Chem 222

#6 Ch 6Sep 9, 2004

Announcement

♦ Answers for Quiz 1 at Web site

♦ Quiz 2 will be held on next Thursday (9/16).

We plan to give• One question from Ch3-5• One question from Ch6

Please memorize all the equations highlighted by red and solve home work in this section.

♦Submit your report for Exp 3 on next W/R.

Lab Report Procedure♦ Do not hand in your lab notebook to the TA until

the end of the term. ♦ Lab reports will have the following information in

them:

1. Cover sheet with the final results and a statement of what you measured (i.e., Nicotine in unknown tobacco sample #_____) unknown number, average value of your determinations, and both absolute and relative standard deviations. YOU NEED TO PRINT OUT A COVER SHEET TO RECORD YOUR LAB RESULTS at the web site (www.chem.uic.edu/chem222).

2. Photocopy of the experiment description you placed in your notebook before the start of the experiment.

3. Photocopy (or carbon) copies of the signed data pages you recorded in lab. Do not recopy these for neatness.

4. Analysis section showing at least example calculations, as well as computer generated graphs, etc.

♦ If you did not submit the photocopies of your notebook for Exp 2 because of lack of information, submit them to your TA at the beginning of M/T Lab (Sep 13/14). TA will accept them.

Lab report will be graded based on 1) Whether you submitted the report on time?

20-40 %2) How accurate your measurement is? 30-40 %3) Calculations and analysis you demonstrated.

30-40%

• Points will be subtracted from the base points on the basis of your performance. For example, the full mark for (1) will be given only when you submit a satisfactory report with all measurements and calculations completed.

In the lab report, please also include the following information

1) If you have only one or two trials, give justifications and describe why you have less than required.

♦ Describe why you could not obtain the results ♦ Show the result of Q-test if you decide to drop one

result out of three.

Ex. Q = 0.94 for # of observation = 3 12.0 12.1 13.0 Should we discard 13.0 or not?

2) If you know your results are not precise than estimated (large S), discuss how the error was introduced.

Ch6 Chemical Equilibrium

Enzymatic reactionE + S ES E + P

v Enzyme Substrate Product

The reaction rate (v) from ES to E +Pv = k[ES],

where [ES] is the concentration of the ES complex.

In general ♦ aA + bB cC + dD

v v = k[A]a[B]b

♦ A + B C + D V1 = k[A][B]v1

♦ A + 2B C + D V2 = k[A][B]2

v2♦ 3A + 2B 5C + 8D V3 = k[A](Q1)[B]2

v3

←→

Ch6. Chemical Equilibrium

• 6-1 The equilibrium constantv1

aA + bB cC + dD [6-1]v2

v1 = k1[A]a[B]b & v2 = k2[C]c[D]D,where [A] stands for the “concentration” of A.

When the reaction reaches equilibrium, v1 = v2 k1[A]a[B]b = k2[C]c[D]d

Namely, K ≡ k1/k2 = [6-2]←The left side in [6-1]

in the denominator

In evaluating an equilibrium constant, 1. Concentrations of solutes should be expressed

as M (moles/L).2. Concentration of gases should be expressed in

bars3. Concentration of pure solids, pure liquids,

solvents are omitted because they are unity.

←→

ba

dc

BADC][][][][

6-2 Manipulating Equilibrium Constants

Q1.• H2O H+ + OH-

KW =[H+][OH-]; KW = 1.0 × 10-14 at 25 °C.When no other chemicals are added, show[H+] = 1.0 × 10-7.

←→

HA H+ + A- K1 = [H+][A-]/[HA]←→

If the direction of the reaction is reversed, the new equilibrium constant is simply the reciprocal of the original value K

H+ + A- HA K1‘ = [HA] /[H+][A-] = 1/K1Combining reactionsIf the two reactions are addedHA H+ + A- K1H+ + C CH+ K2_________________________HA +C A- + CH+ K3

The overall reaction is the product of n reactions.

←→

Reverse Reaction

213 ]][[][

][]][[

]][[]][[ KK

CHCH

HAAH

CHACHAK === +

+−++−

←→

←→

←→

• Ex. (p101)The equilibrium constant for the reactionH2O H+ + OH-

is called Kw ([H+][OH-] = 10-14) at 25 °C. Given that

NH3(aq) + H2O NH4+ + OH-

KNH3 = 1.8 × 10-5

find the equilibrium constant (K3) for the reaction

NH4+ NH3 + H+ K3

wNH

KKOH

HOHOHNHOHNH

NHHNHK

324

23

4

33

1][]][[

]][[]][[

][]][[

===+−

−++

+

←→

←→

←→

= 1.0×10-14/[Q1]

= 5.6 ×10-10

6-2 Equilibrium and Thermodynamics

EnthalpyThe enthalpy change, ∆H, for a reaction is

the heat absorbed or relased when the reaction takes place.

HCl(g) H+(aq) + Cl-(aq) ∆H° = -74.85 kJ/molH1 H2 ∆H = H2 – H1

Heat is released by 74.85 kJ/mol____

H1 ↑∆H↓ ____ H2 If ∆H is positive, heat is

absorbed.

∆H° denotes ∆H at 25 °C

←→

Entropy• The measure of “disorder” for a

system: SKCl(s) K+(aq) + Cl-(aq)S1 S2

∆S ≡ S2 – S1 = 76.4J/(K·mol)∆S is the change is entropy. ∆S > 0 More disordered∆S < 0 Less disordered

←→

Free Energy∆Chemical reaction is favored when ∆ H < 0 or ∆ S >0. ∆G combines the two effects

∆G = ∆ H - T ∆ S (6-5)

When ∆G < 0, the reaction is favored.

