Chem%20162-2011%20lecture%208

# Chem%20162-2011%20lecture%208 - CHEMISTRY 162-2011 LECTURE...

This preview shows pages 1–7. Sign up to view the full content.

Chem 162-2011 Lecture 8 1 CHEMISTRY 162-2011 LECTURE 8 ANNOUNCEMENTS E-MAIL ATTENDANCE Sign in EXAM Wednesday, Feb. 16 th , 2011, 9:40-11:00 PM SEC (Science and Engineering Center?), Frelinghuysen Road, Busch Coverage: H&P, Chapters 12.1 to 13.11. QUIZ

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chem 162-2011 Lecture 8 2 PLAN FOR TODAY : CHAPTER 14 CONTINUATION OF LECTURE 7 Equilibrium calculations CHEMICAL EQUILIBRIUM 14.4 – 14.5 Le Châtelier’s Principle
Chem 162-2011 Lecture 8 3 EQUILIBRIUM (DYNAMIC* EQUILIBRIUM) (*not static equilibrium) ET: At equilibrium, the forward and reverse reaction proceed at equal rates and the concentrations of reactants and products remain constant. Mountain S l u m s a d k 1 = 0.01/s k -1 = 0.10/s town with Palacial happy 100 sad t o w n w i t h p e o p l e / m i 2 1000 happy people/mi 2 ET: At equilibrium, the forward and reverse reaction proceed at equal rates and the concentrations of reactants and products remain constant. Forward rate = k 1 x conc. = 0.01s -1 x 1000 = 10 people/mi 2 s -1 Reverse rate = k -1 x conc. = 0.10s -1 x 100 = 10 people/mi 2 s -1 Equilibrium reached Macroscopically : You only see 1000 happy people on the left and 100 sad people on the right. Microscopically : You see 1000 happy people on the left and 100 sad people on the right. You also see a constant equal exchange of 10 sad people and 10 happy people going across the bridge: a “dynamic equilibrium”. Equilibrium constant: K c = 100 people/1000 people = 0.10 k 1 /k -1 = k f /k r = 0.01s -1 /0.10s -1 = 0.10 K c = [Products]/[Reactants] = k f /k r = 0.10

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chem 162-2011 Lecture 8 4 CHAPTER 14 - CHEMICAL EQUILIBRIUM A + B C + D (gases or solutes, only) Law of mass action (or molar interaction ): K c = ([C][D])/([A][B]) 2A + 3B 4C + 1D A + A + B + B + B C + C + C + C + D K c = ([C] • [C] • [C] • [C] • [D])/( [A] • [A] • [B] • [B] • [B]) K c = ([C] 4 [D])/([A] 2 [B] 3 ) eA + fB gC + hD K c = ([C] g [D] h )/([A] e [B] f ) e.g., 2 NO + O 2 2 NO 2 K c = ([NO 2 ] 2 )/([NO] 2 [O 2 ]) K A B [B]/[A] = K 1 C D [D]/[C] = K 2 A + C B + D ([B][D])/([A][C]) = K 1 X K 2 Addition of ([B]/[A]) + ([D]/[C]) = (([B][C] + [A][D]))/([A][C]) Equations K(s) Add Multiply Reverse Invert Double Square Halve Square root K p inatm = K c inM (RT) ngas
Chem 162-2011 Lecture 8 5 SOLVING EQUILIBRIA PROBLEMS Step 1: Write balanced equilibrium equation Step 2: Fill in and solve ICE table Step 3: Write equilibrium expression Step 4: Solve equilibrium expression

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chem 162-2011 Lecture 8 6 CHEM162-2010 5th WEEK RECITATION CHAPTER 14 - NON ACID-BASE CHEMICAL EQUILIBRIA NON-ACID-BASE EQUILIBRIUM CALCULATIONS ET: Discuss ICE table ET: Simplest equilibrium problem 25. At a particular temperature, a 3.0-L flask contains 2.4 mol Cl 2 , 1.0 mol NOCl, and 4.5 x 10 -3 mol NO at equilibrium. Calculate K at this temperature for the following reaction: 2NOCl(g) 2NO(g) + Cl 2 (g) [Cl 2 ] = 2.4 mol/3.0L = 0.8M at equilibrium [NOCl] = 1.0 mol/3.0L = 0.333M at equilibrium
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 24

Chem%20162-2011%20lecture%208 - CHEMISTRY 162-2011 LECTURE...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online