Equilibrium

Chapter 15

At room temperature colorless N2O4 decomposes to brown NO2.

N2O4(g) 2NO2(g)

(colorless) (brown)

• We expect the forward reaction rate to slow over time and the rate of the reverse reaction rate to increase over time.

• At some time, the color stops changing which indicates and we have an equilibrium mixture of N2O4 and NO2.

• Chemical equilibrium is the point at which the concentrations (partial pressures of gases) of all species are constant.

• A dynamic equilibrium exists when the rates of the forward and reverse reactions are equal. – The reaction does NOT stop, but there is no

NET change in the reactant or product concentrations.

• Use a double arrow () to imply that the process is an equilibrium.

Haber process:

• Industrial preparation of ammonia from nitrogen and hydrogen:

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

• Process is carried out at high temperature (500oC) and pressure (200 atm).

• Much of the NH3 produced industrially is used as a fertilizer.

N2(g) + 3H2(g) 2NH3(g)• If we start with a mixture of nitrogen and hydrogen (in

any proportions), the reaction will reach equilibrium with constant concentrations of nitrogen, hydrogen and ammonia.

• If we start with just ammonia, the reaction will reach equilibrium the same equilibrium with constant concentrations of nitrogen, hydrogen and ammonia.

• No matter what the starting composition of reactants and products is, the equilibrium mixture contains the same relative concentrations of reactants and products.– Equilibrium can be reached from either direction.

• Consider a simple reaction.A(g) B(g)

• Assume that it is an elementary process.We can write rate expressions for each reaction.

Forward reaction: A B Rate = kf[A] kf = rate constant (forward reaction)

Reverse reaction: B ARate = kr[B] kr = rate constant (reverse reaction)

In equilibrium the forward and reverse rates are equalRearranging: kf[A]= kr[B]

[ ]

[ ]f

cr

kBK

A k

This is called theEquilibrium constant!“c” because it’s based on concentrations

• For gaseous substances we can use the ideal gas equation to convert between concentration and pressure:

• PV = nRT so M = (n/V) = (P/RT)• For substances A and B:• [A] = (PA/RT) and [B] = (PB/RT)• Ratefwd = kf PA/RT and Raterev = krPB/RT• At equilibrium: kf PA/RT = krPB/RT• Rearranging, we get:

fBp

A

B

A r

kPK

P k

PRT

PRT

This is called theEquilibrium constant!“p” because it’s based on pressure

15.2 The Equilibrium Constant• We can write an expression for the relationship between the

concentration of the reactants and products at equilibrium. • For a general reaction: aA + bB cC + dD• The equilibrium expression (law of mass action)• is given by:

*equilibrium concentration/pressure of (g) and (aq) only!• Where Kc and Kp are equilibrium constants (Keq).

c dC D

p a bA B

P PK

P P[ ] [ ]

[ ] [ ]

c d

c a b

C DK

A B * P in atm

Converting between Kp and Kc

Kp=Kc(RT)Δn

Δn = (sum of coefficients of gaseous product) – (sum of coefficients of gaseous reactant)

R is the gas constant (0.0821)

T must be Kelvin

When Δn is zero Kp=Kc

• Keq depends on stoichiometry.

• Keq varies with temperature.

• Keq does NOT depend on initial concentrations (pressures)

• Keq does NOT depend on the reaction mechanism.

• We generally omit the units of Keq

The Magnitude of Equilibrium Constants

• The equilibrium constant, Keq, is the ratio of products to reactants.– the larger Keq the more products are present at

equilibrium.– the smaller Keq the more reactants are present at

equilibrium.• If Keq >> 1, then products dominate at

equilibrium and “equilibrium lies to the right”• If Keq << 1, then reactants dominate at

equilibrium and “equilibrium lies to the left”

The Direction of the Chemical Equation and K

• An equilibrium can be approached from any direction.• Consider the reaction: N2O4(g) 2NO2(g)

• The equilibrium constant (at 100oC) is:• However, when we write the equilibrium expression for

the reverse reaction: 2NO2(g) N2O4(g)

• The equilibrium constant (at 100oC) is:• The equilibrium constant for a reaction in one direction is

the reciprocal of the equilibrium constant of the reaction in the opposite direction.

A + B 2 C K 2 C A + B 1/K

2

2 4

2

6.49NOeq

N O

PK

P

2 4

2

N O

eq 2NO

PK 0 154

P .

• The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number. A + B 2 C K

2 A + 2 B 4 C K2 • The equilibrium constant for a net reaction made

up of two or more steps is the product of the equilibrium constants for the individual steps.

A + B 2 C K1

+ C + A D K2

2 A + B C + D K1K2

• Homogeneous equilibria - Equilibria in which all reactants and products are present in the same phase

• Heterogeneous equilibria - Equilibria in which one or more reactants or products are present in a different phase

Heterogeneous equilibria• Consider the decomposition of calcium carbonate:

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

• Experimentally, the amount of CO2 does not depend on the amounts of CaO and CaCO3. Why?

• The concentration of a pure solid or pure liquid equals its density divided by its molar mass.

• Neither density nor molar mass is a variable. • Since the concentrations of solids and liquids are

constant, they are not included in the equilibrium expression.

• Since the concentration of a pure solid or pure liquid is not included in the equilibrium constant expression. We anticipate that the amount of CO2 formed will not depend on the amounts of CaO and CaCO3 present.

*As long as CaO andCaCO3 are both presentAt equilibrium!

• Note: Although the concentrations of these species are not included in the equilibrium expression, they do participate in the reaction and must be present for an equilibrium to be established!

