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Chapter 6 Chapter 6 Chemical Equilibrium Chemical Equilibrium

Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

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Page 1: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Chapter 6Chapter 6

Chemical EquilibriumChemical Equilibrium

Page 2: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Chapter 6: Chemical Equilibrium

6.1 The Equilibrium Condition6.2 The Equilibrium Constant6.3 Equilibrium Expressions Involving Pressures6.4 The Concept of Activity6.5 Heterogeneous Equilibria6.6 Applications of the Equilibrium Constant6.7 Solving Equilibrium Problems6.8 LeChatelier’s Principle6.9 Equilibria Involving Real Gases

Page 3: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Nitrogen dioxide shown immediately after expanding

Page 4: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Figure 6.1: Reaction of 2NO2(g) and N2O4(g) over time in a closed vessel

Page 5: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Reddish brown nitrogen dioxide, NO2 (g)

Page 6: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Reaching Equilibrium on the Macroscopic and Molecular Level

N2O4 (g) 2 NO2 (g)

Colorless Brown

Page 7: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

The State of Equilibrium

At equilibrium: ratefwd = raterev

ratefwd = kfwd[N2O4] raterev = krev[NO2]2

For the Nitrogen dioxide - dinitrogen tetroxide equilibrium:

N2O4 (g, colorless) = 2 NO2 (g, brown)

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

krev [N2O4]= = Keq

1) Small k N2 (g) + O2 (g) 2 NO(g) K = 1 x 10 -30

2) Large k 2 CO(g) + O2 (g) 2 CO2 (g) K = 2.2 x 1022

3) Intermediate k 2 BrCl(g) Br2 (g) + Cl2 (g) K = 5

Page 8: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 9: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Reaction Direction and the Relative Sizes of Q and K

Page 10: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Initial and Equilibrium Concentrations for theN2O4-NO2 System at 100°C

Initial Equilibrium Ratio

[N2O4] [N2O4] [N2O4][NO2] [NO2] [NO2]2

0.1000 0.0000 0.0491 0.1018 0.211

0.0000 0.1000 0.0185 0.0627 0.212

0.0500 0.0500 0.0332 0.0837 0.211

0.0750 0.0250 0.0411 0.0930 0.210

Page 11: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Figure 6.2: Changes in concentration with time for the reaction

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

Page 12: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Molecular model: When equilibrium is reached, how many molecules of

H2O, CO, H2, and CO2 are present?

Page 13: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Figure 6.3: H2O and CO are mixed in equal numbers

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

Page 14: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Figure 6.4: Changes with time in the rates of forward and reverse reactions

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

Page 15: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Figure 6.5: Concentration profile for the reaction

Page 16: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Like Example 6.1 (P 195) - IThe following equilibrium concentrations were observed for the Reaction between CO and H2 to form CH4 and H2O at 927oC.

CO(g) + 3 H2 (g) = CH4 (g) + H2O(g)

[CO] = 0.613 mol/L [CH4] = 0.387 mol/L [H2] = 1.839 mol/L [H2O] = 0.387 mol/L

a) Calculate the value of K at 927oC for this reaction.b) Calculate the value of the equilibrium constant at 927oC for: H2O(g) + CH4 (g) = CO(g) + 3 H2 (g)

c) Calculate the value of the equilibrium constant at 927oC for: 1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g)

Solution:a) Given the equation above:

K = = = ______L2/mol2 [CO] [H2]3

[CH4] [H2O] (0.387 mol/L) (0.387 mol/L)(0.613 mol/L) (1.839 mol/L)3

Page 17: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Like Example 6.1 (P 195) - IIb) Calculate the value of the equilibrium constant at 927oC for:

H2O(g) + CH4 (g) = CO(g) + 3 H2 (g)

K = = = 25.45 mol2/L2[CO] [H2]3

[CH4][H2O]

(0.613 mol/L) (1.839 mol/L)3

(0.387 mol/L) (0.387 mol/L)

This is the reciprocal of K:1K

= = 25.45 mol2/L21

0.0393 L2/mol2

c) Calculate the value of the equilibrium constant at 927oC for:

1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g)

K = =[CO]1/3 [H2]

[H2O]1/3 [CH4]1/3 (0.387mol/L)1/3 (0.387 mol/L)1/3

(0.613 mol/L)1/3 (1.839 mol/L)

K = = 0.340 L2/3/mol2/3 = (0.0393L2/mol2)1/3(0.729) (0.729)

(0.850)(1.839)

Page 18: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Summary: Some Characteristics of the Equilibrium Expression

The equilibrium expression for a reaction written in reverse is the reciprocal of that for the original reaction.

