1 / 1

Overview

2 / 1

The Equilibrium State

All reactions are reversible and under suitable conditions will reach astate of equilibrium.

At equilibrium, the concentrations of products and reactants nolonger change because the rates of the forward and reverse reactionsare equal.

At equilibrium: rateforward = ratereverse

Chemical equilibrium is a dynamic state because reactions continueto occur, but because they occur at the same rate, no net change isobserved on the macroscopic level.

3 / 1

The Equilibrium State

All reactions are reversible and under suitable conditions will reach astate of equilibrium.

At equilibrium, the concentrations of products and reactants nolonger change because the rates of the forward and reverse reactionsare equal.

At equilibrium: rateforward = ratereverse

Chemical equilibrium is a dynamic state because reactions continueto occur, but because they occur at the same rate, no net change isobserved on the macroscopic level.

4 / 1

The Equilibrium State

All reactions are reversible and under suitable conditions will reach astate of equilibrium.

At equilibrium, the concentrations of products and reactants nolonger change because the rates of the forward and reverse reactionsare equal.

At equilibrium: rateforward = ratereverse

Chemical equilibrium is a dynamic state because reactions continueto occur, but because they occur at the same rate, no net change isobserved on the macroscopic level.

5 / 1

The Equilibrium State

All reactions are reversible and under suitable conditions will reach astate of equilibrium.

At equilibrium, the concentrations of products and reactants nolonger change because the rates of the forward and reverse reactionsare equal.

At equilibrium: rateforward = ratereverse

Chemical equilibrium is a dynamic state because reactions continueto occur, but because they occur at the same rate, no net change isobserved on the macroscopic level.

6 / 1

Reaching equilibrium on the macroscopic andmolecular levels.

N2O4(g) −−⇀↽−− 2NO2(g)

7 / 1

The Equilibrium ConstantConsider the reaction: N2O4(g) −−⇀↽−− 2 NO2(g)

At equilibrium

rateforward = ratereverse

so kfwd [N2O4]eq = krev [NO2]2eq

thenkfwd

krev=

[NO2]2eq

[N2O4]eq

The ratio of constants gives a new constant, the equilibrium constant:

K =kfwd

krev=

[NO2]2eq

[N2O4]eq

8 / 1

K and the extent of reaction

K reflects a particular ratio of product concentrations to reactantconcentrations for a reaction.

K therefore indicates the extent of a reaction, i.e., how far a reactionproceeds towards the products at a given temperature.

A small value for K indicates that the reaction yields little productbefore reaching equilibrium. The reaction favors the reactants.

A large value for K indicates that the reaction reaches equilibriumwith very little reactant remaining. The reaction favors the products.

9 / 1

K and the extent of reaction

K reflects a particular ratio of product concentrations to reactantconcentrations for a reaction.

K therefore indicates the extent of a reaction, i.e., how far a reactionproceeds towards the products at a given temperature.

A small value for K indicates that the reaction yields little productbefore reaching equilibrium. The reaction favors the reactants.

A large value for K indicates that the reaction reaches equilibriumwith very little reactant remaining. The reaction favors the products.

10 / 1

K and the extent of reaction

K reflects a particular ratio of product concentrations to reactantconcentrations for a reaction.

K therefore indicates the extent of a reaction, i.e., how far a reactionproceeds towards the products at a given temperature.

A small value for K indicates that the reaction yields little productbefore reaching equilibrium. The reaction favors the reactants.

A large value for K indicates that the reaction reaches equilibriumwith very little reactant remaining. The reaction favors the products.

11 / 1

In generalFor the generic reaction aA + bB −−⇀↽−− pP + qQ, we can write anexpression for the reaction quotient Q as

Q =(cP)

p (cQ)q

(cA)a (cB)

b

• Q gives the ratio of product concentrations to reactantconcentrations at any point in a reaction.

• At equilibrium: Q = K• For a particular system and temperature, the same equilibrium

state is attained regardless of starting concentrations. The valueof Q indicates how close the reaction is to equilibrium, and inwhich direction it must proceed to reach equilibrium.

