Principles of chemical equilibria

Yongsik Lee

Gibbs energy change for reaction

For chemical reactions

G = n G (products) - n G (reactants)

and

Grxn = H

rxn - T Srxn

rxnG and position of equilibrium

Consider the reaction: A B

rxnG = GB - G

A

If GA> G

B , rxnG is <0 spontaneous reaction exergonic reaction

At equilibrium, rxnG = 0.

ie. Not all A is converted into B; stops at equilibrium point.

Equilibrium diagram

For non-spontaneous rxn. GB > GA

Gas phase reactions

Consider the reaction in the gas phase: N2(g) + 3H2(g) → 2NH3(g)

Composition dependence of the μ μJ = μ°J + RT ln aJ

For gases, aJ = partial pressure of J

rG = {2 μNH3 } - {3 μH2 + μN2 }= rG + RT lnQ

Gas phase reactions

N2(g) + 3H2(g) → 2NH3(g) Q =( pNH3 / p)2 /( ( pN2/ p) (pH2/ p)3 )

where :Q = rxn quotientp = partial pressurep = standard pressure = 1 bar

Q is dimensionless because units of partial pressure cancelled by p .

Reactions at equilibrium At equilibrium:

rG = 0 and Qeq = K rG° = -RT ln K Qeq = K = (( pNH3 / p)2 / ( pN2/ p) (pH2/ p)3

)e

q

K > 1 if rG° < 0 Abundant products when K > 103

K < 1 if rG° > 0

Thermodynamic criteria of spontaneity

K > 1 if rG° < 0 rG° = rH° - T rS°

Case I Exothermic reaction rH° < 0 rS° > 0 Then rG° <0 at all temperature Spontaneous reaction at all temperature

Case II Endothermic & rS° < 0 Spontaneous reaction - Impossible

Standard reaction Gibbs energy

At 1 atm, Super cooled water spontaneously freezing at constant temperature -20℃. sysG < 0 (spontaneous change) sysS < 0 (liquid to solid) sysH < 0 (exothermic) surrS = - sysH/T > 0

Standard Gibbs energy of formation

Reference state Elements at 1.0 bar, Usually at 25 ℃

fG° = G° for elements in reference state to a compound

Thermodynamically unstable compound fG ° > 0 Ozone +163 kJ/mol Benzene +124 kJ/mol Solw decomposition!

Coupled reactions

ATP(aq) + H2O(l)→ ADP(aq) + Pi-(aq) + H+(aq)

Adenosine triphosphate Adenosine diphosphate Exergonic reaction (

Response of equilibria Le Chatelier’s Principl

e When a system at equi

librium is subjected to a disturbance,

the composition of the system adjusts so as to tend to minimize the effect of the disturbance

Presence of a catalyst

Catalyst Accelerates a reaction without itself appearing i

n the overall chemical equation Faster route but same reactants and products

lnK = -RT rxnG°

rxnG° is same The presence of a catalyst does not change the e

quilibrium constant of a reaction

Since Grxn = - RT ln K = H

rxn - TSrxn

ln K = - Grxn / RT = - H

rxn/RT + Srxn/R

ln K1 = - Grxn / RT1 = - H

rxn/RT1 + Srxn/R

ln K2 = - Grxn / RT2 = - H

rxn/ RT2 + Srxn/ R

ln K1 – ln K2 = - H

rxn / R ( 1/ T1 - 1/ T2 ) 0r

ln ( K1/ K2) = - Hrxn / R ( 1/ T1 - 1/ T2 )

van’t Hoff equation

Effect of temperature on K

흡열반응의 온도를 증가시키면 A → B with rxnH >0 (endothermic)

If T is increased, (T’>T) Then 1/T > 1/T’ ln ( K/ K’) = - H

rxn / R ( 1/ T - 1/ T’) <0 K < K’ 평형이 오른쪽으로 이동한다

Pressure dependence

lnK = -RT rxnG° rxnG° is independent of pressure But… PJ is changing!

Q=aNH32 /aN2aH2

3

압력이 두 배가 되면 , 모든 분압도 두 배 Q’ = (4/16)Q 따라서 평형은 오른쪽으로 이동해야 같은 K 기체 몰 수를 줄이는 쪽으로 이동

references http://www.whfreeman.com/ECHEM/INDEX.