When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is theoriginal expression raised to the nth power. Thus Knew = (Koriginal)n

The apparent units for K are determined by the powers of the various concentration terms. The (apparent) units for K therefore depend on the reaction being considered. We will have more to say about the units for K in section 6.4.

Page 19: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 20: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Expressing K with Pressure Units

For gases, PV=nRT can be rearranged to give: P = RTnV

or: =n PV RT

Since = Molarity, and R is a constant if we keep the temperature constant then the molar concentration is directly proportional to the pressure.

nV

Therefore for an equilibrium between gaseous compounds we can express the reaction quotient in terms of partial pressures.

For: 2 NO(g) + O2 (g) 2 NO2 (g)

Qp = P 2

NO2

P 2NO x PO2

If there is no change in the number of moles of reactants and products then n = 0 then Kc = Kp , or if there is a change in the number of moles of reactants or products then:

Kp = Kc(RT) ngas

Page 21: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Figure 6.6: Position of the equilibrium

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

Page 22: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Writing the Reaction Quotient or Mass-Action Expression

Q = mass-action expression or reaction quotient

Q = Product of the Reactant ConcentrationsProduct of the Product Concentrations

For the general reaction: a A + bB cC + dD

Q =[C]c [D]d

[A]a [B]b

Example: The Haber process for ammonia production:

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

Q =[NH3]2

[N2][H2]3

Page 23: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Reaction Direction and the Relative Sizes of Q and K

Page 24: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Writing the Reaction Quotient from the Balanced Equation

Problem: Write the reaction quotient for each of the following reactions: (a) The thermal decomposition of potassium chlorate: KClO3 (s) = KCl(s) + O2 (g)

(b) The combustion of butane in oxygen: C4H10 (g) + O2 (g) = CO2 (g) + H2O(g)

Plan: We first balance the equations, then construct the reaction quotientas described by equation 17.4.Solution: (a) 2 KClO3 (s) 2 KCl(s) + 3 O2 (g) Qc =

[KCl]2[O2]3

[KClO3]2

(b) 2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O(g)

Qc =[CO2]8 [H2O]10

[C4H10]2 [O2]13

Page 25: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Writing the Reaction Quotient for an Overall Reaction–I

Problem: Oxygen gas combines with nitrogen gas in the internal combustion engine to produce nitric oxide, which when out in theatmosphere combines with additional oxygen to form nitrogen dioxide.

(1) N2 (g) + O2 (g) 2 NO(g) Kc1 = 4.3 x 10-25 (2) 2 NO(g) + O2 (g) 2 NO2 (g) Kc2 = 6.4 x 109

(a) Show that the overall Qc for this reaction sequence is the same as the product of the Qc’s for the individual reactions.(b) Calculate Kc for the overall reaction.Plan: We first write the overall reaction by adding the two reactions together and write the Qc. We then multiply the individual Kc’s for the total K. (1) N2 (g) + O2 (g) 2 NO(g)

(2) 2 NO(g) + O2 (g) 2 NO2 (g)

overall: N2 (g) + 2 O2 (g) 2 NO2 (g)

Page 26: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Writing the Reaction Quotient for an Overall Reaction–II

Qc (overall) =[NO]2

[N2][O2]2

For the individual steps:

(1) N2 (g) + O2 (g) 2 NO(g) Qc1 =

(2) 2 NO(g) + O2 (g) 2 NO2 (g) Qc2 =

[NO]2

[N2] [O2]

[NO]2 [O2][NO2]2

Qc1 x Qc2 = x =[NO]2

[N2] [O2][NO2]2

[NO]2 [O2][NO2]2

[N2][O2]2

(a) cont.

