1. Chem. 1B – 9/6 Lecture

2. Announcements I Adding: Possibly spots open in Sect. 2, 3, and 5 (based on MySacState listing – some of these openings were filled – students not turning in add slips yet – and adding will be based on lab instructor’s waitlists) 1 st Quiz – this Wed. and Thurs. in lab –On lecture + related homework + book chapters –Also on review topics (pre-lab material + nomenclature) –See class website for nomenclature requirements

3. Announcements II SacCT –Will set up this week –Plan to use Sect. 1 for posting solutions and Lab Sections for posting grades (note: I will try to keep the grades updated, but it is better if you calculate your grades independently) Mastering Chemistry –About 75% signed in now –A few emails regarding problems (not certain of difficulty – but use course ID in syllabus) –First assignment due a week from today

4. Announcements III Today’s Lecture – cont. –K P vs. K C –Equilibrium Problems: STARTING AT EQUILIBRIUM –Equilibrium Problems: STARTING AT INITIAL CONDITIONS –Reaction Quotient and Direction (if time)

5. Chem 1B - Equilibrium Manipulating Equations – Another Example Given: (1)Cu(OH) 2 (s) ↔ Cu 2+ (aq) + 2OH - (aq) K = 4.8 x 10 -20 (2) HC 2 H 3 O 2 (aq) ↔ H + (aq) + C 2 H 3 O 2 - (aq) K = 1.75 x 10 -5 (3) H 2 O (l) ↔ H + (aq) + OH - (aq) K = 1.00 x 10 -14 (4)Cu 2+ (aq) + 2C 2 H 3 O 2 - (aq) ↔ Cu(C 2 H 3 O 2 ) 2 (aq) K = 4.27 x 10 3 Determine K for Cu(OH) 2 (s) + 2HC 2 H 3 O 2 (aq) ↔ Cu(C 2 H 3 O 2 ) 2 (aq) + 2H 2 O (l)

6. Chem 1B - Equilibrium Equilibrium Constants – K P vs. K C Example: 2NO 2 (g) ↔ N 2 O 4 (g) K P = P N2O4 /P NO2 2 Or K P = ([N 2 O 4 ]RT)/([NO 2 ]RT) 2 = K C (RT) -1 General Rule: K P = K C (RT)  n where  n = change in number of moles (moles gas product – moles gas reactants) = 1 – 2 = -1 in above example

7. Chem 1B - Equilibrium Most Common Types of Problems Conditions givenQuestion Asking for K Question Asking for Equilibrium Concentrations Only at equilibriumRequires knowledge of concentrations (or pressures) of all species Requires knowledge of K and concentrations of all but one species Initial conditions (requires use of ICE table) Usually requires equilibrium concentration of at least 1 species Requires knowledge of K

8. 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.

9. 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 ?

10. Chem 1B - Equilibrium Equilibrium Problems – AT EQUILIBRIUM 3 rd Example Problem: (your turn to do) Syngas (CO + H 2 ) can be used to generate fuels. From the reaction 2CO(g) + 4H 2 (g) ↔ C 2 H 6 O(g) + H 2 O(g) (K P = 2.8 x 10 6 ), calculate P C2H6O if P CO = 2.0 x 10 -2 atm, P H2 = 9.2 x 10 -2 atm, and P H2O = 0.21 atm (at equilibrium)

11. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions In this case an initial concentration or pressure is given (typically of reactants) The reaction then proceeds to equilibrium The student calculates K or the 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

12. 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 (covered in Chem 1A) Example: 1.00 mol/L H 2 + 1.00 mol/L O 2 going to H 2 O in a container at 200°C. reaction 2H 2 (g) + O 2 (g) → 2H 2 O(g) initial conc. 1.00 mol/L 1.00 mol/L 0 change -1.00 mol/L -0.50 mol/L +1.00 mol/L completion0 mol/L0.50 mol/L 1.00 mol/L limiting reagentremember: for every 2 mol H 2 we use 1 mol O 2 mol/L O 2 lost = (1.00 H 2 mol /L)(1 mol O 2 /2 mol H 2 )

13. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions 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 5.0 L container and sets T. He measures [NO 2 ]. When the concentration stops increasing, he finds [NO 2 ] = 0.0028 M. What is K?

14. 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 2 is measured. At equilibrium (when I 2 (g) doesn’t change), [I 2 (g)] is found to be 0.015 M. Calculate K C

15. 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)

16. 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 ?

17. 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

18. 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

19. 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)

20. 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