8–1 John A. Schreifels Chemistry 212 Chapter 15-1 Chapter 15 Chemical Equilibrium

John A. SchreifelsChemistry 212

Chapter 15-1

8–1

Chapter 15

Chemical Equilibrium


Chapter 15-2

8–2

Overview

• Describing Chemical Equilibrium– Chemical Equilibrium – A Dynamic Equilibrium (the link to Chemical

Kinetics)

– The Equilibrium Constant.

– Heterogeneous Equilibria; solvents in homogeneous equilibria.

• Using the Equilibrium Constant– Qualitatively Interpreting the Equilibrium Constant

– Predicting the direction of a Reaction

– Calculating Equilibrium Concentrations

• Changing Reaction Conditions; Le Châtelier’s Principle.– Removing Products or Adding Reactants

– Changing the Pressure or Temperature

– Effect of a Catalyst.


Chapter 15-3

8–3

Chemical Equilibrium – A Dynamic Equilibrium

• Upon addition of reactants and/or products, they react until a constant amount of reactants and products are present = equilibrium.

• Equilibrium is dynamic since product is constantly made (forward reaction), but at the same rate it is consumed (reverse reaction).


Chapter 15-4

8–4

The Equilibrium State

• Not all reactants are completely converted to product.• Reaction equilibria deal with the extent of reaction.• Arrows between reactants and products separate them and

qualitatively indicate the extent of reaction.– Single arrow points to dominant side: H2(g) + O2(g) H2O(g)

– Double arrow indicates both reactants and products present after equilibrium obtained: N2O4 (g) 2NO2(g).

• Equilibrium exists when rates of forward and reverse reaction are the same.

E.g. When rate of N2O4 decomposition equal the rate of formation of N2O4, reaction at equilibrium

N2O4 (g) 2NO2(g).

• Equilibrium can be obtained from any mixture of reactants and products.


Chapter 15-5

8–5

The Link Between Chemical Equilibrium and Chemical Kinetics

Reactions with one elementary step:

• At equilibrium: Rf = Rr.

E.g. decomposition of N2O4:

• kf[N2O4] = kr[NO2]2. or

where Kc is called the equilibrium constant.

• At equilibrium the ratio of concentrations equals a constant. • A generalized form of this expression that describes the

equilibrium condition:

aA + bB + cC + ... mM + nN + oO ....

]ON[

]NO[

k

kK

42

22

r

fc

cba

onm

c[C][B][A]

[O][N][M] K


Chapter 15-6

8–6

The Link Between Chemical Equilibrium and Chemical Kinetics II

Reactions with more than one elementary step • Use the rate limiting step and • “Steady State Approximation”: at equilibrium the rates of the

forward and reverse reactions are equal.• Eliminate intermediates.• E.g.: Determine the equilibrium expression for the two step

decomposition of ozone.

1. (g) O (g)O (g)O 2

k

k3

1,f

1,r

Fast

2. (g)2O O(g) (g)O 2

k

k3

2,f

2,r

Slow

Sum 2O3 (g) 3O2.

• Step 2 (rate limiting step)


Chapter 15-7

8–7

Equilibrium Constant: Multi-step

• Rate equations lead to:

• Remove intermediate, O:

• Substitute into the first equation to get:

• Conclusion: Form of equilibrium expression is independent of mechanism.

]O][O[

]O[

k

k

]O[k]O][O[k

3

22

2,r

2,f

222,r32,f

]O[

]O[

k

k]O[

]O][O[k]O[k

2

3

1,r

1,f

21,r31,f

23

32

2,r

2,f

1,r

1,fc

]O[

]O[

k

k

k

kK


Chapter 15-8

8–8

The Equilibrium Constant Kc

• Equilibrium constant of reverse reaction:

aA + bB cC + dD

cC + dD aA + bB

E.g.1 Determine the equilibrium constant for the reaction: ½N2(g) + 3/2 H2(g) NH3(g)

given N2(g) + 3H2(g) 2NH3(g) Kc = 1.7x102

E.g.2 determine the equilibrium constant for the formation of HI(g) if the equilibrium concentration of H2, I2 and HI are 0.0060 M, 0.106 M, and 0.189 M, respectively.