12 / 1

In generalFor the generic reaction aA + bB −−⇀↽−− pP + qQ, we can write anexpression for the reaction quotient Q as

Q =(cP)

p (cQ)q

(cA)a (cB)

b

• Q gives the ratio of product concentrations to reactantconcentrations at any point in a reaction.

• At equilibrium: Q = K• For a particular system and temperature, the same equilibrium

state is attained regardless of starting concentrations. The valueof Q indicates how close the reaction is to equilibrium, and inwhich direction it must proceed to reach equilibrium.

13 / 1

In generalFor the generic reaction aA + bB −−⇀↽−− pP + qQ, we can write anexpression for the reaction quotient Q as

Q =(cP)

p (cQ)q

(cA)a (cB)

b

• Q gives the ratio of product concentrations to reactantconcentrations at any point in a reaction.

• At equilibrium: Q = K• For a particular system and temperature, the same equilibrium

state is attained regardless of starting concentrations. The valueof Q indicates how close the reaction is to equilibrium, and inwhich direction it must proceed to reach equilibrium.

14 / 1

In generalFor the generic reaction aA + bB −−⇀↽−− pP + qQ, we can write anexpression for the reaction quotient Q as

Q =(cP)

p (cQ)q

(cA)a (cB)

b

• Q gives the ratio of product concentrations to reactantconcentrations at any point in a reaction.

• At equilibrium: Q = K• For a particular system and temperature, the same equilibrium

state is attained regardless of starting concentrations. The valueof Q indicates how close the reaction is to equilibrium, and inwhich direction it must proceed to reach equilibrium.

15 / 1

In generalFor the generic reaction aA + bB −−⇀↽−− pP + qQ, we can write anexpression for the reaction quotient Q as

Q =(cP)

p (cQ)q

(cA)a (cB)

b

• Q gives the ratio of product concentrations to reactantconcentrations at any point in a reaction.

• At equilibrium: Q = K• For a particular system and temperature, the same equilibrium

state is attained regardless of starting concentrations. The valueof Q indicates how close the reaction is to equilibrium, and inwhich direction it must proceed to reach equilibrium.

16 / 1

Variation of Q with the reaction

17 / 1

Heterogeneous Reactions

18 / 1

Alternative forms of the equilibrium constant

1. If a chemical equation is multiplied by a constant, its equilibriumconstant should be raised to that number as exponent.

2. The equilibrium constant of the reverse reaction is the inverse ofthe equilibrium constant of the forward reaction

3. If a chemical equation can be expressed as the sum of two ormore equations, the equilibrium constant of the overall reaction isthe product of the equilibrium constant of the individual “steps”.

19 / 1

Exercise

The following equilibrium constants are given at 500. K

H2(g) + Br2(g) � 2HBr(g) K = 7.9× 1011

H2(g) � 2H(g) K = 4.8× 10−41

Br2(g) � 2Br(g) K = 2.2× 10−15

Calculate K for the reaction of H atoms and Br atoms to give HBr

H(g) + Br(g) � HBr(g)

Ans: 2.7×1033

20 / 1

Exercise

The following equilibrium constants are given at 500. K

H2(g) + Br2(g) � 2HBr(g) K = 7.9× 1011

H2(g) � 2H(g) K = 4.8× 10−41

Br2(g) � 2Br(g) K = 2.2× 10−15

Calculate K for the reaction of H atoms and Br atoms to give HBr

H(g) + Br(g) � HBr(g)

Ans: 2.7×1033

21 / 1

Expressions for Calculating Q and K

22 / 1

The Equilibrium Constant as a Function of thePartial Pressures

For reactions involving only gases, we can write an expression for theequilibrium constant as a function of the partial pressures instead ofthe concentrations: For

2NO(g) + O2(g) � 2NO2(g)

we can write

Kp =pNO2

2

(p2NO)(pO2)

23 / 1

Relationship between Kc and Kp

In the general expression for the constant Kp, we can replace thepartial pressures an equivalent expression based on the ideal gaslaw:

