equilibrium in chemical reaction

EquilibriumIn Chemical Reaction

The double arrow tells us thatthis reaction can go in both directions:

N2 + 3H2 2NH3

1) Reactants react to become products,

N2 + 3H2 2NH3

N2 + 3H2 2NH3

(‘forward’ reaction)

1) Reactants react to become products,

N2 + 3H2 2NH3

while simultaneously,

N2 + 3H2 2NH3

(‘forward’ reaction)

2) Products react to become reactantsN2 + 3H2 2NH3

(‘reverse’ reaction)

N2 + 3H2 2NH3

In a closed system, where no reactants, products, or energycan be added to or removed from the reaction,a reversible reaction will reach equilibrium.

N2 + 3H2 2NH3

At equilibrium, the rate of the forward reaction becomes equal to the rate of the reverse reaction, and so, like our escalator metaphor, the two sides, reactants and products, will have constant amounts, even though the reactions continue to occur.

N2 + 3H2 2NH3

However (like the metaphor), the equilibrium amounts of reactants and products are usually not equal, they just remain unchanged.

N2 + 3H2 2NH3

reverse

forward

etc!the reactions go on continuously in both directions.

reverse

forward

Changes in the concentrations of the reactants and products can be graphed; the graph indicates when equilibrium has been reached.

conc

entr

atio

n

time

For N2 + 3H2 2NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M

conc

entr

atio

n

time


conc

entr

atio

n

time

[N2]

[H2]

[NH3]


conc

entr

atio

n

timeQuestion 3: at what point

has equilibrium been established?

[N2]

[H2]

[NH3]


conc

entr

atio

n

time

Question 4: what does the graph tell you about theconcentration of each species once equilibrium is established?

[N2]

[H2]

[NH3]


conc

entr

atio

n

timeQuestion 5: what might a rate vs time graph

look like for the above reaction?

[N2]

[H2]

[NH3]

rate

timeQuestion 5: what might a rate vs time graph

look like for the above reaction?

For and still beginning withN2 + 3H2 2NH3

N2 = 1 M, H2 = 1 M, and NH3 = 0 M

For and still beginning with

rate

time

N2 + 3H2 2NH3

N2 = 1 M, H2 = 1 M, and NH3 = 0 M

For and still beginning with

rate

time

N2 + 3H2 2NH3

N2 = 1 M, H2 = 1 M, and NH3 = 0 M

Question 6: at what pointhas equilibrium been established?

forward

reve

rse

rate

time

Question 7: describe how the two graphs are related.

conc

entr

atio

n

time

N2

H2

NH3

forward

reve

rse

rate

time

Question 8: do either of the two graphsindicate if Keq >1 or Keq <1?

conc

entr

atio

n

time

N2

H2

NH3

forward

reve

rse

58

Equilibrium: the extent of a reaction

In stoichiometry we talk about theoretical yields, and the many reasons actual yields may be lower.

Another critical reason actual yields may be lower is the reversibility of chemical reactions: some reactions may produce only 70% of the product you may calculate they ought to produce.

Equilibrium looks at the extent of a chemical reaction.

59

The Concept of EquilibriumThe Concept of Equilibrium Consider colorless frozen N2O4. At room temperature, it

decomposes to brown NO2:

N2O4(g) 2NO2(g).

At some time, the color stops changing and we have a mixture of N2O4 and NO2.

Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant.

Using the collision model: as the amount of NO2 builds up, there is a chance that two NO2

molecules will collide to form N2O4.

At the beginning of the reaction, there is no NO2 so the reverse reaction (2NO2(g) N2O4(g)) does not occur.

60

As the substance warms it begins to decompose:

N2O4(g) 2NO2(g)

When enough NO2 is formed, it can react to form N2O4:

2NO2(g) N2O4(g).

At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4

The double arrow implies the process is dynamic.N2O4(g) 2NO2(g)A B

61

As the reaction progresses– [A] decreases to a constant,– [B] increases from zero to a constant.– When [A] and [B] are constant,

equilibrium is achieved.

A BA B

62

• No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium.

• For a general reaction

the equilibrium constant expression is

where Kc is the equilibrium constant.

aA + bB(g) pP + qQ

ba

qp

cKBA

QP

63

Kc is based on the molarities of reactants and products at equilibrium.

We generally omit the units of the equilibrium constant.

Note that the equilibrium constant expression has products over reactants.