HTML http://www.schaft.org/eri/people.html http://cwx.prenhall.com/bookbind/pubbooks

/hillchem3/medialib/media_portfolio/17.html Hill’s general chemistry

exercises

7-4, 7-5, 7-7, 7-28, 7-38

Consequences of equilibrium

Yongsik Lee

contents

Proton transfer equilibria Salts in water Solubility equilibria

Acid and Base

Bronensted-Lowry theory Acid = Proton donor Base = proton acceptor

Acid and Base

산의 정의 양성자를 내놓는 물질 수용액에서 양성자 이온을 만드는 물질 비공유 전자쌍을 받는 물질

염기의 정의 OH- 를 내놓는 물질 수용액에서 양성자 이온을 받아주는 물질 비공유 전자쌍을 주는 물질

What Is an Acid?

산의 성질 수용액에서 H+ 이온을 내놓는 물질 . 산의 묽은 수용액은 신맛이 있다 . 지시약에서 색깔이 변한다 . 산의 수용액은 전류를 흐르게 한다 . Carbonate( 달걀껍질 , 대리석 등 ) 을 포함하는

물질과 화학반응을 한다 . 금속과 반응하여 수소기체가 발생한다 .

산과 금속의 반응 염산과 마그네슘의 반응

Mg + 2H+ + 2Cl- → Mg2+ + 2Cl + H2↑

염산과 아연의 반응 Zn + 2H+ + 2Cl- → Zn2+ + 2Cl + H2↑

염산과 철의 반응 Fe + 2H+ + 2Cl- → Fe2+ + 2Cl + H2↑

산의 세기 강한 산

이온화 정도가 큰 산 수용액 속에서 대부분 이온화 → H+ 을 많이 냄 염산 (HCl), 황산 (H2SO4), 질산 (HNO3), HBr, HI, HClO4

약한 산 이온화 정도가 작은 산 수용액 속에서 일부만 이온화 → H+ 를 적게 냄 붕산 (H3BO4), 탄산 (H2CO3), 아세트산 (CH3COOH) 등

대부분의 산

중요한 산 : 염산 (HCl) 진한 염산과 묽은 염산

비중 1.18( 진한염산 ), 35% 이상의 염화수소 기체가 남아있음

묽은 염산 : 10% 이하의 염화수소가 녹아있는 수용액 제법

2NaCl + H2SO4 → NaSO4 + 2HCl 염화수소 : HCl + NH3 → NH4Cl( 흰연기 )

수용액에서 대부분 이온화 되므로 강한 산이다 .

황산 (H2SO4) 제법

S + O2 → SO2, SO3 + H2O → H2SO4

진한황산 농도가 98% 이며 물보다 무거운 점성이 큰 액체이다 . 물에 녹을 때 많은 열이 발생한다 . 수분을 흡수하는 성질이 매우 강해서 건조제로 사용한다 . 탈수 작용을 한다 . 수분이 거의 없어 이온화 되지 못하므로 산성을 나타내지 않는다 .

묽은 황산 강한 산성 금속과 반응하면 수소 기체가 발생

질산 (HNO3)

제법 2NaNO3 + H2SO4 → Na2SO4 + 2HNO3

진한 질산 무색의 발연성 액체 빛이나 열을 받으면 분해되기 떄문에 갈색병에 넣어

어두운 곳에 보관 . 묽은 질산

금속과 반응하면 수소가 발생 일산화질소 (NO) 와 이산화질소 (NO2) 가 함께 발생하므로

순수한 수소를 얻는 데는 부적당

다른 산 아세트산 (CH3COOH)

빙초산 ( 어는첨 17℃) 이라고도 하며 4~6% 수용액을 사용한다 .