Kp =Pc

C PdD

PaA Pb

B

since PJ =nJ R T

Vwe get

Kp =

(nC R T

V

)c (nD R T

V

)d

(nA R T

V

)a (nB R T

V

)b

24 / 1

Relationship between Kc and Kp

In the general expression for the constant Kp, we can replace thepartial pressures an equivalent expression based on the ideal gaslaw:

Kp =Pc

C PdD

PaA Pb

B

since PJ =nJ R T

Vwe get

Kp =

(nC R T

V

)c (nD R T

V

)d

(nA R T

V

)a (nB R T

V

)b

25 / 1

Relationship between Kc and KReplacing cJ =

nJ

Vthis becomes

Kp =(cC R T )c (cD R T )d

(cA R T )a (cB R T )b

Grouping the R T terms and applying the properties of exponents

Kp = (R T )c+d−(a+b) (cC)c (cD)

d

(cA)a (cB)

b

where the term in the right contains the expression for Kc

And we conclude that

Kp = (R T )∆n Kc26 / 1

Relationship between Kc and KReplacing cJ =

nJ

Vthis becomes

Kp =(cC R T )c (cD R T )d

(cA R T )a (cB R T )b

Grouping the R T terms and applying the properties of exponents

Kp = (R T )c+d−(a+b) (cC)c (cD)

d

(cA)a (cB)

b

where the term in the right contains the expression for Kc

And we conclude that

Kp = (R T )∆n Kc27 / 1

Relationship between Kc and KReplacing cJ =

nJ

Vthis becomes

Kp =(cC R T )c (cD R T )d

(cA R T )a (cB R T )b

Grouping the R T terms and applying the properties of exponents

Kp = (R T )c+d−(a+b) (cC)c (cD)

d

(cA)a (cB)

b

where the term in the right contains the expression for Kc

And we conclude that

Kp = (R T )∆n Kc28 / 1

Exercise 1

At 127 ◦C, the equilibrium constant for N2O4(g) � 2NO2(g) is Kp =47.9. What is the value of Kc at that temperature? Ans:

Kc = (R T )−∆n Kp

= (0.08314bar · L ·mol−1 · K−1 × 400.K )−1 × 47.9

= 1.44

29 / 1

Exercise 1

At 127 ◦C, the equilibrium constant for N2O4(g) � 2NO2(g) is Kp =47.9. What is the value of Kc at that temperature? Ans:

Kc = (R T )−∆n Kp

= (0.08314bar · L ·mol−1 · K−1 × 400.K )−1 × 47.9

= 1.44

30 / 1

Determining the Direction of a Reaction

The value of Q indicates the direction in which a reaction mustproceed to reach equilibrium.

If Q < K , the reactants must increase and the products decrease;reactants −−→ products until equilibrium is reached.

If Q > K , the reactants must decrease and the products increase;products −−→ reactants until equilibrium is reached.

If Q = K , the system is at equilibrium and no further net changetakes place.

31 / 1

Determining the Direction of a Reaction

The value of Q indicates the direction in which a reaction mustproceed to reach equilibrium.

If Q < K , the reactants must increase and the products decrease;reactants −−→ products until equilibrium is reached.

If Q > K , the reactants must decrease and the products increase;products −−→ reactants until equilibrium is reached.

If Q = K , the system is at equilibrium and no further net changetakes place.

32 / 1

Determining the Direction of a Reaction

The value of Q indicates the direction in which a reaction mustproceed to reach equilibrium.

If Q < K , the reactants must increase and the products decrease;reactants −−→ products until equilibrium is reached.

If Q > K , the reactants must decrease and the products increase;products −−→ reactants until equilibrium is reached.

If Q = K , the system is at equilibrium and no further net changetakes place.

33 / 1

Determining the Direction of a Reaction

The value of Q indicates the direction in which a reaction mustproceed to reach equilibrium.

If Q < K , the reactants must increase and the products decrease;reactants −−→ products until equilibrium is reached.