64

Write the equilibrium expression for the following reaction:

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

65

The Equilibrium Constant The Equilibrium Constant in Terms of Pressurein Terms of Pressure

•If KP is the equilibrium constant for reactions involving gases, we can write:

•KP is based on partial pressures measured in atmospheres.

ba

qp

PPP

PPK

BA

QP

66

The Magnitude The Magnitude

of Equilibrium Constantsof Equilibrium Constants

Therefore, the larger K the more products are present at equilibrium.

Conversely, the smaller K the more reactants are present at equilibrium.

If K >> 1, then products dominate at equilibrium and equilibrium lies to the right.

If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.

An equilibrium can be approached from any direction.

Example:

N2O4(g) 2NO2(g)

212.0

ONNO

42

22 cK

However,

The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

2NO2(g) N2O4(g)

72.4212.01

NO

ON2

2

42 cK

Heterogeneous EquilibriaHeterogeneous Equilibria• When all reactants and products are in one

phase, the equilibrium is homogeneous.• If one or more reactants or products are in a

different phase, the equilibrium is heterogeneous.

• Consider:

– experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

CaCO3(s) CaO(s) + CO2(g)

Heterogeneous EquilibriaHeterogeneous Equilibria

Heterogeneous EquilibriaHeterogeneous Equilibria• Neither density nor molar mass is a variable, the

concentrations of solids and pure liquids are constant. (You can’t find the concentration of something that isn’t a solution!)

• We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions.

• The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present.

Kc = [CO2] Kp = [pCO2]

CaCO3(s) CaO(s) + CO2(g)

72

Applications of Equilibrium ConstantsApplications of Equilibrium Constants

Predicting the Direction of ReactionPredicting the Direction of Reaction

We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium

as

where [A], [B], [P], and [Q] are molarities at any time.

Q = K only at equilibrium.

aA + bB(g) pP + qQ

ba

qpQ

BA

QP

73

Predicting the Direction of ReactionPredicting the Direction of Reaction• If Q > K then the reverse reaction must occur to

reach equilibrium (go left)• If Q < K then the forward reaction must occur to

reach equilibrium (go right)

Note the moles into a 10.32 L vessel stuff ... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M

2 HI H2 + I2

222

][

]][[

HI

IHKeq Initial:

Change:Equil:

0.242 M 0 0

-2x +x +x

0.242-2x x x

32

2

21026.1

]2242.0[]2242.0[

]][[

xx

x

x

xxKeq

What we are asked for here is the equilibrium concentration of H2 ... ... otherwise known as x. So, we need to solve this beast for x.

32

2

1026.1]2242.0[

xx

x

232 ]2242.0[1026.1 xxx

]4968.00586.0[1026.1 23 xxx

2335 1004.51022.11038.7 xxxxx

01038.71022.1995.0 532 xxxx

And yes, it’s a quadratic equation. Doing a bit of rearranging:

a

acbbx

2

42

x = 0.00802 or –0.00925Since we are using this to model a real, physical system,we reject the negative root.The [H2] at equil. is 0.00802 M.

This type of problem is typically tackled using the “three line” approach:

2 NO + O2 2 NO2

Initial:

Change:

Equil.:

Approximating

If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants.

0.20 – x is just about 0.20 is x is really dinky.

If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.

77

Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M

I2 2 I

Initialchangeequil:

0.20 0-x +2x0.20-x 2x

102

10

2

2

1094.2]20.0[

]2[

1094.2][

][

xx

x

xI

IKeq

With an equilibrium constant that small, whatever x is, it’s near dink, and 0.20 minus dink is 0.20 (like a million dollars minus a nickel is still a million dollars).

0.20 – x is the same as 0.20

102

1094.220.0

]2[ xx

x = 3.83 x 10-6 M

More than 3orders of mag.between thesenumbers. The simplification willwork here.

Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M

I2 2 I

Initial:Change:equil:

0.20 0-x +2x0.20-x 2x 209.0

]20.0[

]2[

209.0][

][

2

2

2

x

x

I

IKeq

These are too close toeach other ... 0.20-x will not betrivially close to 0.20here.

Looks like this one has to proceed through the quadratic ...

Le Chatelier’s Principle: if you disturb an equilibrium, it will shift to undo the disturbance.

Remember, in a system at equilibrium, come what may, the concentrations will always arrange themselves to multiply and divide in the Keq equation to give the same number (at constant temperature).

