1. Chem. 1B – 9/8 Lecture correction on slide #8

2. Announcements I Lab Announcements (see p. 9 of syllabus) for Wed. and Thurs. labs –Quiz I (on review topics, lab I relevant topics, and equilibrium lecture topics) –Review topics (nomenclature and aqueous reactions including net ionic equations) – besides text pages given previously, also see p. 13 – p. 24 + p. 221 – 223 in lab manual –Pre-lab questions (p. 30 – p. 37) This week is last to add (normally) SacCT – set up

3. Announcements II Mastering Chemistry –Some (43 students) not yet signed in as of Monday –first assignment due next Tuesday (9/15) Last lecture follow up –state symbols – include in all reactions Today’s Lecture – note: can solve problems for K (14.6) or for unknown concentrations (14.8) –Equilibrium Problems: STARTING AT EQUILIBRIUM –Equilibrium Problems: STARTING AT INITIAL CONDITIONS –Reaction Quotient and Reaction Direction (if time)

4. Chem 1B - Equilibrium Equilibrium Problems – AT EQUILIBRIUM In this case the equilibrium equation is used with concentrations (or pressures) given AT EQUILIBRIUM These types of problems are very important for environmental chemistry, but underemphasized in text For example, an atmospheric chemist measured high NO in air near fresh lava. He wondered if it came from the N 2 (g) + O 2 (g) ↔ 2NO(g) reaction. If K P (T = 1000 K) = 7 x 10 -9, calculate P NO in equilibrium with N 2 and O 2 in air.

5. Chem 1B - Equilibrium Equilibrium Problems – AT EQUILIBRIUM 2 nd Example Problem: A rich chemist wants to measure K C for the reaction: N 2 O 4 (g) ↔ 2NO 2 (g) He puts N 2 O 4 in a container at the temperature he wants to measure K C. He measures [NO 2 ] and [N 2 O 4 ] (using an expensive mass spectrometer) until the concentrations stop changing. He finds [NO 2 ] = 0.0311 M and [N 2 O 4 ] = 0.000170 M. What is K C ?

6. Chem 1B - Equilibrium Equilibrium Problems – FROM INITIAL CONDITIONS In this case initial concentration(s) or pressure(s) are given (typically of reactants) The reaction then proceeds to equilibrium The student calculates K (equilibrium information needed) or the equilibrium concentration of a reactant or product An important part of working out this problem is to make an ICE table ICE stands for initial change equilibrium The ICE table accounts for rxn stoichiometry

7. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions To understand how an ICE table works, let’s start with a reaction that goes 100% to completion Example: 2.00 mol/L H 2 + 2.00 mol/L O 2 going to H 2 O in a container at 200°C. (Note: for gases we also could use atm in place of mol/L) reaction 2H 2 (g) + O 2 (g) → 2H 2 O(g) initial conc. 2.00 mol/L 2.00 mol/L 0 change -2.00 mol/L -1.00 mol/L +2.00 mol/L completion0 mol/L1.00 mol/L 2.00 mol/L limiting reagentremember: for every 2 mol H 2 we use 1 mol O 2 mol/L O 2 lost = (2.00 H 2 mol /L)(1 mol O 2 /2 mol H 2 )

8. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions Now, let’s look at the same type problem (bad example in this case) which goes to some equilibrium state where [H 2 (g)] ≠ 0 Since we don’t know the final conc., we must change the table, but keep the stoichiometry reaction 2H 2 (g) + O 2 (g) → 2H 2 O(g) I = initial conc. 2.00 mol/L 2.00 mol/L 0 C = change -2x -x +2x E = equil 2.00 – 2x 2.00 – x 2x In this case, we would need to know K or a final concentration to solve this problem

9. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions Next Example Problem: The rich chemist lost his research grant and had his mass spectrometer repossessed. He still has a UV-Visible spectrometer to measure [NO 2 ] (it’s a brown gas – while N 2 O 4 is invisible). –Can he still calculate K? –Yes, but we need to define the experiment more carefully –Initially, the chemist puts 0.0100 mol N 2 O 4 into a 5.00 L container and sets T. He measures [NO 2 ]. When the concentration stops increasing, he finds [NO 2 ] = 0.00281 M. What is K?

10. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions Similar Example Problem: In the following reaction, the concentration of I 2 can be measured (it is purple in color) H 2 (g) + I 2 (g) ↔ 2HI(g) –A reaction starts with 0.100 mol of H 2 and 0.100 mol I 2 in a 1.00 L flask. As the reaction proceeds, I is measured. At equilibrium (when I 2 (g) doesn’t change), [I 2 (g)] is found to be 0.015 M. Calculate K C

11. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions Q. Can the now not rich chemist measure K without any chemical measurements? A. Actually, he can. He could put N 2 O 4 into a flask and measure the initial pressure (P Measured = P N2O4 0 ). As the reaction occurs, because 1 mol N 2 O 4 = 2 mol NO 2, the pressure increases: N 2 O 4 (g) ↔ 2NO 2 (g) I P N2O4 0 0 C-x+2x EP N2O4 0 – x2x P 0 = P N2O4 0 P equil = P N2O4 + P NO2 = P N2O4 0 + x or  P = P equil – P 0 = x and K P = (2x) 2 /(P N2O4 0 - x)

12. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions Determination of Equilibrium Concentrations –This is usually more difficult than determining K or determining equilibrium concentrations when given concentrations AT EQUILIBRIUM –Sometimes we can get an algebraic expression for the answer, but it is difficult to solve (e.g. a third order polynomial requires a cubic equation)

13. 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 ?

14. 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

15. Chem 1B - Equilibrium Equilibrium Problems – Overview Does problem as 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

16. 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)

17. 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 Q = 0), reaction proceeds toward products

18. 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?