If Q > K , the reactants must decrease and the products increase;products −−→ reactants until equilibrium is reached.

If Q = K , the system is at equilibrium and no further net changetakes place.

34 / 1

Determining the Direction of a Reaction

The value of Q indicates the direction in which a reaction mustproceed to reach equilibrium.

If Q < K , the reactants must increase and the products decrease;reactants −−→ products until equilibrium is reached.

If Q > K , the reactants must decrease and the products increase;products −−→ reactants until equilibrium is reached.

If Q = K , the system is at equilibrium and no further net changetakes place.

35 / 1

Using concentrations to determine reaction direction

For the reaction N2O4(g) −−⇀↽−− 2 NO2(g) , Kc = 0.21 at 100 ◦C

At a point during the reaction, cN2O4= 0.12 mol·L−1 and cNO2

= 0.55mol·L−1. Is the reaction at equilibrium? If not, in which direction is itprogressing?

We write an expression for Qc, find its value by substituting the givingconcentrations, and compare the value with the given Kc.

Qc > Kc , therefore the reaction is not at equilibrium and will proceedfrom products to reactants, until Qc = Kc .

36 / 1

Using concentrations to determine reaction direction

For the reaction N2O4(g) −−⇀↽−− 2 NO2(g) , Kc = 0.21 at 100 ◦C

At a point during the reaction, cN2O4= 0.12 mol·L−1 and cNO2

= 0.55mol·L−1. Is the reaction at equilibrium? If not, in which direction is itprogressing?

We write an expression for Qc, find its value by substituting the givingconcentrations, and compare the value with the given Kc.

Qc > Kc , therefore the reaction is not at equilibrium and will proceedfrom products to reactants, until Qc = Kc .

37 / 1

Using concentrations to determine reaction direction

For the reaction N2O4(g) −−⇀↽−− 2 NO2(g) , Kc = 0.21 at 100 ◦C

At a point during the reaction, cN2O4= 0.12 mol·L−1 and cNO2

= 0.55mol·L−1. Is the reaction at equilibrium? If not, in which direction is itprogressing?

We write an expression for Qc, find its value by substituting the givingconcentrations, and compare the value with the given Kc.

Qc > Kc , therefore the reaction is not at equilibrium and will proceedfrom products to reactants, until Qc = Kc .

38 / 1

Calculating the constant from concentration dataIn order to study hydrogen halide decomposition, a researcher fills anevacuated 2.00-L flask with 0.200 mol of HI gas and allows thereaction to proceed at 453 ◦C. 2 HI(g) −−⇀↽−− H2(g) + I2(g)

At equilibrium cHI = 0.078 mol·L−1. Calculate Kc

Plan: First find the concentration of the starting material and then findthe amount of each component, reactants and products, atequilibrium.

A B C DInitial

ChangeEquilibrium

39 / 1

Calculating the constant from concentration dataIn order to study hydrogen halide decomposition, a researcher fills anevacuated 2.00-L flask with 0.200 mol of HI gas and allows thereaction to proceed at 453 ◦C. 2 HI(g) −−⇀↽−− H2(g) + I2(g)

At equilibrium cHI = 0.078 mol·L−1. Calculate Kc

Plan: First find the concentration of the starting material and then findthe amount of each component, reactants and products, atequilibrium.

A B C DInitial

ChangeEquilibrium

40 / 1

Calculating the constant from concentration dataIn order to study hydrogen halide decomposition, a researcher fills anevacuated 2.00-L flask with 0.200 mol of HI gas and allows thereaction to proceed at 453 ◦C. 2 HI(g) −−⇀↽−− H2(g) + I2(g)

At equilibrium cHI = 0.078 mol·L−1. Calculate Kc

Plan: First find the concentration of the starting material and then findthe amount of each component, reactants and products, atequilibrium.