Le Châtelier’s PrincipleLe Châtelier’s Principle

Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• Adding a reactant or product shifts the equilibrium

away from the increase.• Removing a reactant or product shifts the equilibrium

towards the decrease.• To optimize the amount of product at equilibrium, we

need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier).

• We illustrate the concept with the industrial preparation of ammonia

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

82

• Consider the Haber process

• If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier).

• That is, the system must consume the H2 and produce products until a new equilibrium is established.

• Therefore, [H2] and [N2] will decrease and [NH3] increases.

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

83

Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• The unreacted nitrogen and hydrogen are recycled

with the new N2 and H2 feed gas.

• The equilibrium amount of ammonia is optimized because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added.

Effects of Volume and PressureEffects of Volume and Pressure• As volume is decreased pressure increases.• Le Châtelier’s Principle: if pressure is increased the

system will shift to counteract the increase.a

• Consider the production of ammonia

• As the pressure increases, the amount of ammonia present at equilibrium increases.

• As the temperature decreases, the amount of ammonia at equilibrium increases.

• Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

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

85

Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations

87

Effects of Volume and PressureEffects of Volume and Pressure• The system shifts to remove gases and

decrease pressure.• An increase in pressure favors the direction

that has fewer moles of gas.• In a reaction with the same number of

product and reactant moles of gas, pressure has no effect.

• Consider N2O4(g) 2NO2(g)

88

Effects of Volume and PressureEffects of Volume and Pressure• An increase in pressure (by decreasing the

volume) favors the formation of colorless N2O4.

• The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased.

• The system moves to reduce the number moles of gas (i.e. the reverse reaction is favored).

• A new equilibrium is established in which the mixture is lighter because colorless N2O4 is favored.

N2O4(g) 2NO2(g)

Effect of Temperature ChangesEffect of Temperature Changes• The equilibrium constant is temperature

dependent.• For an endothermic reaction, H > 0 and heat

can be considered as a reactant.• For an exothermic reaction, H < 0 and heat can

be considered as a product.• Adding heat (i.e. heating the vessel) favors

away from the increase:– if H > 0, adding heat favors the forward reaction,– if H < 0, adding heat favors the reverse reaction.

90

Effect of Temperature ChangesEffect of Temperature Changes• Removing heat (i.e. cooling the vessel), favors

towards the decrease:– if H > 0, cooling favors the reverse reaction,– if H < 0, cooling favors the forward reaction.

• Consider

for which H > 0.

– Co(H2O)62+ is pale pink and CoCl4

2- is blue.

Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)

Penggabungan Rumus Tetapan Kesetimbangan

Jika diketahui:

N2(g) + O2(g) 2NO(g)Kc = 4,1 x 10-31

N2(g) + ½ O2(g) N2O(g) Kc = 2,4 x 10-18

Bagaimana Kc reaksi:

N2O(g) + ½ O2(g) 2NO(g) Kc = ?Kita dapat menggabungkan persamaan diatas

N2(g) + O2(g) 2NO(g) Kc = 4,1 x 10-31

N2O(g) N2(g) + ½ O2(g) Kc = 1/(2,4 x 10-18) = 4,2 x 1017

N2O(g) + ½ O2(g) 2NO(g) Kc = ?

131731

2/122

2

2

2/122

22

2

107,1102,4101,4)2()1()(

)(]][[

][

][

]][[

]][[

][

xxxxKxKbersihK

netKOON

NO

ON

ONx

ON

NO

ccc

c

Tetapan kesetimbangan untuk reaksi bersih adalah hasil kali tetapan kesetimbangan untuk reaksi-reaksi terpisah yang digabungkan

Soal Latihan

1.Untuk reaksi NH3 ↔ ½ N2 + 3/2 H2 Kc = 5,2 x 10-5 pada 298 K. Berapakah nilai Kc pada 298 K untuk reaksi:

N2 + 3H2 ↔ 2NH3

2. Senyawa ClF3 disiapkan melalui 2 tahap reaksi fluorinasi gas klor sebagai berikut

(i) Cl2(g) + F2(g) ClF(g)

(ii) ClF(g) + F2(g) ClF3(g)– Seimbangkan masing-masing reaksi diatas dan tuliskan

reaksi overallnya!– Buktikan bahwa Kc overall sama dengan hasil kali Kc

masing-masing tahap reaksi ?

Hubungan Tetapan Kesetimbangan Kc dan Kp

• Tetapan kesetimbangan dalam sistem gas dapat dinyatakan berdasarkan tekanan parsial gas, bukan konsentrasi molarnya

• Tetapan kesetimbangan yang ditulis dengan cara ini dinamakan tetapan kesetimbangan tekanan parsial dilambangkan Kp.