H2(g) + I2(g) 2HI(g) Kc = ?E.g. 3 Determine the equilibrium constant for the reaction below:

½ H2(g) + ½ I2(g) HI(g) Kc = ?

f,cr,c K

1K

ba

cd

f,c]B[]A[

]C[]D[K

cd

ba

r,c]C[]D[

]B[]A[K


Chapter 15-9

8–9

Equilibrium Constant Kc, Kp

• For gases Kc and Kp are used. • KP same format as Kc except pressures used instead of

concentrations.

E.g. Write out the equilibrium expression for KP using the reaction below:

N2(g) + 3H2(g) 2NH3(g) KP = ?

E.g.2 What is equilibrium expression (KP) for the reaction below?

½ N2(g) + 3/2 H2(g) NH3(g) KP = ?

E.g.3 Determine the equilibrium constant (KP) for the formation of one mole of ammonia if at 500K, PNH3 = 0.15 atm, PN2 = 1.2 atm. and PH2 = 0.81 atm.


Chapter 15-10

8–10

Relationship between Kc and Kp

• Relationship between concentration and pressure obtained from the ideal gas law.

– Recall PV = nRT or

– Substitute for P in equilibrium expression. Consider the reaction:aA + bB cC + dD

– Use this relationship to relate KP and Kc

E.g. Determine Kp for the synthesis of ammonia at 25°C;N2(g) + 3H2(g) 2NH3(g) Kc = 4.1x108

E.g. 2 Determine Kp for the synthesis of HI at 458°C;½ H2(g) + ½ I2(g) HI(g) Kc = 0.142.

RT]A[

RTV

nP A

A

nc

)ba()dc(cP

baba

dcdc

ba

dc

bB

aA

dD

cC

P

RTKRTKK

RT]B[]A[

RT]D[]C[

RT]B[RT]A[

RT]D[RT]C[

PP

PPK


Chapter 15-11

8–11

Heterogeneous Equilbria

• Do not include a solvent or solid in the equilibrium expression. Their composition is constant and included in the equilibrium constant.– Water concentration is 55.5 M; very high compared with

reactants and products.– The concentration of a solid such as CaCO3 stays the same

as long as some solid is present.

E.g.: Determine the equilibrium expression for the reaction:

CaCO3(s) + C(gr) CaO(s) + 2CO(g). E.g.2 Determine the equilibrium expression for the reaction of acetic acid with water.

CH3COOH(aq) + H2O(l) CH3COO(aq) + H3O+(aq)


Chapter 15-12

8–12

Reaction Sequence

• As with H in thermo. Kc can be determined from a reaction sequence.

• Consider the reactions to the right. The third reaction is the result of the sum of the other two (called a reaction sequence..

aA + bB cC + dD

ba

cd

1]B[]A[

]C[]D[K

cC + dD eE + fF

dc

fe

2]D[]C[

]F[]E[K

aA + bB eE + fF

21

dc

fe

ba

dc

ba

fe

3

KK

]D[]C[

]F[]E[

]B[]A[

]D[]C[

]B[]A[

]F[]E[K

E.g.: Determine the Kp for the reaction: CaCO3(s) + C(gr) CaO(s) + 2CO(g).

Given:CaCO3(s) CaO(g) + CO2(g) , K1 = 0.039 atm. = PCO2, and C(g) + CO2(g) 2CO(g), K2 = 1.9 atm.


Chapter 15-13

8–13

Applications of the Equilibrium Constant

• Extent of reaction: The magnitude of the equilibrium constant allows us to predict the extent of the reaction. – Very large K (e.g. 1010) mostly products. – Very small K (e.g. 1010) mostly reactants. – When K is around 1, a significant amount of reactant and

product present in the equilibrium mixture.