약산 수용액 ( 농도 [A]) 의 pH 계산 Ka = x2/(A-x) x = (Ka[A])1/2

-log x = -(1/2)log Ka – (1/2)log [A] pH = (1/2)pKa - (1/2)log[A] Fraction deprotonated (f) = x /[A]

What is Base? 염기

물에 녹아서 수산화이온 (OH-) 을 낼 수 있는 물질 NaOH, KOH, Ca(OH)2, Ba(OH)2, Cu(OH)2, NH4OH CH3OH( 메탄올 ), C2H5OH( 에탄올 ) → 중성 ( 염기 X)

공통적 성질 쓴맛 , 미끈미끈 단백질 용해 : tissue 와 textile 의 손상 수용액은 전해질 용액 지시약 변화

염기의 세기 강한 염기

OH- 를 많이 내는 물질 ( 수용액 상태에서 ) NaOH, KOH, Ca(OH)2

약한 염기 이온화가 잘 안되어 OH- 를 조금밖에 내놓지

않음 . NH4OH, Mg(OH)2, Cu(OH)2

중요한 염기 : 수산화나트륨 (NaOH) 제법

2NaCl + 2H2O → 전기분해 → 2NaOH + H2↑ + Cl2↑ 성질

조해성 – 공기 중에서 수증기와 반응해서 스스로 녹는 현상 (ex. CaCl2)

단백질을 용해 : 털실이나 명주실로 된 옷을 비누로 세탁해서는 안됨 .

공기중 CO2 와 반응해서 흰 고체 생성 2NaOH + CO2 → Na2CO3 + H2O

보관 : 건조제를 넣은 플라스틱 병에 보관 ( 유리부식 ) 용도 : 양재물 , 비누 , 펄프 , 섬유 , 유리 등의 제조 원료

Ca(OH)2 – 수산화칼슘 ( 소석회 ) 제법

CaO + H2O → Ca(OH)2

성질 물에 잘 안 녹지만 이온화가 잘 되어 강한 염기 수용액 (석회수 ) 에 CO2 통과 → 뿌옇게 흐려져 CO2 검출

Ca(OH)2 + CO2 → CaCO3↓ + H2O 용도

석회벽돌 , 표백분의 원료 산성토양의 중화

NH3( 암모니아 )

약한 염기 OH- 이온이 없음에도 불구하고 염기

물에 잘 녹는 기체 ( 자극성 냄새 ) NH3 + H2O → NH4

+ + OH-

염산과 반응해서 흰 연기 발생 NH3 + HCl → NH4Cl( 염화암모늄 – 흰연기 )

용도 비료의 원료 냉동기의 냉매 . 프레온 이전부터 사용됨 .

제산제 = 중화 반응 중화반응

산의 H+ 와 염기의 OH- 가 만나서 물 (H2O) 이 되는 반응

H+ + OH- → H2O ( 중성 ) 중화반응의 예

염산 + 수산화나트륨 ( 산의 농도 = 염기의 농도 )

HCl + NaOH → H2O + Na+ + Cl- ( 중성 ) 질산 + 수산화칼륨

HNO3 + KOH → H2O + K+ + NO3- ( 중성 )

Conjugate Acid/Base

Conjugate acid/base HA (aq) + H2O (aq) → H3O+ (aq) + A-(aq)

K = [H3O+][A-]/[HA][H2O] Ka = [H3O+][A-]/[HA]

B(aq) + H2O (aq) → BH+(aq) + OH-(aq) K = [BH+][OH-]/[B][H2O] Kb = [BH+][OH-]/[B]

pH 척도

pH 수소이온의 농도를 나타내는 지수 pH = -log aH3O+

산성이 더 강한 용액일수록 pH 값은 작아진다 . [H+] >10-7 , pH < 7 용액은 산성이다 [H+] <10-7 , pH > 7 용액은 염기성이다

수소이온 몰농도와 그에 대응하는 pH 값

[H+] pH

1.0 =100 0

0.1 =10-1 1

0.01 =10-2 2

0.001 =10-3 3

0.00001 =10-5 5

0.0000001 =10-7 7

0.000000001 =10-9 9

일반적인 물질들의 pH

여러가지 물질의 pH

pH 미터

Autoprotolysis equilibrium Proton transfer between water

2H2O (aq) → H3O+ (aq) + OH-(aq) Autoprotolysis constant

K = [H3O+][OH-]/[H2O]2

Kw = [H3O+][OH-] pKw = -log Kw pKw = pH + pOH

Fraction protonated pKa + pKb = pKw

As the strength of a base decreases, the strength of its conjugate acid increases

Fraction deprotonated Fraction of acid molecules that have donated a

proton Fraction protonated

Fraction of base molecules that have protonated

Polyprotic acid Polyprotic acid

Donate more than one proton H2A(aq) + H2O(l) = H3O+(aq) + HA-(aq)

Ka1 = [H3O+][HA-]/[H2A]