A B C DInitial

ChangeEquilibrium

41 / 1

Calculating the constant from concentration data

Concentrations in mol·L−1

HI H2 I2Initial 0.100 0 0

Change -2x +x +xEquilibrium 0.100 -2x x x

Kc =(0.011) (0.011)

(0.078)2 = 0.020

42 / 1

Calculating the constant from concentration data

Concentrations in mol·L−1

HI H2 I2Initial 0.100 0 0

Change -2x +x +xEquilibrium 0.100 -2x x x

Kc =(0.011) (0.011)

(0.078)2 = 0.020

43 / 1

Exercise

The initial partial pressures of nitrogen and hydrogen in a rigid sealedvessel are 0.010 and 0.020 bar, respectively. The mixture is heated toa temperature at which K = 0.11 for

N2(g) + 3H2(g) � 2NH3(g)

What are the equilibrium partial pressures of the substances in themixture.

44 / 1

Exercise 2The initial partial pressures of nitrogen and hydrogen in a rigid sealedvessel are 0.010 and 0.020 bar, respectively. The mixture is heated toa temperature at which K = 0.11 for

N2(g) + 3H2(g) � 2NH3(g)

What are the equilibrium partial pressures of the substances in themixture.

Answer:

N2 H2 NH3Initial 0.010 0.020 0

Change -x -3x +2xEquilibrium 0.010-x 0.020-3x 2x

K =(2x)2

(0.010− x)(0.020− 3x)3 = 0.11

Assuming x << 0.010, then x = 4.7×10−5 and PNH3= 9.4×10−5 barWithout the assumption, use TI-89/Mathematica/Excel to getx=4.6×10−5

45 / 1

Exercise 2The initial partial pressures of nitrogen and hydrogen in a rigid sealedvessel are 0.010 and 0.020 bar, respectively. The mixture is heated toa temperature at which K = 0.11 for

N2(g) + 3H2(g) � 2NH3(g)

What are the equilibrium partial pressures of the substances in themixture.

Answer:

N2 H2 NH3Initial 0.010 0.020 0

Change -x -3x +2xEquilibrium 0.010-x 0.020-3x 2x

K =(2x)2

(0.010− x)(0.020− 3x)3 = 0.11

Assuming x << 0.010, then x = 4.7×10−5 and PNH3= 9.4×10−5 barWithout the assumption, use TI-89/Mathematica/Excel to getx=4.6×10−5

46 / 1

Exercise 2The initial partial pressures of nitrogen and hydrogen in a rigid sealedvessel are 0.010 and 0.020 bar, respectively. The mixture is heated toa temperature at which K = 0.11 for

N2(g) + 3H2(g) � 2NH3(g)

What are the equilibrium partial pressures of the substances in themixture.

Answer:

N2 H2 NH3Initial 0.010 0.020 0

Change -x -3x +2xEquilibrium 0.010-x 0.020-3x 2x

K =(2x)2

(0.010− x)(0.020− 3x)3 = 0.11

Assuming x << 0.010, then x = 4.7×10−5 and PNH3= 9.4×10−5 barWithout the assumption, use TI-89/Mathematica/Excel to getx=4.6×10−5

47 / 1

Exercise 2The initial partial pressures of nitrogen and hydrogen in a rigid sealedvessel are 0.010 and 0.020 bar, respectively. The mixture is heated toa temperature at which K = 0.11 for

N2(g) + 3H2(g) � 2NH3(g)

What are the equilibrium partial pressures of the substances in themixture.

Answer:

N2 H2 NH3Initial 0.010 0.020 0

Change -x -3x +2xEquilibrium 0.010-x 0.020-3x 2x

K =(2x)2

(0.010− x)(0.020− 3x)3 = 0.11

Assuming x << 0.010, then x = 4.7×10−5 and PNH3= 9.4×10−5 barWithout the assumption, use TI-89/Mathematica/Excel to getx=4.6×10−5

48 / 1

Exercise 2The initial partial pressures of nitrogen and hydrogen in a rigid sealedvessel are 0.010 and 0.020 bar, respectively. The mixture is heated toa temperature at which K = 0.11 for

N2(g) + 3H2(g) � 2NH3(g)

What are the equilibrium partial pressures of the substances in themixture.