The same!

(b) K = Kc1 x Kc2 = (4.3 x 10-25)(6.4 x 109) = ______________

Page 27: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

The Form of Q for a Forward and Reverse Reaction

The production of sulfuric acid depends upon the conversion of sulfurdioxide to sulfuric trioxide before the sulfur trioxide is reacted with water to make the sulfuric acid.

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

Qc(fwd) =[SO3]2

[SO2]2[O2]

2 SO3 (g) 2 SO2 (g) + O2 (g)For the reverse reaction:

Qc(rev) = =[SO2]2[O2][SO3]2

1Qc(fwd)

and: Kc(fwd) = = = _____________

at 1000K Kc(fwd) = 261

1Kc(rev)

1261

Page 28: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Ways of Expressing the Reaction Quotient, Q

Form of Chemical Equation Form of Q Value of K

Reference reaction: A B Q(ref) = K(ref) =

Reverse reaction: B A Q = = K =

Reaction as sum of two steps:

[B][A]

[B]eq

[A]eq

1 [A]Q(ref) [B]

1K(ref)

(1) A C

(2) C BQoverall = Q1 x Q2 = Q(ref) Koverall = K1 x K2

= x = = K(ref)

[C] [B] [B]

[A] [C] [A]

Coefficients multiplied by n Q = Qn(ref) K = Kn

(ref)

[A] [C][C] [B]Q1 = ; Q2 =

Reaction with pure solid or Q’ = Q(ref)[A] = [B] K’ = K(ref)[A] = [B] liquid component, such as A(s)

Page 29: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Example 6.2 (P 202) - IFor the synthesis of ammonia at 500oC, the equilibrium constant is 6.0 x 10-2 L2/mol2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases.a) [NH3]0 = 1.0 x 10-3 M; [N2]0= 1.0 x 10-5 M; [H2]0=2.0 x 10-3 Mb) [NH3]0 = 2.00 x 10-4 M; [N2]0= 1.50 x 10-5 M; [H2]0= 3.54 x 10-1 Mc) [NH3]0 = 1.0 x 10-4 M; [N2]0= 5.0 M; [H2]0= 1.0 x 10-2 MSolutiona) First we calculate the Q:

Q = =

= ____________________ L2/mol2

[NH3]02

[N2]0[H2]03

(1.0 x 10-3 mol/L)2

(1.0 x 10-5 mol/L)(2.0 x 10-3 mol/L)3

Since K = 6.0 x 10-2 L2/mol2, Q is much greater than K. For thesystem to attain equilibrium, the concentrations of the products must be decreased and the concentrations of the reactants increased. The system will shift to the left:

Page 30: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Example 6.2 (P 202) - II

b) We calculate the value of Q:

[NH3]02

[N2]0[H2]03

Q = =

= 6.01 x 10-2 L2/mol2

(2.00 x 10-4 mol/L)2

(1.50 x 10-5 mol/L) (3.54 x 10-1 mol/L)3)

In this case Q = K, so the system is at equilibrium. No shift will occur.

c) The value of Q is:

[NH3]02

[N2]0[H2]03

Q = =

= _________________ L2/mol2

(1.0 x 10-4 mol/L)2

(5.0 mol/L) (1.0 x 10-2 mol/L)3

Here Q is less than K, so the system will shift to the right, attainingequilibrium by increasing the concentration of the product anddecreasing the concentrations of the reactants. More Ammonia!