E.g. For each of the following decide which species will predominate at equilibrium.

AgCl(s) Ag+(aq) + Cl (aq) Ksp = 1.8x10 10

Ag+(aq) + 2NH3(aq) 23 )Ag(NH 7

f 10x7.1K

H2CrO4(aq) + H2O(l) 4HCrO (aq) +

H3O+(aq)

Ka1 = 0.15


Chapter 15-14

8–14

Using the Equilibrium Constant

• Direction of Reaction: the reaction quotient can be used to determine the direction of a reaction with certain initial concentrations.

• Reaction Quotient: For the general reaction,aA + bB cC + dD,.

where the t refers to the time that concentrations of the mixture are measured; not necessarily at equilibrium.

• Comparison of Qc with Kc reveals direction of reaction. • When only reactants Qc = 0; leads to

– If Qc < Kc, products will form. • When only products present, Qc .

– If Qc > Kc, reactants will form. • When Qc = Kc, no net reaction.

E.g. Determine direction of reaction: H2(g) + I2(g) 2HI(g). Assume that [H2]o = [I2]o = [HI]o = 0.0020M at 490oC for which Kc = 46.

ba

ct

dt

c[B][A]

[C][D] Q

tt


Chapter 15-15

8–15

Calculating Equilibrium Concentrations

• Using initial concentrations, stoichiometry and Kc, equilibrium concentrations of all components can be determined. E.g. Determine equilibrium amounts of hydrogen and iodine if you started with exactly 1.00 mol H2(g) and 2.00 mol I2 in a 1.00 L flask; after equilibrium had been attained, [HI] = 1.86 M.

H2(g) + I2(g) 2HI(g)

E.g.2 [H2]o = [I2]o = 24 mM, were mixed and heated to 490oC in a container. Calculate equil. composition. Given that Kc = 46. Solution: set up an equilibrium table and solve for unknown after substitution into equilibrium expression.

H2(g) +I2(g)2HI(g) H2(g) + I2 2HI(g)

Init. Conc. 0.024 M 0.024 M 0 to Equil. x x +2x

Equil. 0.024 M x 0.024 M x +2x


Chapter 15-16

8–16

Using the Equilibrium Constant (cont)

• E.g.2 For the reaction: PCl5(g) PCl3(g) + Cl2(g), Kc = 0.800 M at 340oC. Find the equilibrium amounts of these compounds if they all start out at 0.120 M.

• Solution: Determine Qc so that you know which direction the reaction

will proceed. Qc = 0.120 < Kc reaction proceeds to products. PCl5(g) Cl2 + PCl3(g)

Init. Conc. 0.120M 0.120M 0.120M

to Equil. x +x +x Equil. 0.120 M x 0.120 M + x 0.120 M + x

• Equilibrium concentrations of each component are in last row. Substitute into equilibrium expression and solve for x.

• It may be necessary to rearrange so that quadratic equation can be used.

.Rearrange to ax2 + bx + c = 0; determine a, b, c and substitute into quadratic equation:

x M 0.120

x M 0.120K

2

c

a2

ac4bbx

2


Chapter 15-17

8–17

Factors that Alter the Composition of an Equilibrium Mixture

• A change to the system, which is initially at equilibrium, can cause a change in the equilibrium composition.

• Le Châtelier’s Principle: “If a stress is applied to a reaction mixture at equilibrium, reaction occurs in the direction that relieves the stress.”

• Types of stress on equilibrium:– Concentration of reactants or products. You can add or

remove one or more components in a reaction mixture. – With gases changing the pressure or volume is a way of

changing the concentrations of all components in the mixture.

– Change temperature.


Chapter 15-18

8–18

Changes in Concentrations

• Addition or removal of either reactant or product shifts the equilibrium to reduce the excess compound.