Answer:

N2 H2 NH3Initial 0.010 0.020 0

Change -x -3x +2xEquilibrium 0.010-x 0.020-3x 2x

K =(2x)2

(0.010− x)(0.020− 3x)3 = 0.11

Assuming x << 0.010, then x = 4.7×10−5 and PNH3= 9.4×10−5 barWithout the assumption, use TI-89/Mathematica/Excel to getx=4.6×10−5

49 / 1

Exercise 2The initial partial pressures of nitrogen and hydrogen in a rigid sealedvessel are 0.010 and 0.020 bar, respectively. The mixture is heated toa temperature at which K = 0.11 for

N2(g) + 3H2(g) � 2NH3(g)

What are the equilibrium partial pressures of the substances in themixture.

Answer:

N2 H2 NH3Initial 0.010 0.020 0

Change -x -3x +2xEquilibrium 0.010-x 0.020-3x 2x

K =(2x)2

(0.010− x)(0.020− 3x)3 = 0.11

Assuming x << 0.010, then x = 4.7×10−5 and PNH3= 9.4×10−5 barWithout the assumption, use TI-89/Mathematica/Excel to getx=4.6×10−5

50 / 1

Exercise 3Bromine monochloride decomposes into bromine and chlrine andreaches the equilibrium

2BrCl(g) � Br2(g) + Cl2(g)

for which K = 32 at 500. K. If initially BrCl is present at a pressure of3.30×10−3 bar, what is its partial pressure at equilibrium?

BrCl Br2 Cl2Initial 3.30×10−3 0 0

Change -2x +x +xEquilibrium 3.30×10−3-2x x x

K =x2

(3.30× 10−3 − 2x)2 = 32

x=.001516 and PBrCl ,eq = 2.7×10−4 bar51 / 1

Exercise 3Bromine monochloride decomposes into bromine and chlrine andreaches the equilibrium

2BrCl(g) � Br2(g) + Cl2(g)

for which K = 32 at 500. K. If initially BrCl is present at a pressure of3.30×10−3 bar, what is its partial pressure at equilibrium?

BrCl Br2 Cl2Initial 3.30×10−3 0 0

Change -2x +x +xEquilibrium 3.30×10−3-2x x x

K =x2

(3.30× 10−3 − 2x)2 = 32

x=.001516 and PBrCl ,eq = 2.7×10−4 bar52 / 1

Exercise 3Bromine monochloride decomposes into bromine and chlrine andreaches the equilibrium

2BrCl(g) � Br2(g) + Cl2(g)

for which K = 32 at 500. K. If initially BrCl is present at a pressure of3.30×10−3 bar, what is its partial pressure at equilibrium?

BrCl Br2 Cl2Initial 3.30×10−3 0 0

Change -2x +x +xEquilibrium 3.30×10−3-2x x x

K =x2

(3.30× 10−3 − 2x)2 = 32

x=.001516 and PBrCl ,eq = 2.7×10−4 bar53 / 1

Exercise 3Bromine monochloride decomposes into bromine and chlrine andreaches the equilibrium

2BrCl(g) � Br2(g) + Cl2(g)

for which K = 32 at 500. K. If initially BrCl is present at a pressure of3.30×10−3 bar, what is its partial pressure at equilibrium?

BrCl Br2 Cl2Initial 3.30×10−3 0 0

Change -2x +x +xEquilibrium 3.30×10−3-2x x x

K =x2

(3.30× 10−3 − 2x)2 = 32

x=.001516 and PBrCl ,eq = 2.7×10−4 bar54 / 1

The Response of Equilibria toChanges in Conditions

Dynamic equilibria respond to the change in conditions. If we add orremove reactants or products, the reaction will be displaced fromequilibrium and there will be then a direction of spontaneous changeuntil de system is restored to the equilibrium position.

Also changes in the pressure and the temperature will displace theposition of the equilibrium.

55 / 1

LeChatelier’s Principle

When a chemical system atequilibrium is subjected to astress, the system will return toequilibrium by shifting to reduceeffect of the stress.