Page 31: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Like Example 6.3 (P203-5) - ILook at the equilibrium example for the formation of Hydrogen Chloridegas from Hydrogen gas and Chlorine gas. Initially 4.000 mol of H2, and4.000 mol of Cl2, are added to 2.000 mol of gaseous HCl in a 2.000 liter flask. H2 (g) + Cl2 (g) 2 HCl(g)

K = 2.76 x 102 = [Cl2] = [H2] = 4.000mol/2.000L = 2.000M [HCl] = 2.000 mol/2.000L = 1.000M

Initial Concentration Change Equilibrium Conc. (mol/L) (mol/L) (mol/L)

[H2]o = 2.000M -x [H2] = 2.000-x

[Cl2]o = 2.000M -x [Cl2] = 2.000-x

[HCl]o = 1.000M +2x [HCl] = 1.000 + 2x

[HCl]2

[H2] [Cl2]

Page 32: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Like Example 6.3 (P203-5) - II

K = 2.76 x 102 = = = [HCl]2

[H2] [Cl2] (1.000 + 2x)2

(2.000 –x)(2.000 – x)(1.000 + 2x)2

(2.000 – x)2

Take the square root of each side:

16.61 =(1.000 + 2x) (2.000 – x)

33.22 – 16.61x = 1.000 + 2x Therefore: [H2] = 0.269 M32.22 = 18.61x [Cl2] = 0.269 M x = 1.731 [HCl] = 4.462 M

Check:

= = 276 OK! [HCl]2

[H2] [Cl2] (4.462)2

(0.269)(0.269)

Page 33: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Summary: Solving Equilibrium Problems

Write the balanced equation for the reaction.

Write the equilibrium expression using the law of mass action.

List the initial concentrations.

Calculate Q and determine the direction of the shift to equilibrium.

Determine the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations.

Substitute the equilibrium concentrations into the equilibriumexpression, and solve for the unknown.

Check your calculated equilibrium concentrations by making sure thatthey give the correct value of K.

Page 34: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Determining Equilibrium Concentrations from K–I

Problem: One laboratory method of making methane is from carbon disulfide reacting with hydrogen gas, and K this reaction at 900°C is 27.8.

CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g)

At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 molCS2, 1.10 mol of H2, and 0.45 mol of H2S, how much methane was formed?Plan: Write the reaction quotient, and calculate the equilibrium concentrations from the moles given and the volume of the container. Use the reaction quotient and solve for the concentration of methane.Solution: CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g)

K = = 27.8[CH4] [H2S]2

[CS2] [H2]4

[CS2] = 0.250 mol

4.70 L

[CS2] = ____________ mol/L

Page 35: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Determining Equilibrium Concentrations from K–II

Solution cont.

[H2] = = 0.23404 mol/L1.10 mol4.70 L

[H2S] = = 0.095745 mol/L0.450 mol4.70 L

[CH4] = =Kc [CS2] [H2]4

[H2S]2

(27.8)(0.05319)(0.23404)4

(0.095745)2

[CH4] = = 0.485547 mol/L = 0.486 M0.0091670.004436

Check: Substitute the concentrations back into the equation for K and make sure that you get the correct value of K

K = = = 27.81875[CH4] [H2S]2

[CS2] [H2]4

(0.485547 M)(0.095745 M)2

(0.05319 M)(0.23404 M)4OK!

Page 36: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Determining Equilibrium Concentrations from Initial Concentrations and K –I

Problem: Given the that the reaction to form HF from molecular hydrogen and fluorine has a reaction quotient of 115 at a certain temperature. If 3.000 mol of each component is added to a 1.500 L flask,calculate the equilibrium concentrations of each species.

H2 (g) + F2 (g) 2 HF(g)

Plan: Calculate the concentrations of each component, and then figure the changes, and solve the equilibrium equation to find the resultant concentrations.Solution:

K = = 115[HF]2

[H2] [F2]

[H2] = = 2.000 M3.000 mol

3.000 mol

3.000 mol

[F2] = = 2.000 M

[HF] = = 2.000 M1.500 L

1.500 L

1.500 L

Page 37: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Determining Equilibrium Concentrations from Initial Concentrations and K–II

Concentration (M) H2 F2 HF

Initial 2.000 2.000 2.000Change -x -x +2xFinal 2.000-x 2.000-x 2.000+2x

K = = 115 = =[HF]2

[H2][F2](2.000 + 2x)2

(2.000 - x)(2.000 - x)(2.000 + 2x)2

(2.000 - x)2

Taking the square root of each side we get:

(115)1/2 = =10.7238(2.000 + 2x)(2.000 - x)

x = 1.528

[H2] = 2.000 - 1.528 = 0.472 M

[F2] = 2.000 - 1.528 = 0.472 M

[HF] = 2.000 + 2(1.528) = 5.056 M

K = = [HF]2

[H2][F2](5.056 M)2

(0.472 M)(0.472 M)

K = 115check:

Page 38: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Calculating K from Concentration Data–I

Problem: Hydrogen iodide decomposes at moderate temperatures by thereaction below:

When 4.00 mol HI was placed in a 5.00 L vessel at 458°C, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of Kc ?Plan: First we calculate the molar concentrations, and then put them intothe equilibrium expression to find it’s value.Solution: To calculate the concentrations of HI and I2, we divide the amounts of these compounds by the volume of the vessel.

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

Starting conc. of HI = = 0.800 M4.00 mol 5.00 L

Equilibrium conc. of I2 = = 0.0884 M0.442 mol 5.00 L

Conc. (M) 2HI(g) H2 (g) I2 (g)

Starting 0.800 0 0Change - 2x x xEquilibrium 0.800 - 2x x x = 0.0884

Page 39: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Calculating K from Concentration Data–II

[HI] = M = (0.800 - 2 x 0.0884) M = 0.623 M

[H2] = x = 0.0884 M = [I2]

Kc = = = ____________[H2] [I2]

[HI]2

( 0.0884)(0.0884)(0.623)2

Therefore the equilibrium constant for the decomposition of HydrogenIodide at 458°C is only 0.0201 meaning that the decomposition does notproceed very far under these temperature conditions. We were given the initial concentrations, and that of one at equilibrium, and found the others that were needed to calculate the equilibrium constant.

Page 40: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Using the Quadratic Formula to Solve for the Unknown

Given the Reaction between CO and H2O:

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

Initial 2.00 1.00 0 0Change -x -x +x +xEquilibrium 2.00-x 1.00-x x x

Qc = = = = 1.56[CO2][H2]

[CO][H2O](x) (x)

(2.00-x)(1.00-x)x2

x2 - 3.00x + 2.00

We rearrange the equation: 0.56x2 - 4.68x + 3.12 = 0 ax2 + bx + c = 0

quadratic equation:x = - b + b2 - 4ac

2a

x = = 7.6 M and 0.73 M

4.68 + (-4.68)2 - 4(0.56)(3.12)2(0.56)

[CO] = 1.27 M[H2O] = 0.27 M[CO2] = 0.73 M [H2] = 0.73 M

Page 41: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Predicting Reaction Direction and Calculating Equilibrium Concentrations –I

Problem: Two components of natural gas can react according to the following chemical equation:

CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)

In an experiment, 1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S, and 2.00 mol H2 are mixed in a 250 mL vessel at 960°C. At this temperature, K = 0.036. (a) In which direction will the reaction go?(b) If [CH4] = 5.56 M at equilibrium, what are the concentrations of theother substances?Plan: The find the direction, we calculate Qc using the calculated concentrations from the data given, and compare it with Kc. (b) Based upon (a), we determine the sign of each component for the reaction tableand then use the given [CH4] at equilibrium to determine the others.Solution:[CH4] = = 4.00 M1.00 mol

0.250 L

[H2S] = 8.00 M, [CS2] = 4.00 Mand [H2 ] = 8.00 M

Page 42: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Predicting Reaction Direction and Calculating Equilibrium Concentrations –II

Q = = = 64.0[CS2] [H2]4

[CH4] [H2S]2

4.00 x (8.00)4

4.00 x (8.00)2

Comparing Q and K: Q > K (64.0 > 0.036, so the reaction goes to the left. Therefore, reactants increase and products decrease their concentrations.(b) Setting up the reaction table, with x = [CS2] that reacts, which equalsthe [CH4] that forms.Concentration (M) CH4 (g) + 2 H2S(g) CS2(g) + 4 H2(g)

Initial 4.00 8.00 4.00 8.00Change +x +2x -x - 4xEquilibrium 4.00 + x 8.00 + 2x 4.00 - x 8.00 -4x