• Removal of product or addition of reactant has the same effect; they shift the equilibrium to the right.E.g. Consider the formation of NH3(g); the equilibrium conditions are listed in the first row; then half of the NH3(g) is removed. We calculate Qc and compare it with Kc. Qc < Kc reaction goes to right to form more NH3(g).

E.g.2 Instead lets add N2 so that its concentration increases 10x. This produces a similar effect. Since hydrogen is a reactant (in denominator), Qc will become much smaller than Kc. This means that product must form.

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

Equil. Conc. 0.50 M 3.00 M 1.98 M Kc = 0.291 After NH3(g) 0.50 M 3.00 M 0.99 M Qc = 0.073

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

Equil. Conc. 0.50 M 3.00 M 1.98 M Kc = 0.291 After N2(g) 5.0 M 3.00 M 1.98 M Qc = 0.0291


Chapter 15-19

8–19

Changes in Concentrations 2

• Removal of reactant or addition of product shifts the equilibrium in the opposite direction – to the left.

• E.g. Half of N2(g) is removed. We calculate Qc and compare it with Kc. Notice that it is greater than the equilibrium constant. Therefore the reaction proceeds to the left to form more N2(g).

• E.g.2 Add NH3 so that its concentration increases by a factor of 2. This produces a similar effect. Since ammonia is a product (in numerator), Qc > Kc. Reaction proceeds to left since there is too much product. N2(g) + 3 H2 2 NH3(g)

Equil. Conc. 0.50 M 3.00 M 1.98 M Kc = 0.291

After NH3 0.50 M 3.00 M 3.96 M Qc = 1.162

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

Equil. Conc. 0.50 M 3.00 M 1.98 M Kc = 0.291 After N2 0.25 M 3.00 M 1.98 M Qc = 0.582


Chapter 15-20

8–20

Changes in Pressure and Volume

• Equilibria containing gases are sensitive to changes in pressure (volume).– Since pressure and volume are inversely proportional, an increase

in pressure will give same result as a decrease in volume.

– A decrease in the volume increases concentration of each of the components in the reaction mixture.

– A decrease in V shifts equilibrium to side with least number of moles of gas.

• E.g. Determine the effect of a 2.00x increase in pressure (volume decreases by factor of 2;Vf = ½ VI). Concentrations of reactants and products increase by a factor of 2. Since Qc < Kc, the reaction proceeds to the right to form more NH3(g).

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

Equil. Conc. 0.50 M 3.00 M 1.98 M Kc = 0.291 After P 1.00 M 6.00 M 3.96 M Qc = 0.073


Chapter 15-21

8–21

Changes in Pressure and Volume 2

E.g.2 Determine the effect of a factor of 2.00 decrease in the pressure (volume increases by a factor of 2;Vf = 2 Vi).

• We need to be vigilant so that we do not include any solids or solvents in this, since they do not affect the equilibrium.

• E.g. Determine the ratio of Qc/Kc when the pressure increases by a factor of two and from this discuss the effect on the equilibrium: C(s) + CO2(g) 2CO(g).

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

Equil. Conc. 0.50 M 3.00 M 1.98 M Kc = 0.291

After P 0.25 M 1.50 M 0.99 M Qc = 1.16


Chapter 15-22

8–22

Changes in Temperature

T Qc. T produces Kc • Whether Kc increases or decreases depends upon two factors,

T and Hrxn. – An increase in temperature causes the equilibrium to shift towards

the endothermic side; – A decrease in temperature causes the equilibrium to shift in the

opposite direction.

E.g. Will an increase or a decrease in temperature increase the amount of CO produced from the following reaction

2CO2(g)2 CO(g) +O2(g) H = 566 kJ

The Effect of a Catalyst• It has no effect on the equilibrium. • It speeds up attainment of equilibrium. • Kc related to the H and not Ea

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8–1 John A. Schreifels Chemistry 212 Chapter 15-1 Chapter 15 Chemical Equilibrium