HA-(aq) + H2O(l) = H3O+(aq) + A2-(aq) Ka2 = [H3O+][A2-]/[HA-]

Usually Ka1 ≈ 1000 Ka2

Second proton is more difficult to remove Partly on account of HA-

Some polyprotic acids

H2S Hydrogen sulfide

H2SO4 Sulfuric acid

H2SO3 Sulfurous acid

H3PO4 Phosphoric acid

H2C2O4 Oxalic acid

H2CO3 Carbonic acid

H2C3H2O4 Malonic acid

Amino acid lysine(Lys) Lys can accept two proton

s on its nitrogen atoms Donate one from its carbo

xyl group H3Lys2+

H2Lys+

Hlys pKa1=2.18 pKa2 = 8.95 pKa3 = 10.53

Fractional composition of Lys Lys can accept two proton

s on its nitrogen atoms Donate one from its carbo

xyl group H3Lys2+

H2Lys+

Hlys pKa1=2.18 pKa2 = 8.95 pKa3 = 10.53

Lysine Lysine is an essential amino acid

a basic building block of all protein. first isolated in 1889 from casein

Required for growth and bone development in children,

assists in calcium absorption and maintaining the correct nitrogen balance in the body and maintaining lean body mass.

to produce antibodies, hormones, enzymes, collagen formation as well as repair of tissue. 

Since it helps with the building of muscle protein, it is useful for patients recovering from injuries and recovery after operations

Lysine biosynthesis

Amphiprotic systems Amphiprotic species

Can both accept and donate protons

예 - amino acids Zwitterion

Double ion form

pH of a 0.100 M NaHSO4 solution Sulfuric acid is a strong acid, and the pKa2 of HSO4

- is 1.92. What is the pH of a 0.100 M NaHSO4 solution?

The salt is completely ionized in its solution. NaHSO4 → Na+ + HSO4

- The anion further ionizes. If [H+] = x, then the equilibrium concentrations of various

species are:   HSO4- = H+ + SO4

2-         Ka2 = 10-1.92 = 0.01200.100 - x   x,       x

Ka2 = x2 / (0.100-x) = 0.0120[H+] = x    = {-0.120 + (0.0122 + 4*0.00120)1/2} / 2    = 0.0292 M

Thus, pH = - log 0.0292 = 1.54

Is the NaHSO4 salt solution acidic?

Although no concentration is stated, such a solution is acidic because of the acidity of HSO4

-

When salt NH4Cl is added Weak acid NH4

+

Very weak base Cl- The solution is acidic

Salt MHA solution M+ + HA-

Amphiprotic HA-

pH = (1/2)(pKa1 + pKa2)

Acid-Base titrations

Yongsik Lee

Acid-base titrations Stoichiometric point

The stage at which a stoichiomtrically equivalent amount of acid has been added to a given amount of base

Analyte The solution being analyz

ed Titrant

The solution in the burret

To fill a buret To fill a buret, close th

e stopcock at the bottom and use a funnel.

You may need to lift up on the funnel slightly, to allow the solution to flow in freely.

Using disposable pipet You can also fill a buret us

ing a disposable transfer pipet. This works better than a funnel for the small, 10 mL burets.

Be sure the transfer pipet is dry or conditioned with the titrant, so the concentration of solution will not be changed.

Condition the buret Before titrating, condition the b

uret with titrant solution and check that the buret is flowing freely.

To condition a piece of glassware, rinse it so that all surfaces are coated with solution, then drain.

Conditioning 2-3 times will insure that the concentration of titrant is not changed by a stray drop of water.

Air bubble Check the tip of the bur

et for an air bubble. volume readings may be i

n error. To remove an air bubbl

e, whack the side of the buret tip while solution is flowing.

Before taking an initial vol reading

Rinse the tip of the buret with water from a wash bottle and dry it carefully.

After a minute, check for solution on the tip to see if your buret is leaking.

The tip should be clean and dry before you take an initial volume reading.

Initial volume reading

A buret reading card with a black rectangle can help you to take a more accurate reading.

Read the bottom of the meniscus. Be sure your eye is at the level of meniscus, not above or b

elow. Reading from an angle, rather than straight on, results in a parallax error.

Record in notebook

Take an initial volume reading and record it in your notebook.

Before beginning a titration, you should always calculate the expected endpoint volume.