(1850-1936)

56 / 1

LeChatelier’s Principle

Thus for the reaction:

4NH4(g) + 3O2(g) � 2N2(g) + 6H2O(g)

We can predict the effect of adding or subtracting each of thereactants and products.

57 / 1

LeChatelier’s Principle

Thus for the reaction:

4NH4(g) + 3O2(g) � 2N2(g) + 6H2O(g)

We can predict the effect of adding or subtracting each of thereactants and products.

58 / 1

LeChatelier’s PrincipleIt can be done also in quantitative terms: NH3,N2andH2 havereached equilibrium in a container of volume 1.00 L at 298 K. Theequilibrium partial pressures in the reaction

N2(g) + 3H2(g) � 2NH3(g)

are 0.080 bar N2, 0.050 bar H2 and 2.57 bar NH3, resulting in Kp =6.6 ×105 bar−2. Calculate the new equilibrium partial pressures ifhalf of the ammonia is removed from the container and theequilibrium is reestablished.

N2 H2 NH3Initial 0.08 0.05 1.28

Change -x -3x +2xEquilibrium 0.08-x 0.05-3x 1.28+2x

59 / 1

LeChatelier’s PrincipleIt can be done also in quantitative terms: NH3,N2andH2 havereached equilibrium in a container of volume 1.00 L at 298 K. Theequilibrium partial pressures in the reaction

N2(g) + 3H2(g) � 2NH3(g)

are 0.080 bar N2, 0.050 bar H2 and 2.57 bar NH3, resulting in Kp =6.6 ×105 bar−2. Calculate the new equilibrium partial pressures ifhalf of the ammonia is removed from the container and theequilibrium is reestablished.

N2 H2 NH3Initial 0.08 0.05 1.28

Change -x -3x +2xEquilibrium 0.08-x 0.05-3x 1.28+2x

60 / 1

LeChatelier’s PrincipleIt can be done also in quantitative terms: NH3,N2andH2 havereached equilibrium in a container of volume 1.00 L at 298 K. Theequilibrium partial pressures in the reaction

N2(g) + 3H2(g) � 2NH3(g)

are 0.080 bar N2, 0.050 bar H2 and 2.57 bar NH3, resulting in Kp =6.6 ×105 bar−2. Calculate the new equilibrium partial pressures ifhalf of the ammonia is removed from the container and theequilibrium is reestablished.

N2 H2 NH3Initial 0.08 0.05 1.28

Change -x -3x +2xEquilibrium 0.08-x 0.05-3x 1.28+2x

Ans: x = 0.0058696, new ammonia pressure 1.29 bar

61 / 1

Compression in an equilibrium system

62 / 1

Exercise: Predicting the Effect of a Change inVolume (Pressure) on the Equilibrium Position

How would you change the volume of each of the following reactionsto increase the yield of the products.

CaCO3(s) � CaO(s) + CO2(g) (1)S(s) + 3F2(g) � SF6(g) (2)Cl2(g) + I2(g) � 2ICl(g) (3)

63 / 1

Effect of the temperature

The effect of the temperature onan equilibrium depends on whichis the direction heat is absorbed(endothermic) or given away(exothermic).In an equilibrium, an increase inthe temperature will favor thedirection of the endothermicreaction.This changes happen through achange in the value of theequilibrium constant.

64 / 1

The Reaction QuotientIn order to find a quantitative indication of how far a system is fromequilibrium, we define the reaction quotient Q.

It has the same algebraic expression that is used for the equilibriumconstant, but the concentrations (or partial pressures) substituted arearbitrary and not necessarily those of the equilibrium condition.

For the generic reaction

aA + bB � cC + dD

the reaction quotient is

Q =cc

CcdD

caAcb

B

65 / 1

The Reaction Quotient

If Q = K, the reaction is in equilibrium.

If Q > K, the system is not in equilibrium. It will move spontaneouslytowards equilibrium by making more reactants, the reverse reaction isfavored.

If Q < K, the system is not in equilibrium. It will move spontaneouslytowards equilibrium by making more product, the forward reaction isfavored.

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