Lecture 40: Equilibrium 1

Transcription

1 Lecture 40: Equilibrium 1 Read: BLB HW: BLB 15:13,14,21 Sup 15:1 4 Know: concept of equilibrium equilibrium constant (K eq ) derivation of K p /K c relationship what magnitude of K eq means FINAL SKILL CHECK TEST DEADLINE: MONDAY, APRIL 27 Need help?? Get help!! TAs in CRC (211 Whitmore) and SI hours on Chem 110 website; my office hours (Mon 12:30-2 & Tues 10:30-12 in 324 Chem Bldg [or 326 Chem]) Final Exam: MONDAY, May 4, 12:20 pm. locations under exam schedule ; no pds, ipods, graphing calculators, etc. Only non-text programmable calculators allowed. Bring PSU ID and pencils Concept Final review session with Sheets: Thursday 6pm in 108 Forum. Please work through the Concept Exam (on lecture note page) before the review session & bring it with you, along with any questions you may have. This review is meant to complement to the review sessions that your TAs will be holding in which they will go over the practice exams. Sheets Page 1 Lecture 40

2 Dynamic equilibrium dynamic equilibrium: no net change, BUT change is occurring on molecular level; this equilibrium can be a physical change or a chemical change forward rate = backward rate dynamic equilibria seen earlier physical changes: vapor pressure (liquid vapor coexistence), dissolution in a saturated solution now, focus on dynamic chemical equilibria reactants products opposing reactions proceed at equal rates at equilibrium, of reactants & products do not change over time (macroscopic) but a lot of action on the molecular level; dynamic balance reaching equilibrium may be sslllloooooowww or fast NOTE: vs. Sheets Page 2 Lecture 40

3 weʼve been writing: Chemical equilibrium 2H 2 (g) + O 2 (g) 2H 2 O (g)!h = 484 kj but in reality: 2H 2 (g) + O 2 (g) 2H 2 O (g) Sheets Page 3 Lecture 40

4 Haber process N 2 (g) + 3 H 2 (g) 2 NH 3 (g) initial state: reactants only initial state: products only BUT final state: same for both! of products to reactants is the relationship between the concentrations of products or reactants at equilibrium will be the same, regardless of starting conditions; it is defined by the equilibrium constant K eq NOTE: CATALYSTS DO NOT affect equilibrium Sheets Page 4 Lecture 40

5 Equilibrium constant K eq (equilibrium constant): equilibrium point of any reaction is characterized by a single number (NO units, see BLB p 637 for more info) in general: aa + bb cc + dd K eq = C A [ ] c [ D] d [ ] a [ B] b Example 2 NO 2 (g) N 2 O 4 (g) K eq = for this reaction: the ratio of equilibrium concentrations will be constant K eq does NOT depend on initial concentrations reaction mechanism BUT it is a function of temperature only, & it depends on the stoichiometry Sheets Page 5 Lecture 40

6 Haber process (cont.) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) What is the equilibrium constant for the Haber process? K eq =? A. K eq = N 2 NH 3 [ ][ H 2 ] B. [ [ ] 2 K eq = N 2 ][ H 2 ] 3 [ ] 2 NH 3 C. K eq = [ NH 3 ] 2 D. [ N 2 ][ H 2 ] 3 K eq = [ NH 3 ] 2 [ N 2 ] 2 [ H 2 ] E. K eq = [ NH 3 ] [ N 2 ][ H 2 ] Sheets Page 6 Lecture 40

7 Equilibrium notation K eq = K c = [ C ] c D [ ] d [ A] a [ B ] b if [x] in units of M K eq = K p = P C c P D d P A a PB b if P x in units of atm; partial pressure NOTE: K p may NOT equal K c!!!! so, when do you use K c or K p for K eq?? Help!?!! use problem context: if concentrations given in M, then K eq = K c if concentrations given in atm, then K eq = K p Sheets Page 7 Lecture 40

8 Derivation of the K p /K c relationship Sheets Page 8 Lecture 40

9 Back to the Haber process N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Equilibrium concentrations of NH 3, N 2, and H 2 were determined at 472 C. Find K c and K p. [H 2 ] = M [N 2 ] = M [NH 3 ] = M Sheets Page 9 Lecture 40

10 Equilibrium math if K eq» 1, then equilibrium lies to right; dominate AND if K eq «1, then equilibrium lies to left; dominate Example: At a certain temperature, K c for the following reaction is 16. H 2 (g) + I 2 (g) 2 HI(g) K c = 16 Which is more prevalent at equilibrium? reactants or products? Sheets Page 10 Lecture 40

11 More equilibrium math K eq of the reverse reaction is the inverse of the K eq in the forward direction K eq of a reaction that has been multiplied by a number is the K eq (of original chemical equation) raised to a power equal to that number K eq for net reaction is equal to the product of the individual steps Sheets Page 11 Lecture 40

12 Example: At a given temperature, H 2 (g) + I 2 (g) 2 HI(g) K c = 16 At this same temperature, what is K c for the following reaction? A. 1/16 B. 4 C. 1/4 D. 16 HI(g) 1/2 H 2 (g) + 1/2 I 2 (g) K c =?? E. There is not enough information to answer this question. Sheets Page 12 Lecture 40

13 Before next class: Read: BLB HW: BLB 15:27,28,29,30,32,35,37,39,46 Sup 15:5 10 Know: homogeneous & heterogeneous equilibria equilibrium apps calculating equilibrium constants math refresher in BLB Appendix A Work through the concept final (on lecture notes website). Good luck studying for finals! Please start now. Answers: p. 6: C p. 9: Kc = ; Kp = p. 12: C Sheets Page 13 Lecture 40