kesetimbangan kimia

KESETIMBANGAN KIMIAREVIEW (4 TO 6)JURUSAN TL-ITS

Kesetimbangan Persamaan reaksi hanya menyatakan

hubungan jumlah (kuantitas) dari zat‐zat yang bereaksi dengan zat‐zat hasil reaksi secara stoikiometri.

Kinetika serta termodinamika reaksi mempelajari berapa lama suatu reaksi akan berlangsung dan ke arah mana yang paling mungkin terjadiKesetimbangan ini akan terjadi jika jumlahPembentukan sama dengan jumlah yang teruraiEQUILIBRIUM.

Con’t A dynamic equilibrium consists of a forward

reaction, in which substances react to give products, and a reverse reaction, in which products react to give the original reactants.

Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.

Copyright © Houghton Mifflin Company.All rights reserved.

Presentation of

Lecture

Outlines, 15–4

A Problem to Consider Applying Stoichiometry to an Equilibrium

Mixture.

– What is the composition of the equilibrium mixture if it contains 0.080 mol NH3?

– Suppose we place 1.000 mol N2 and 3.000 mol H2 in a reaction vessel at 450 oC and 10.0 atmospheres of pressure. The reaction is

(g)2NH )g(H3)g(N 322


Presentation of

Lecture

Outlines, 15–5

A Problem to Consider

– The equilibrium amount of NH3 was given as 0.080 mol. Therefore, 2x = 0.080 mol NH3 (x = 0.040 mol).

Using the information given, set up the following table.

(g)2NH )g(H3 )g(N 322

Starting 1.000 3.000 0Change -x -3x +2x

Equilibrium1.000 - x 3.000 - 3x 2x = 0.080 mol


Presentation of

Lecture

Outlines, 15–6

A Problem to Consider Using the information given, set up the

following table.

Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2

Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2

Equilibrium amount of NH3 = 2x = 0.080 mol NH3

(g)2NH )g(H3 )g(N 322

Starting 1.000 3.000 0Change -x -3x +2x

Equilibrium1.000 - x 3.000 - 3x 2x = 0.080 mol


Presentation of

Lecture

Outlines, 15–7

The Equilibrium Constant Every reversible system has its own

“position of equilibrium” under any given set of conditions.– The ratio of products produced to

unreacted reactants for any given reversible reaction remains constant under constant conditions of pressure and temperature.

– The numerical value of this ratio is called the equilibrium constant for the given reaction.


Presentation of

Lecture

Outlines, 15–8

The Equilibrium Constant The equilibrium-constant expression for a

reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation.

ba

dc

c ]B[]A[]D[]C[K

– For the general equation above, the equilibrium-constant expression would be:

dDcC bBaA


Presentation of

Lecture

Outlines, 15–9

The Equilibrium Constant The equilibrium-constant expression for a

reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation.

ba

dc

c ]B[]A[]D[]C[K

– The molar concentration of a substance is denoted by writing its formula in square brackets.

dDcC bBaA


Presentation of

Lecture

Outlines, 15–10

The Equilibrium Constant The equilibrium constant, Kc, is the

value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted.– A large Kc indicates large concentrations of products

at equilibrium.– A small Kc indicates large concentrations of

unreacted reactants at equilibrium.


Presentation of

Lecture

Outlines, 15–11

Obtaining Equilibrium Constants for Reactions

Equilibrium concentrations for a reaction must be obtained experimentally and then substituted into the equilibrium-constant expression in order to calculate Kc.


Presentation of

Lecture

Outlines, 15–12


Consider the reaction below

O(g)H (g)CH (g)H 3 )g(CO 242

– Suppose we started with initial concentrations of CO and H2 of 0.100 M and 0.300 M, respectively.


Presentation of

Lecture

Outlines, 15–13

– When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows.

Reactants Products[CO] = 0.0613 M[H2] = 0.1893 M

[CH4] = 0.0387 M[H2O] = 0.0387 M


Consider the reaction belowO(g)H (g)CH (g)H 3 )g(CO 242


Presentation of

Lecture

Outlines, 15–14

– The equilibrium-constant expression for this reaction is:

32

24c ]H][CO[

]OH][CH[K


Consider the reaction below (see Figure 15.5).