Problem 6-7 (b) A favorable enthalpy change occurs when ∆H

Free energy and Equilibrium

• ∆G is important in chemical reactions, since the equilibrium constant K depends on ∆G as

K = exp(-∆G/RT),where R is the gas constant [= 8.31 J/K·mol]

For exampleHCl(g) H+(aq) + Cl-(aq) ∆H = -74.85 kJ/mol & ∆S = -130.4 J/K ·mol

∆G = -74.85 ×103 J/mol + (298.15 K) (-130.4 J/K ·mol) = -35.97 kJ/mol

K = exp(-∆G/RT) = 2.00×106 = [H+(aq)][Cl-(aq)]/[HCl]

←→

Le Châtelier’s Principle• If a system is changed from an

equilibrium state, the system moves to a direction that partially cancels the change.

Q1. Suppose the above reaction is at equilibrium.When the concentration of Cr2O72- is changed from 0.10 to 0.20 M, in what direction will the reaction proceed to reach the equilibrium?

Q ≡ [Br-][Cr2O72-][H+]8/{[BrO3-][Cr3+]2} = 2K > K (Q reaction quotient) [Br-][Cr2O72-][H+]8 will be reduced.

Q2. When the temperature is raised from equilibrium, in what direction will the reaction proceed to reach the equilibrium?

How equilibrium is affected by T

K = exp(-∆G/RT) = exp{-(∆H-T∆S)/RT}= exp(-∆H/RT + ∆S/R)= exp(-∆H/RT)exp(∆S/R)

T up If ∆H > 0 K up If ∆H < 0 K down

∆H > 0A + B C + D – heat∆H < 0A + B C + D + heat

←→

←→

6-11 answer

K = exp(-∆H/RT)exp(∆S/R) (6-8)K’ = exp(-∆H/RT’)exp(∆S/R)

K/K’ = exp(-∆H/RT)exp(∆H/RT’)= exp{-(∆H/R)(1/T – 1/T’)}

Ln(K/K’) = -(∆H/R)(1/T – 1/T’)

Note: T and T’ are in K not in °C.

6-7 Protic Acids and Bases

• In aqueous chemistry, an acid is a substance that increases [H3O+] (hydronium ions) when added to water.

• Conversely, a base decreases [H3O+].

• Because Kw =[H3O+][OH-] = const., a decease of [H3O+] always increases [OH-].

a base increases [OH-].

♦ The word protic refers chemistry involving transfer of H+ from one molecules to another. ♦ H+ is also called a proton. H+ and H3O+ are often interchangeably used.

6-8 pH (p113)

• Autoprotolysis of water

H2O H+ + OH-←→

The autoprotolysis constant for water has the special symbol Kw.

Kw = [H+][OH-]

At 25 °C, Kw = 1.01 ×10-14

Ex. How much is [OH-] if [H+] is 1.0 × 10-3 M at 25 °C?

pH = -log[H+]γH+ ~ -log[H+] (γH+ ~ 1)(True definition will be given in (8-10)

Ex2. How much is [OH-] if pH is 5 at 25 °C?

pH & pOH• pOH = -log[OH-]• pH + pOH = -log[H+][OH-]

= -logKW = 14.00 at 25 °C

So when pH = 5, pOH = 14 -5 = 9.

[OH-] = 10-9

Solution is acidic if [H+] > [OH-]Solution is basic if [H+] < [OH-]

Brønsted-Lowry Acids and Bases

DefinitionsBrønsted-Lowry Acids: Proton DonorBrønsted-Lowry Bases: Proton Acceptor

acid

For the reminder of the text, we will use Brønsted-Lowry’s definitions when we speak of acids and bases.

SaltsAny ionic solids, such as a mmoniumchloride, is called a salt. In a formal sense, a a salt can be thought of as the product of an acid-base reactions.

The nature of H+ and OH-

Conjugated Acids and Bases

AutoprotolysisH2O + H2O H3O+ + OH-←→

(base) (acid) (acid) (base)

H2O H+ + OH-←→

(6-16)

(6-17)

(6-16) and (6-17) mean the same.

Other protic solvents undergo autoprotolysis

Strengths of Acids and BasesStrong Acids and BasesA strong acid or base is completely

dissociated in aqueous solution.HCl(aq) H+ + Cl- (~ 100%)KOH(ac) K+ + OH-(~ 100%)

Remember all the acids and bases in Table 6-2. Weak acids and bases

HA H+ + A- a weak acid

Ka = [H+][A-]/[HA]

B + H2O BH+ + OH- a weak baseKb = [BH+][OH-]/[B]

←→

←→

←→

←→

Problem

(a)[H+] = 0.010 M = 1.0× 10-[Q1]pH = -log(1.0 × 10-[Q1])

= -log(1.0) + 2 = [Q2] (See p48)

(b) [OH-] = 3.5 × 10-2[H+] = Kw/[Q3] = 1.0 × 10-14/(3.5 × 10-2)

= 2.85 × 10-13

pH = - log[H+] = -log(2.85) –log10-13= -0.455 +13= 12.545 = 12.54

Lewis Acids and Bases (p108)

adduct

Home Work

• Read Ch 6 (6-3, 6-4, 6-5, 6-6)• Read Ch 7 & Ch 12 for Exp7, 8

ProblemsCh6: 6-B, 6-C, 6-D, 6-6, 6-14, 6-

15, 6-17, 6-19, 6-21a, 6-21b, 6-41, 6-46, 6-47, 6-48, 6-49, 6-50, 6-51, 6-52, 6-53