Misalkan suatu reaksi

2SO2(g) + O2(g) 2SO3(g) Kc = 2,8 x 102 pd 1000 K

][][

][

22

2

23

OSO

SOK c

Sesuai dengan hukum gas ideal, PV = nRT

Dengan mengganti suku-suku yang dilingkari dengan konsentrasi dalam Kc akan diperoleh rumus;

RT

P

V

nO

RT

P

V

nSO

RT

P

V

nSO OOSOSOSOSO 222233 ][][][ 223

RTxPP

P

RTPRTP

RTPK

OSO

SO

OSO

SO

c )()(

)(

)/()/(

)/(

22

3

22

3

2

2

2

2

Terlihat ada hubungan antara Kc dan Kp yaitu:

1)( RTKRT

KKdanRTxKK c

cppc

Jika penurunan yang sama dilakukan terhadap reaksi umum:

aA(g) + bB(g) + … gG(g) + hH(g) + …

Hasilnya menjadi Kp = Kc (RT)n

Dimana n adalah selisih koefisien stoikiometri dari gas hasil reaksi dan gas pereaksi yaitu n = (g+h+…) – (a+b+…).Dalam persamaan reaksi pembentukan gas SO3 diatas kita lihat bahwa n = -1

Soal Latihan

Hitunglah nilai Kp reaksi kesetimbangan berikut:

• PCl3(g) + Cl2(g) PCl5(g) Kc = 1,67 (at 500 K)

• N2O4(g) 2NO2(g); Kc = 6,1 x 10-3 (298 K)

• N2(g) + 3H2(g) 2NH3(g) Kc = 2,4 x 10-3 (at 1000 K).

Kesetimbangan yang melibatkan cairan dan padatan murni (Reaksi Heterogen)

• Persamaan tetapan kesetimbangan hanya mengandung suku-suku yang konsentrasi atau tekanan parsialnya berubah selama reaksi berlangsung

• Atas dasar ini walaupun ikut bereaksi tapi karena tidak berubah, maka padatan murni dan cairan murni tidak diperhitungkan dalam persamaan tetapan kesetimbangan.

C(s) + H2O(g) CO(g) + H2(g) ][

]][[

2

2

OH

HCOK c

CaCO3(s) CaO(s) + CO2(g) Kc = [CO2(g)]

atau jika dituliskan dalam bentuk tekanan parsial menjadi

Kp = PCO2 Kp = Kc(RT)

Latihan:

Calculate Kc for the following reactionCaCO3(s) CaO(s) + CO2(g) Kp = 2,1 x 10-4 (at 1000 K)

Arti Nilai Tetapan Kesetimbangan

Soal Latihan

1) Reaksi: CO(g) + H2O(g) ↔ CO2(g) + H2(g)

pada suhu 1100 K nilai Kc = 1,00. Sejumlah zat berikut dicampur pada suhu tersebut dan dibiarkan bereaksi: 1,00 mol CO, 1,00 mol H2O, 2,00 mol CO2 dan 2,00 mol H2. Kearah mana reaksi akan berjalan dan bagaimana komposisi akhirnya?

2) Klorometana terbentuk melalui reaksi:CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)

pada 1500 K, konstanta kesetimbangan Kp = 1,6 x 104. Didalam campuran reaksi terdapat P (CH4) = 0,13 atm, P(Cl2) = 0,035 atm, P(CH3Cl) = 0,24 atm dan P(HCl) = 0,47 atm. Apakah reaksi diatas menuju kearah pembentukan CH3Cl atau pembentukan CH4.

Study Check

• In a study of hydrogen halide decomposition; a researcher fills an evacuated 2,00 L flask with 0,2 mol HI gas and allows the reaction to proceed at 453oC.2HI(g) H2(g) + I2(g)At equilibrium [HI] = 0.078 M, calculate Kc

• In study of the conversion of methane to other fuels a chemical engineer mixes gaseous CH4 and H2O in a 0,32 L flask at 1200 K. at equilibrium, the flask contains 0,26 mol CO; 0,091 mol H2 and 0,041 mol CH4. What is [H2O] at equilibrium? Kc = 0,26 for the equationCH4(g) + H2O CO(g) + 3H2(g)

• The decomposition of HI at low temperature was studied by injecting 2,50 mol HI into a 10,32-L vessel at 25oC. What is [H2] at equilibrium for the reaction 2HI(g) H2(g) + I2(g); Kc = 1.26 x 10-3?