Prepare analyte

Prepare the analyte by placing it in a clean Erlenmeyer flask or beaker.

If your sample is a solid, make sure it is completely dissolved.

Put a magnetic stirrer and add indicator.

titration

Use the buret to deliver a stream of titrant to within a couple of mL of your expected endpoint.

You will see the indicator change color when the titrant hits the solution in the flask, but the color change disappears upon stirring.

Approaching the endpoint

Approach the endpoint more slowly and watch the color of your flask carefully.

Use a wash bottle to rinse the sides of the flask and the tip of the buret, to be sure all titrant is mixed in the flask.

Partial drop of titrant

As you approach the endpoint, you may need to add a partial drop of titrant.

You can do this with a rapid spin of a teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle.

For phenolphthalein, the endpoint is the first permanent pale pink. The pale pink fades in 10 to 20 minutes.

If you think you might have reached the endpoint, you can record the volume reading and add another partial drop.

Gone too far…

Record the final volume

When you have reached the endpoint, read the final volume in the buret and record it in your notebook.

calculate the number of moles of reactant in your analyte solution.

pH curve

Henderson-Hasselbalch equation pH ≈ pKa – log{[acid]/[base]} Half way to the stoichiometric point,

[acid] = [base] pH = pKa – log 1 = pKa

Buffer range? Log function

Buffer action

Buffer action Slow variation of the pH Ability of a solution to oppose changes in

pH when small amounts of acids and bases are added

Buffer solution Weak acid + conjugate base Weak base + conjugate acid

Indicator

Indicators

지시약 pH 에 따라 색이 달라지는 물질 – 냄새 ? 맛 ? 지시약 자신은 약한 염기나 약한 산성을 띰 ( 액성에

따라 색깔이 다름 ) 지시약으로 쓰이는 천연 물질 : 과일즙 , 꽃잎

Indicator Large, water soluble, organic molecule with acid(HIn) and its conjugate base(In-) forms tha

t differ in color

Color change

용액의 성질

지시약의 종류산성 중성 염기성

리트머스 종이 붉은색 보라색 푸른색

BTB 용액 노란색 녹색 푸른색

페놀프탈레인 무색 무색 붉은색

메틸오렌지 붉은색 주황색 노란색

천연물 지시약 소량의 레몬 주스를 한 컵의 차에 넣으면 차의 색깔이 갈색에서 연한 노란색이 된다 . 이것은 주스의 시트로산에 의해서 차 용액의 pH 가 감소할 때 , 차에 들어 있는 색깔 성분이 지시약으로 작용하기 때문이다 .

붉은 배추의 즙도 천연 지시약이다 . 산성 용액에서 붉은 색인 이 지시약은 염기의 첨가에 따라 초록과 파랑을 포함하는 다른 색깔을 낸다 .

Color change

Equilibrium in solution HIn(aq) + H2O(l) = H3O+(aq) + In-(aq)

KIn = [H3O+][In-]/[HIn] Table 8.4 for Ka

KIn /[H3O+]= [In-]/[HIn] pH – pKIn = log{ [In-]/[HIn]}

Color changes of indicator During titration

HA, H+, A-, Hin Hin is weaker than HA Na+, OH-

When HA is used up No more HA present Hin → H+, In- Na+, OH-

Solubility constant, Ks

For sparingly soluble ion compounds Solubility constant is a K

K = [A+(aq)][B-(aq)]/[AB(s)] Ks = [A+(aq)][B-(aq)] Solubility product (constant) Ksp

Molar solubility (S) Moles/L Concentration in saturated solution

Solubility product

The Common-ion effect Reduction of the solubility of a sparingly

soluble salt by the presence of a common ion

화학과전공설명회

4 월 28 일 목요일

전공설명회오후 4 시 30 분공학관 148 호실

오픈 랩오후 5 시공학관 B 동 4 층 전체

4/28 목요일 17:00 화학과 오픈 랩피자 /맥주 스탠딩 파티 !

경고 : 파티 후유증

Exercises

8-3, 8-7, 8-8, 8-18, 8-20

References http://www.whfreeman.com/ECHEM/INDEX.HTML http://www.anyvitamins.com/lysine-

info.htm http://www.dartmouth.edu/

~chemlab/techniques/titration.html http://leroy.uwaterloo.ca/c123/Ch18/

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