O(g)H (g)CH (g)H 3 )g(CO 242


Presentation of

Lecture

Outlines, 15–15

– If we substitute the equilibrium concentrations, we obtain:

93.3)M1839.0)(M0613.0()M0387.0)(M0387.0(K 3c


Consider the reaction below (see Figure 15.5).

O(g)H (g)CH (g)H 3 )g(CO 242


Presentation of

Lecture

Outlines, 15–16

The Equilibrium Constant, Kp In discussing gas-phase equilibria, it is

often more convenient to express concentrations in terms of partial pressures rather than molarities– It can be seen from the ideal gas equation that the

partial pressure of a gas is proportional to its molarity.

MRTRTVnP )(


Presentation of

Lecture

Outlines, 15–17

The Equilibrium Constant, Kp If we express a gas-phase equilibria in

terms of partial pressures, we obtain Kp.– Consider the reaction below.

O(g)H (g)CH (g)H 3 )g(CO 242

– The equilibrium-constant expression in terms of partial pressures becomes:

3HCO

OHCHp

2

24

KPP

PP


Presentation of

Lecture

Outlines, 15–18

The Equilibrium Constant, Kp In general, the numerical value of Kp

differs from that of Kc.– From the relationship n/V=P/RT, we can show that

where Dn is the sum of the moles of gaseous products in a reaction minus the sum of the moles of gaseous reactants.

ncp )RT(KK D


Presentation of

Lecture

Outlines, 15–19


– We know that

ncp )RT(KK D

From the equation we see that Dn = -1. We can simply substitute the given reaction temperature and the value of R (0.08206 L.atm/mol.K) to obtain Kp.

Consider the reaction(g)SO 2 )g(O)g(SO2 322


Presentation of

Lecture

Outlines, 15–20


– Since

ncp )RT(KK D

3.4 K) 1000 08206.0( 108.2K 1-Kmol

atmL2p

Consider the reaction

(g)SO 2 )g(O)g(SO2 322


Presentation of

Lecture

Outlines, 15–21

Equilibrium Constant for the Sum of Reactions Similar to the method of combining

reactions that we saw using Hess’s law, we can combine equilibrium reactions whose Kc values are known to obtain Kc for the overall reaction.– With Hess’s law, when we reversed reactions or

multiplied them prior to adding them together, we had to manipulate the DH’s values to reflect what we had done.

– The rules are a bit different for manipulating Kc.


Presentation of

Lecture

Outlines, 15–22 2. If you multiply each of the coefficients in an equation by

the same factor (2, 3, …), raise Kc to the same power (2, 3, …).

3. If you divide each coefficient in an equation by the same factor (2, 3, …), take the corresponding root of Kc (i.e., square root, cube root, …).

4. When you finally combine (that is, add) the individual equations together, take the product of the equilibrium constants to obtain the overall Kc.

1. If you reverse a reaction, invert the value of Kc.

Equilibrium Constant for the Sum of Reactions


Presentation of

Lecture

Outlines, 15–23

For example, nitrogen and oxygen can combine to form either NO(g) or N2O (g) according to the following equilibria.

NO(g) 2 )g(O)g(N 22

O(g)N )g(O)g(N 2221

2

Kc = 4.1 x 10-31

Kc = 2.4 x 10-18

(1)

(2)

Kc = ?

– Using these two equations, we can obtain Kc for the formation of NO(g) from N2O(g):

NO(g) 2 )g(O)g(ON 221

2 (3)



Presentation of

Lecture

Outlines, 15–24

To combine equations (1) and (2) to obtain equation (3), we must first reverse equation (2). When we do we must also take the reciprocal of its Kc value.

