium constant and for above reaction, note: for this reaction, [C] = conc. C (but not necessarily at equilibrium conditions)

8. Chem 1B – Equilibrium The Reaction Quotient and Reaction Direction When Q > K, we are too heavy on products, so reaction would proceed toward reactants (loss of C and D and gain of A and B) When Q < K (e.g. initial conditions if A and B are mixed and [C] = [D] = 0 or Q = 0), reaction proceeds toward products Q = K indicates we are at equilibrium

9. Chem 1B - Equilibrium The Reaction Quotient and Reaction Direction Example: An air resource board employee is studying the effects of car exhaust pipe length on pollution concentrations Air leaving the engine has both NO and NO 2 (NO 2 is a worse pollutant) In the exhaust pipe, the reaction can continue toward equilibrium: 2NO (g) + O 2 (g) ↔ 2NO 2 (g) with K P = 4.2 x 10 8 The gas partial pressures are measured just leaving the engine (start of exhaust pipe) and found to be: P NO = 1.0 x 10 -4 atm, P O2 = 0.030 atm, and P NO2 = 2.2 x 10 -7 atm. In which direction will this reaction proceed?

10. Chem 1B - Equilibrium Equilibrium Problems – Large K Value Questions The example covered last time for calculation of Q is an example of this type In the exhaust pipe (if long enough), the reaction could reach equilibrium: 2NO (g) + O 2 (g) ↔ 2NO 2 (g) with K P = 4.2 x 10 8 From the initial gas partial pressures: P NO = 1.0 x 10 -4 atm, P O2 = 0.030 atm, and P NO2 = 2.2 x 10 -7 atm, let’s calculate equilibrium partial pressures show why use of standard ICE table fails and how to add full right and backwards to get it to work (NOTE: THIS IS NOT COVERED IN CHAPTER 14 – BUT YOU NEED TO LEARN THIS LATER)

11. Chem 1B - Equilibrium Le Châtelier’s Principle Le Châtelier’s Principle is used to determine how a reaction at equilibrium will shift due to a change in conditions Continuing the past example (reaction in a car’s exhaust pipe: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) If we can change conditions (e.g. fuel/oxygen ratio or temperature) we may be able to limit formation of NO 2 in the exhaust pipe Le Châtelier’s Principle: When a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance

12. Chem 1B - Equilibrium Le Châtelier’s Principle Changes in Conditions – Types: –Changes in reactant or product concentrations (or partial pressures) –Effect of a change in volume (compression/expansion or dilution/concentration) –Change in temperature Top two changes affect Q; bottom change affects K, in all cases with the “stress” Q ≠ K Given the above changes, we should be able to determine if, under new conditions, the system will re- establish equilibrium by shifting to reactants or products Can take intuitive (all) or mathematical approaches (1 st two changes) to solving problems

13. Chem 1B - Equilibrium Le Châtelier’s Principle Intuitive Method –Addition to one side results in switch to other side –Example: Mathematical Method AgCl(s) ↔ Ag + (aq) + Cl - (aq) Addition of Ag + When Q>K, reaction goes toward reactants When Q<K, reaction goes toward products Example: Q = [Ag + ][Cl - ] As [Ag + ] increases, Q>K reaction shifts to reactants (more AgCl(s))

14. Chem 1B - Equilibrium Le Châtelier’s Principle Stress Number 1 Reactant/Products: Addition of reactant: shifts toward product Removal of reactant: shifts toward reactant Addition of product: shifts toward reactant Removal of product: shifts toward product

15. Chem 1B - Equilibrium Le Châtelier’s Principle – Product/Reactant Stress Example: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) reaction in tailpipe Change:Expectation –Increase O 2 (“lean” conditions) –Decrease O 2 (fuel rich conditions) –Increase NO (run engine hotter) –Remove NO 2 (NO 2 trap??) Right pointing arrow means more NO 2 produced (not desired)

16. Chem 1B - Equilibrium Le Châtelier’s Principle – Volume Stress Example: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) –Mathematical explanation: –Initially at equilibrium K C = 10 5 and [NO] = 0.0010 M, [O 2 ] = 0.0010 M and [NO 2 ] = 0.010 M –Now we reduce the volume from 10.0 to 1.00L –New concentrations: [NO] = 0.010 M, [O 2 ] = 0.010 M and [NO 2 ] = 0.10 M (same number of moles in 1/10 th the volume so 10X more concentrated) –Q = (0.10 M) 2 /[(0.010 M) 2 (0.010 M)] = 10 4 < K C, so products favored

17. Chem 1B - Equilibrium Le Châtelier’s Principle – Volume Stress Note: in aqueous solutions, dilution works in the same way (increase in space due to dilution favors side with more moles) a 1 M HC 2 H 3 O 2 (acetic acid) solution is diluted by adding an equal volume of water. How does this reaction change? HC 2 H 3 O 2 (aq) ↔ H + (aq) + C 2 H 3 O 2 - (aq)

18. Chem 1B - Equilibrium Le Châtelier’s Principle – Temperature Stress Note: change in T changes K (while initial K becomes Q) If ΔH>0, as T increases, products favored - this also means K increases with T If ΔH<0, as T increases, reactants favored Easiest to remember by considering heat a reactant or product Example: OH - + H + ↔ H 2 O(l) + heat (reaction  H < 0) Increase in T

19. Chem 1B - Equilibrium Le Châtelier’s Principle Looking at the reaction below, that is initially at equilibrium, AgCl(s) ↔ Ag + (aq) + Cl - (aq) (ΔH°>0) determine the direction (toward products or reactants) each of the following changes will result in a)increasing the temperature b)addition of water (dilution of system) c)addition of AgCl(s) d)addition of NaCl