Solving for x at equilibrium: [CH4] = 5.56 M = 4.00 M + xx = ____________ M

Page 43: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Predicting Reaction Direction and Calculating Equilibrium Concentrations –III

x = 1.56 M = [CH4]

Therefore:

[H2S] = 8.00 M + 2x = 8.00 M + 2(1.56 M) = _________ M

[CS2] = 4.00 M - x = 4.00 M - 1.56 M = __________ M

[H2] = 8.00 M - 4x = 8.00 M - 4(1.56 M) = __________ M

[CH4] = __________ M

Page 44: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Le Chatelier’s Principle

“If a change in conditions (a “stress”) is imposed on a systemat equilibrium, the equilibrium position will shift in a directionthat tends to reduce that change in conditions.”

A + B C + D + Energy

For example: In the reaction above, if more A or B is added youwill force the reaction to produce more product, if they are removed,it will force the equilibrium to form more reactants. If C or D is added you will force the reaction to form more reactants, if they areRemoved from the reaction mixture, it will force the equilibrium toForm more products. If it is heated, you will get more reactants, and if cooled, more products.

Page 45: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Henri Louis Le Chatelier

Source: Photo Researchers

Page 46: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 47: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Blue Anhydrous cobalt(II) chloride

CoCl2 (s) + 6 H2O(g) CoCl2 6 H2O(s)

Page 48: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Figure 6.7: Equilibrium mixture

Page 49: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

The Effect of a Change in Concentration–I

Given an equilibrium equation such as :

CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)

If one adds ammonia to the reaction mixture at equilibrium, it will force the reaction to go to the right producing more product. Likewise, if one takes ammonia from the equilibrium mixture, it will force the reaction back to produce more reactants by recombining H2 and HCN to givemore of the initial reactants, CH4 and NH3.

CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)

CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)

Add NH3

Forces equilibrium to produce more product.

Forces the reaction equilibrium to go back to the left and produce more of the reactants.

Remove NH3

Page 50: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

The Effect of a Change in Concentration–II

CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)

If to this same equilibrium mixture one decides to add one of the products to the equilibrium mixture, it will force the equilibrium backtoward the reactant side and increase the concentrations of reactants.Likewise, if one takes away some of the hydrogen or hydrogen cyanidefrom the product side, it will force the equilibrium to replace it.

CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)

CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)

Add H2Forces equilibrium to go toward the reactant direction.

Remove HCNForces equilibrium to make more produce and replace the lost HCN.

Page 51: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

The Effect of a Change in Pressure (Volume)

Pressure changes are mainly involving gases as liquids and solidsare nearly incompressible. For gases, pressure changes can occur in three ways: Changing the concentration of a gaseous component Adding an inert gas (one that does not take part in the reaction)

Changing the volume of the reaction vessel

When a system at equilibrium that contains a gas undergoes a changein pressure as a result of a change in volume, the equilibrium position shifts to reduce the effect of the change. If the volume is lower (pressure is higher), the total number of gas molecules decrease. If the volume is higher (pressure is lower), the total number of gas molecules increases.

Page 52: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Figure 6.8: A mixture of NH3(g), N2(g), and H2(g) at equilibrium

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

Page 53: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Figure 6.9: Brown NO2(g) and colorless N2O4(g) at equilibrium in a syringe

Source: Ken O’Donoghue

2 NO2 (g) N2O4 (g)

Brown Colorless

Page 54: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 55: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 56: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

The Effect of a Change in Temperature

Only temperature changes will alter the equilibrium constant, and that iswhy we always specify the temperature when giving the value of Kc.

The best way to look at temperature effects is to realize that temperature is a component of the equation, the same as a reactant, or product. For example, if you have an exothermic reaction, heat (energy) is on the product side of the equation, but if it is an endothermic reaction, it will be on the reactant side of the equation.

O2 (g) + 2 H2 (g) 2 H2O(g) + Energy = Exothermic

Electrical energy + 2 H2O(g) 2 H2 (g) + O2 (g) = Endothermic

A temperature increase favors the endothermic direction and a temperature decrease favors the exothermic direction.