#1 Suppose that the gas phase reactions A → B and B → A are both elementary processes with rate constants of

3.8 10-2 s-1 and 3.1 10-1 s-1 respectively.

(a) What is the value of the equilibrium constant for the equilibrium A B?

(b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

(a)

(b) [A] Kc is less than 1 so reactants are favored

fc

r

kK = =

k

-2 -1

-1 -1

3.8 x 10 s =

3.1 x 10 s 0.12

#2 Write the expression for Kc for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous.

(a) 3 NO(g) N2O(g) + NO2(g)

(b) Ni(CO)4(g) Ni(s) + 4 CO(g)

(c) HF(aq) H+1(aq) + F-1(aq)

(d) 2 Ag(s) + Zn+2(aq) 2 Ag+1(aq) + Zn(s)

(a) and (c) are homogeneous

22c 3

[N O][NO ](a) K =

[NO]

2

2

+

c

[Ag ](d) K =

[Zn ]

4

4

c

[CO](b) K =

[Ni(CO) ]

+

c

[H ][F ](c) K =

[HF]

#3 When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products?

(a) N2 + O2 2 NO Kc = 1.5 10-10

(b) 2 SO2 + O2 2 SO3 Kp = 2.5 109

(a) mostly reactants (Kc is much less than 1)

(b) mostly products (Kp is much greater than 1)

#4 If Kc = 0.042 for PCl3(g) + Cl2(g) PCl5(g)

at 500 K, what is the value of Kp for this reaction at this temperature? n = 1 mol products - 2 mol reactants = -1

np cK = K (RT) =

-10.042 [(0.0821)(500)] =

-3 1.0 x 10

#5 At 1000 K, Kp = 1.85 for the reaction:

SO2(g) + ½ O2(g) SO3(g)

(a) What is the value of Kp for the reaction

SO3(g) SO2(g) + ½ O2(g)?

(b) What is the value of Kp for the reaction

2 SO2(g) + O2(g) 2 SO3(g)?

(a) 1 / 1.85 = 0.541(b) (1.85)2 = 3.42

#6 Mercury (I) oxide decomposes into mercury and oxygen: 2 Hg2O(s) 4 Hg(l) + O2(g)(a) Write the equilibrium constant expression for this reaction in terms of partial pressures(b) Explain why we normally exclude pure solids and liquids from equilibrium constant expressions.

(a)

(b) the concentration/pressure is constant for pure solids and liquids

2p OK = P

#7 Gaseous hydrogen iodide is placed in a closed container at 425oC, where it partially decomposes to hydrogen and iodine:

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

At equilibrium, it is found that

[HI] is 3.53 10-3 M, [H2] is 4.79 10-4 M, and [I2] is 4.79 10-4 M.

What is the value of Kc at this temperature?

2 22c

[H ][I ]K = =

[HI] 2

4 79-4 -4

-3

(4.79 x 10 )( . x 10 ) =

(3.53 x 10 ) 0.0184

#8 The equilibrium: 2 NO(g) + Cl2(g) 2 NOCl(g) is established at 500 K.

An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for NO, Cl2, and NOCl respectively.

Calculate Kp for this reaction at 500 K.

2

2NOCl

p 2NO Cl

PK = =

(P )(P ) 20 095 0 171

2(0.28) = ( . ) ( . )

51

Calculating Equilibrium Constants

• Tabulate initial and equilibrium concentrations (or partial pressures) for all species in the equilibrium.

• If an initial and an equilibrium concentration is given for a species, calculate the change in concentration.

• Use the coefficients in the balanced chemical equation to calculate the changes in concentration of all species.

• Deduce the equilibrium concentrations of all species.• Use these equilibrium concentrations to calculate the

value of the equilibrium constant.

#9 A mixture of 0.10 mol of NO and 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0 L vessel at 300 K. The following equilibrium is established:

2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g) At equilibrium, [NO] = 0.062 M.

(a) Calculate the equilibrium concentrations of H2, N2, and H2O(b) Calculate Kc

(a) calculate the change in [NO], 0.10 – 0.062 = 0.04

2 NO + 2 H2 N2 + 2 H2Oinitial 0.10 0.050 0 0.10change - 0.04 - 0.04 + 0.02 + 0.04equil 0.06 0.01 0.02 0.14

(b)2

2 2c 2 2

2

[N ][H O]K = =

[NO] [H ]

2

2 2

(0.02)(0.14) = (0.06) (0.01) 1000

This is called

an "ICE" chart

#10 A mixture of 0.2000 mol CO2, 0.1000 mol H2, and 0.1600 mol H2O is placed in a 2.000 L vessel. The following equilibrium is established at 500 K:

CO2(g) + H2(g) CO(g) + H2O(g)

(a) Calculate the initial partial pressures of CO2, H2, and H2O

(b) At equilibrium, PH2O = 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO

(c) Calculate Kp for the reaction

(a)

(b) the change in pressure of H2O = 3.51 – 3.28 = 0.23 atm

CO2 + H2 CO + H2Oinitial 4.10 2.05 0 3.28change - 0.23 - 0.23 + 0.23 + 0.23equil 3.87 1.82 0.23 3.51

(c)

2

2

2

CO

H

H O

P = nRT/V

P = (0.2000)(0.0821)(500) / 2.000 = 4.10 atm

P = (0.1000)(0.0821)(500) / 2.000 = 2.05 atm

P = (0.1600)(0.0821)(500) / 2.000 = 3.28 atm

2

2 2

CO H O

pCO H

(P )(P ) (0.23)(3.51)K = = =

(P )(P ) (3.87)(1.82)0.11