NO(g) 2 )g(O)g(ON 221

2

NO(g) 2 )g(O)g(N 22 Kc = 4.1 x 10-31(1)

)g(O (g)N O(g)N 221

22 Kc = (2)

(3)

18-10 4.21

1318

31c 107.1

104.21)101.4()overall(K )(



Presentation of

Lecture

Outlines, 15–25

Heterogeneous Equilibria A heterogeneous equilibrium is an

equilibrium that involves reactants and products in more than one phase.– The equilibrium of a heterogeneous system is

unaffected by the amounts of pure solids or liquids present, as long as some of each is present.

– The concentrations of pure solids and liquids are always considered to be “1” and therefore, do not appear in the equilibrium expression.

Kc - ignoring solids and liquids

Kc calculations involve: 1) Concentrations in mol/L, 2) Gaseous products and reactants

For gases, if mol or L changes so does mol/LQ - If a gas is 1 mol/L, what is it’s new [ ] if it’s

compressed to 1/2 of its original volume?A - Now we have 2 mol/L (e.g. 1 mol / 0.5 L)Conversely, solids and liquids don’t compressQ - If a solid is initially 1 mol/L, what is the

concentration if the volume is reduce to 1/2?A - To get half the volume we must cut it in half … we

still have 1 mol/L (e.g. 0.5 mol / 0.5 L)Concentrations of liquids and solids do not change.

Concentrations of ions and gases do.


Presentation of

Lecture

Outlines, 15–27

Heterogeneous Equilibria Consider the reaction below.

(g)H CO(g) )g(OH)s(C 22 – The equilibrium-constant expression contains terms for only those

species in the homogeneous gas phase…H2O, CO, and H2.

]OH[]H][CO[K

2

2c

Precipitation Pelarutan atau kelarutan endapan Endapan adalah zat atau materi yang

memisahkan diri sebagai fase padat dari sistem larutan.

Fase padat ini dapat terjadi dalam bentuk kristal, bentuk tersuspensi, maupun bentuk koloid.

Centrifuge, Filtrasi, Flotation and Sedimentation.

Faktor yang mempengaruhi... Solubility Zat padat yang ditambahkan tidak

mampu lagi larut dalam pelarutnya - Ksp

Ion sekutu ion‐ion dari zat lain yang merupakan bahan endapan

Ksp Larutan yang sudah tidak mampu lagi

melaruitkan zat terlarut ini dinamakan larutan jenuh, yaitu larutan yang ion‐ionnya telah mencapai kesetimbangan antara melarut dan mengkristal

KspIAP

nRTG ln

D KspIAPSI log

SInRTG 303.2

D

Con’t When SI = 0, hence DG =0, the solution

is in equilibrium. When SI < 0, DG > 0, the solution is

undersaturated and precipitation is impossible in solution

When SI > 0, DG < 0, the solution is supersaturated and precipitation is spontaneous.

Ksp (solubility product) - background

Bottom line: 1) Ksp is similar to Kc, 2) It deals with ions instead of gases, 3) one side of the chem. equation has a solid, which is ignored

We have used Kc in equilibrium problems“K” indicates an equilibrium“c” indicates units are mol/L (in this course it will also indicate we are dealing with gases)

There are other types of K: Ksp, Kw, Ka, KbEach subscript immediately indicates some detail about the equilibriumKw: equilibrium of water, Ka: acid, Kb: baseKsp: equilibrium between solid and ions

Ksp - background

The equilibrium between solids and ions is different from the equilibrium between gases

The equilibrium between solids and ions is a “phase” equilibrium (e.g. NaCl(aq))

NaCl(s)

Na+(aq) + Cl-(aq)

Ksp deals with a phase equilibrium: (s) (aq)

Why use Ksp instead of Kc?