Prinsip Le Chatelier

• Usaha untuk mengubah suhu, tekanan atau konsentrasi pereaksi dalam suatu sistem dalam keadaan setimbang merangsang terjadinya reaksi yang mengembalikan kesetimbangan pada sistem tersebut

Pengaruh perubahan Jumlah spesies yang bereaksi

Kesetimbangan awal Gangguan Kesetimbangan akhir

KcOSO

SOQ

KcOSO

SOQ

][][

][

][][

][

22

2

23

22

2

23

Pengaruh Perubahan Tekanan

Jika tekanan pada campuran kesetimbangan yang melibatkan gas ditingkatkan reaksi bersih akan berlangsung kearah yang mempunyai jumlah mol gas lebih kecil begitupun sebaliknya

Pengaruh Gas Lembam (inert)

• Pengaruh tidaknya gas lembam tergantung pada cara melibatkan gas tersebut

• Jika sejumlah gas helium ditambahkan pada keadaan volume tetap, tekanan akan meningkat, sehingga tekanan gas total akan meningkat. Tetapi tekanan parsial gas-gas dalam kesetimbangan tetap

• Jika gas ditambahkan pada tekanan tetap, maka volume akan bertambah. Pengaruhnya akan sama dengan peningkatan volume akibat penambahan tekanan eksternal.

• Gas lembam mempengaruhi keadaan kesetimbangan hanya jika gas tersebut mengakibatkan perubahan konsentrasi (atau tekanan parsial) dari pereaksi-pereaksinya

Pengaruh Suhu

• Penambahan kalor akan menguntungkan reaksi serap-panas (endoterm)

• Pengurangan kalor akan menguntungkan reaksi lepas-panas (eksoterm)

• Peningkatan suhu suatu campuran kesetimbangan menyebabkan pergeseran kearah reaksi endoterm. Penurunan suhu menyebabkan pergeseran kearah reaksi eksoterm

Pengaruh Suhu pada Kesetimbangan

• Umumnya tetapan kesetimbangan suatu reaksi tergantung pada suhu

• Nilai Kp untuk reaksi oksidasi belerang dioksida diperlihatkan pada tabel berikut

Hubungan pada tabel tersebut dapat dituliskan dengan:

tetapan1

303,2log

TR

HK

o

12

12

1

2

303,2log

TT

TT

R

H

K

K o

Persamaan garis lurus y = m .x + b

Dan jika ada dua keadaan yang berbeda kita dapat menghubungkan dengan modifikasi sederhana hingga diperoleh:

• K2 dan K1 adalah tetapan kesetimbangan pada suhu kelvin T2 dan T1. ∆Ho adalah entalpi (kalor) molar standar dari reaksi. Nilai positif dan negatif untuk parameter ini dimungkinkan dan diperlukan asumsi bahwa ∆Ho tidak tergantung pada suhu

• Menurut prinsip Le Chatelier, jika ∆Ho > 0 (endoterm) reaksi kedepan terjadi jika suhu ditingkatkan, menyiratkan bahwa nilai K meningkat dengan suhu. Jika ∆Ho < 0 (eksoterm) reaksi kebalikan terjadi jika suhu ditingkatkan dan nilai K menurun dengan suhu

• Persamaan diatas menghasilkan nilai kuantitatif yang sesuai dengan pengamatan kualitatif dari prinsip Le Chatelier.

Soal Latihan

• Untuk reaksi N2O4(g) 2NO2(g), ∆Ho = +61,5 kJ/mol dan Kp = 0,113 pada 298K

Berapa nilai Kp pada 0oC? (1,2x10-2)

Pada suhu berapa nilai Kp = 1,00 (326 K)

Pengaruh Katalis pada Kesetimbangan

• Katalis dalam reaksi reversibel dapat mempercepat reaksi baik kekanan atau kekiri. Keadaan kesetimbangan tercapai lebih cepat tetapi tidak mengubah konstanta kesetimbangan dari spesies-spesies yang bereaksi.

• Peranan katalis adalah mengubah mekanisme reaksi agar tercapai energi aktivasi yang lebih rendah.

• Keadaan kesetimbangan tidak bergantung pada mekanisme reaksi

• Sehingga tetapan kesetimbangan yang diturunkan secara kinetik tidak dipengaruhi oleh mekanisme yang dipilih.

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