Sheets Page 1 Lecture 40

Lecture 40: Equilibrium 1 Read: BLB 15.1–15.3 HW: BLB 15:13,14,21 Sup 15:1–4 Know: • concept of equilibrium • equilibrium constant (Keq) • derivation of Kp/Kc relationship • what magnitude of Keq 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]) F inal 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 4/30 @ 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 2 Lecture 40

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 • 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.

reactants products

Sheets Page 3 Lecture 40

Chemical equilibrium • weʼve been writing:

• but in reality:

2H2 (g) + O2 (g) 2H2O (g)

!H = –484 kJ

2H2 (g) + O2 (g) 2H2O (g)

Sheets Page 4 Lecture 40

Haber process

N2(g) + 3 H2(g) 2 NH3(g)

initial state: reactants initial state: products only only BUT… final state: of products to reactants is the same for both! • 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 Keq • NOTE: CATALYSTS DO NOT affect equilibrium

Sheets Page 5 Lecture 40

Equilibrium constant • Keq (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

Keq =C[ ]

c D[ ]d

A[ ]a B[ ]

b Example 2 NO2 (g) N2O4(g) Keq = • for this reaction: the ratio of equilibrium concentrations will be constant • Keq does NOT depend on

⇒ initial concentrations ⇒ reaction mechanism

BUT it is a function of temperature only, & it depends on the stoichiometry

Sheets Page 6 Lecture 40

Haber process (cont.)

N2(g) + 3 H2(g) 2 NH3(g) What is the equilibrium constant for the Haber process? Keq = ?

A.

Keq =N2[ ] H2[ ]NH3[ ]2

B.

Keq =N2[ ] H2[ ]

3

NH3[ ]2

C.

Keq =NH3[ ]2

N2[ ] H2[ ]3 D.

Keq =NH3[ ]2

N2[ ]2 H2[ ]

E.

Keq =NH3[ ]N2[ ] H2[ ]

Sheets Page 7 Lecture 40

Equilibrium notation

Keq = Kc =C[ ]

c D[ ]d

A[ ]a B[ ]

b if [x] in units of M

Keq = Kp =PC

cPDd

PAaPB

b if Px in units of atm;

partial pressure • NOTE: Kp may NOT equal Kc!!!!

• so, when do you use Kc or Kp for Keq?? Help!?!! • use problem context:

if concentrations given in M, then Keq = Kc if concentrations given in atm, then Keq = Kp

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Derivation of the Kp/Kc relationship

Sheets Page 9 Lecture 40

Back to the Haber process…

N2(g) + 3 H2(g) 2 NH3(g) Equilibrium concentrations of NH3, N2, and H2 were determined at 472°C. Find Kc and Kp.

[H2] = 0.1207 M [N2] = 0.0402 M [NH3] = 0.00272 M

Sheets Page 10 Lecture 40

Equilibrium math • if Keq » 1, then equilibrium lies to right;

dominate AND • if Keq « 1, then equilibrium lies to left;

dominate Example: At a certain temperature, Kc for the following reaction is 16. H2(g) + I2(g) 2 HI(g) Kc = 16 Which is more prevalent at equilibrium? reactants or products?

Sheets Page 11 Lecture 40

More equilibrium math • Keq of the reverse reaction is the inverse of the Keq in the forward direction • Keq of a reaction that has been multiplied by a number is the Keq (of original chemical equation) raised to a power equal to that number • Keq for net reaction is equal to the product of the individual steps

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Example: At a given temperature, H2(g) + I2(g) 2 HI(g) Kc = 16 At this same temperature, what is Kc for the following reaction? HI(g) 1/2 H2(g) + 1/2 I2(g) Kc = ??

A. 1/16 B. 4 C. 1/4 D. 16 E. There is not enough information to answer this

question.

Sheets Page 13 Lecture 40

Before next class: Read: BLB 15.4–15.5 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 = 0.1047; Kp = 2.80 × 10–5 p. 12: C