A temperature rise will increase Kc for a system with a positive H0rxn

A temperature rise will decrease Kc for a system with a negative H0rxn

Page 57: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 58: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Shifting the N2O4(g) and 2NO2(g) equilibrium by changing the temperature

Page 59: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 60: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 61: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 62: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 63: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions
Page 64: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Table 6.4 (P 216)

Shifts in the Equilibrium Position for the Reaction N2O4(g) 2 NO2 (g)

Change Shift

Addition of N2O4 (g) More ProductsAddition of NO2 (g) More ReactantsRemoval of N2O4 (g) More ReactantsRemoval of NO2 (g) More ProductsAddition of He(g) NoneDecrease in Container Volume More ReactantsIncrease in Container Volume More ProductsIncrease in Temperature More ProductsDecrease in Temperature More Reactants

Page 65: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Table 6.5 (P 217)

Values of Kpobs at 723 K for the Reaction

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

as a Function of Total Pressure (at equilibrium)

Total Pressure Kpobs

(atm) (atm-2)

10 4.4 x 10-5

50 4.6 x 10-5 100 5.2 x 10-5 300 7.7 x 10-5

600 1.7 x 10-4

1000 5.3 x 10-4

Page 66: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Percent Yield of Ammonia vs. Temperature (°C)

at five different operating pressures.

Page 67: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Key Stages in the Haber Synthesis of Ammonia

Page 68: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Predicting the Effect of a Change in Concentration on the Position of the Equilibrium

Problem: Carbon will react with water to yield carbon monoxide andand hydrogen, in a reaction called the water gas reaction that was used to convert coal into a fuel that can be used by industry.

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

What happens to:(a) [CO] if C is added? (c) [H2O] if H2 is added?(b) [CO] if H2O is added? (d) [H2O] if CO is removed?Plan: We either write the reaction quotient to see how equilibrium will be effected, or look at the equation, and predict the change in direction of the reaction, and the effect of the material desired.Solution: (a) No change, as carbon is a solid, and not involved in the equilibrium, as long as some carbon is present to allow the reaction.(b) The reaction moves to the product side, and [CO] increases.(c) The reaction moves to the reactant side, and [H2O] increases.(d) The reaction moves to the product side, and [H2O] decreases.

Page 69: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Predicting the Effect of Temperature and Pressure

Problem: How would you change the volume (pressure) or temperaturein the following reactions to increase the chemical yield of the products?(a) 2 SO2 (g) + O2 (g) 2 SO3 (g); H0 = 197 kJ(b) CO(g) + 2 H2 (g) CH3OH(g); H0 = -90.7 kJ(c) C(s) + CO2 (g) 2 CO(g); H0 = 172.5 kJ(d) N2(g) + 3 H2(g) 2 NH3(g); H0 = -91.8 kJPlan: For the impact of volume (pressure), we examine the reaction for the side with the most gaseous molecules formed. For temperature, we see if the reaction is exothermic, or endothermic. An increase in volume(pressure) will force a reaction toward fewer gas molecules. Solution: To get a higher yield of the products you should:

(a) Increase the pressure, and increase the temperature.(b) Increase the pressure, and decrease the temperature.(c) A pressure change will not change the yield, an increase in the temperature will increase the product yield.(d) Increase the pressure, and decrease the temperature.

Page 70: Chapter 6 Chemical Equilibrium. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions

Effect of Various Disturbances on an Equilibrium SystemDisturbance Net Direction of Reaction Effect on Value of KConcentration Increase [reactant] Toward formation of product None Decrease [reactant] Toward formation of reactant NonePressure (volume) Increase P Toward formation of lower amount (mol) of gas None Decrease P Toward formation of higher amount (mol) of gas NoneTemperature Increase T Toward absorption of heat Increases if H0

rxn> 0 Decreases if H0

rxn< 0 Decrease T Toward release of heat Increases if H0

rxn< 0 Decreases if H0

rxn> 0Catalyst added None; rates of forward and reverse reactions increase equally None