Read 14.9 (574 - 575), Try PE 13Consider: NaCl Na+(aq) + Cl-(aq)

NaCl(s)

Na+(aq) + Cl-(aq)

Ksp = [Na+(aq)] [Cl-(aq)]We use Ksp because it

indicates that the equilibrium is a phase equilibrium (between a

solid and it’s ions)

K =[Na+(aq)] [Cl-(aq)]

[NaCl(s)]

[NaCl(s)] doesn’t change (it’s a konstant)

K • [NaCl(s)] = [Na+(aq)] [Cl-(aq)]

Example The solubility of silver sulfate is 0.014

mol/L (Ag2SO4). Calculate the Ksp. We just saw that the solubility of nickel (II)

carbonate is 3.7 x 10-4 mol/L. What mass of NiCO3 is needed to prepare 500 mL of saturated solution? Mr=118.72

0.50 L of 1.0 x 10–5 M Pb(OAc)2 is combined with 0.50 L of 1.0 x 10–3 M K2CrO4.

a. Will a ppt occur? Ksp(PbCrO4) = 1.8 x 10–1

b. Find the Eq. conc. of Pb2+ remaining in solution

after the PbCrO4 precipitates.

Calculating the Ksp of silver sulfate

The solubility of silver sulfate is 0.014 mol/L. This means that 0.0144 mol of Ag2SO4 will dissolve to make 1.0 L of saturated solution. Calculate the value of the equilibrium constant, Ksp for this salt.

Ag2SO4 (s) 2 Ag+ (aq) + SO42- (aq)

--- --- + 2s + s 2s s

Ksp = [Ag+]2[SO42-] = (2s)2(s) = (4s2)(s) = 4s3

We know: s = 0.0144 mol/L

Ksp = 4(0.0144)3 = 1.2 x 10-5

Other ways to express solubility... We just saw that the solubility of nickel (II)

carbonate is 3.7 x 10-4 mol/L. What mass of NiCO3 is needed to prepare 500 mL of saturated solution?

0.022 g of NiCO3 will dissolve to make 500 mL solution. Try Problems 9 - 26

g 0.022 NiCO mol 1

g 118.72 x L 0.500 x L1

NiCO mol10x3.7

3

34

39

Problem: 0.50 L of 1.0 x 10–5 M Pb(OAc)2 is combined with 0.50 L of 1.0 x 10–

3 M K2CrO4.

a. Will a ppt occur? Ksp(PbCrO4) = 1.8 x 10–14

Pb(OAc)2(aq) + K2CrO4(aq) PbCrO4(s) + 2KOAc(aq)

then: PbCrO4(s) Pb2+ + CrO42– Ksp= [Pb2+][CrO4

2–]

[Pb2+]:LPb mol

2

0.50 L

LPb(OAc) mol10x 0.1 2

-5

2

2

Pb(OAc) 1Pb 1

1 L5.0 x 10–6

[CrO42-]:

LCrO mol

-24

0.50 LL

CrOK mol10x 1.0 42-3

42

-24

CrOK 1CrO 1 5.0 x 10-4

Q = [5.0 x 10-6][5.0 x 10-4] = 2.5 x 10-9

compare to Ksp: Q > Ksp so a ppt. will occur

1 L

40

b. find the Eq. conc. of Pb2+ remaining in solution after the PbCrO4 precipitates.

Since [Pb2+] = 5.0 x 10-6 and [CrO42-] = 5.0 x 10-4 and there is a 1:1

stoichiometry, Pb2+ is the limiting reactant.

PbCrO4(s) Pb2+ + CrO42–

I. (after ppt.)C.E.

0 5.0 x 10-4 - 5.0 x 10-6 = 5.0 x 10-4

+x +x x 5.0 x 10–4 + x

Ksp = [x][5.0 x 10–4 + x] Try dropping the “+ x” term.

Ksp(PbCrO4) = 1.8 x 10–14

1.8 x 10–14 = x(5.0 x 10-4)x = [Pb2+] = 3.6 x 10–11

This is the concentration of Pb2+ remaining in solution.

Selective PrecipitationPengendapan SulfidaPengendapan dan Kelarutan Hidroksida

Pengedapan Sulfida Merupakan pelarut logam Mempunyai kelarutan yang rendah di air pH mempengaruhi proses : H2S merupakan asam lemah, yang akan

terdisosiasi menjadi ion H+ dan S2-, dengan konsentrasi yang bervariasi sesuai dengan keadaan pH larutan.

Tahap dissosiasinya adalah :

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