Chapter 14 Page 1

CHAPTER 14: CHEMICAL EQUILIBRIUM

Part One: Describing Chemical Equilibrium A. Basic Concepts.

1. All chemical reactions are, in principle, reversible, i.e., they can go both directions.

aA + bB cC + dD 2. The symbol ( ) means reactions are taking place in forward and reverse directions

simultaneously. 3. Chemical equilibrium exists when the two opposite reactions occur simultaneously

at the same rate. 4. At equilibrium the concentrations of reactants and products become constant. 5. Chemical equilibria are nevertheless dynamic equilibria, in that reactions are still

happening even though concentrations are no longer changing. 6. Balance does NOT mean reactant and product concentration will be equal. If products

are more stable, they will be present in greater abundance at equilibrium. 7. Example. Catalytic Methanation:

CO(g) + 3 H2(g) CH4(g) + H2O(g) -start with only CO(g) and H2(g) present. -initially, only reaction taking place will be the forward reaction

Later, as product concentration grows, reverse reaction will be taking place.

Figure 14.3

Chapter 14 Page 2

B. The Equilibrium Constant and the Reaction Quotient. (Section 14.2) 1. Consider the total reaction:

N2 + 3 H2 2 NH3 2. We define the Reaction Quotient Qc for this reaction as:

Q =NH3[ ]2

N2[ ] H2[ ]3

3. Before equilibrium is reached, Qc will be changing in time, as concentrations change. 4. As equilibrium is achieved, Qc converges to a constant and then remains fixed. This is

a consequence of the law of mass action. 5. This constant is Kc, the equilibrium constant for this rxn. Qc → Kc as equilibrium is reached. when Qc ≠ Kc, not at equilib. when Qc = Kc, are at equilib. 6. So we could say:

€

Kc =NH3[ ]

2

N2[ ] H2[ ]3

equilib mixture

7. Example, in the following reaction at a certain temperature:

2 SO2(g) + O2(g) 2 SO3(g) at equilibrium, the concentrations are: [SO2] = 0.344 M; [O2] = 0.172 M; [SO3] = 0.056 M What is the value of the equilib. constant Kc for this reaction (at this T)?

€

Kc =SO3[ ]2

SO2[ ]2 O2[ ]

equilib values

€

=0.056 M( )2

0.344 M( )2 0.172 M( ) = 1.5 x 10-1 = 0.15 (drop units)

Chapter 14 Page 3

8. Note that because Kc < 1, product conc. [SO3] is smaller than reactant conc. [SO2] and [O2].

9. Kc >> 1 implies greater product stability; Kc << 1 implies greater reactant stability. 10. Kc is constant for a given rxn, varying only with T. 11. Kc is independent of initial concentrations of a reacting mixture. 12. Kc has no units. 13. Kc depends on the way the chemical equation is balanced. For example:

€

H2O

€

H2 + 12 O2

€

Kc =H2[ ] O2[ ]

12

H2O[ ]

€

2H2O

€

2H2 +O2

€

′ K c =H2[ ]2 O2[ ]

H2O[ ]2

€

′ K c = Kc( )2 (doubling coefficients squares the Kc)

€

H2 + 12 O2

€

H2O

€

′ ′ K c =H2O[ ]

H2[ ] O2[ ]1

2

€

′ ′ K c =1

Kc (reversing the rxn inverts Kc)

Part Two: Using the Equilibrium Constant A. Calculating Concentrations of Reactants & Products at Equilibrium.

1. Once K has been determined for a rxn, it can be used to compute concentrations at

equilib. given any starting concentrations. 2. Example: For the rxn below, the equilib. constant at 25°C is Kc = 1.9. If a vessel

initially contained [PCl5(g)] = 2.0 M, what will be the equilibrium concentrations?

Chapter 14 Page 4

PCl3(g) + Cl2(g) PCl5(g) initial 0 0 2.0 M change to x x - x reach equilib. at equilib. x x 2.0 - x

€

Kc =2.0 − x( )x( ) x( )

= 1.9

Rearrange and solve for x: 2.0 - x = 1.9 x2 or 0 = 1.9 x2 + x - 2.0 Is of the form of a quadratic: 0 = ax2 + bx + c

Such that:

€

x =−b ± b2 − 4ac

2a a = 1.9; b = 1; c = -2.0

€

x =−1± 12 − 4∗1.9 ∗ -2.0( )

2∗1.9

€

=−1± 16.2

3.8=−1± 4.023.8

x = 0.795 and -1.32 The -1.32 solution is physically impossible, so: [PCl3(g)] = 0.795M [Cl2(g)] = 0.795 M [PCl5(g)] = 2.0 - 0.795 = 1.205 M

Chapter 14 Page 5

B. Factors that Affect Equilibria. 1. Le Chatelier’s Principle = if a change of conditions is applied to a system at

equilibrium, the system responds so as to restore equilibrium. 2. The reaction quotient Qc and its comparison with Kc enables us to understand this

quantitatively. 3. Four types of changes:

a. Concentration perturbation. b. Pressure perturbation. c. Temperature perturbation. d. Introduction of catalysts.

C. Concentration Perturbation.

1. Consider rxn below at some temperature T:

2 NO2(g) N2O4(g) Kc = 0.25 2. Initially in equilibrium, with: [NO2] = 2.0 M [N2O4] = 1.0 M Let’s check to make sure system is in equilibrium:

€

Q =N2O4[ ]NO2[ ]2

=1.02.0( )2

=1.04.0

= 0.25 = Kc

Yes, we’re in equilibrium. 3. Perturb system by dumping in reactant NO2, increasing its concentration to 4.0 M.

How will system respond?

Chapter 14 Page 6

Let’s see how this works: new Q =

1.04.0( )2

=1.0

16.0< 0.25

Now Qc < Kc. System will respond to raise Qc back to 0.25 = Kc.

2 NO2(g) N2O4(g) perturbed 4.0 1.0 system at new 4.0 - 2x 1.0 + x equilib.

€

N2O4[ ]NO2[ ]2

=1.0 + x4.0 − 2x( )2

= 0.25

Solve for x. Turns out to be x = 0.70 New equilib. [NO2] = 4.0 - 2x = 2.6 M [N2O4] = 1.0 + x = 1.7 M

4. Note that perturbing concentrations did not change Kc! It changed Qc so that Qc no longer = Kc. System had to respond to restore equilib.

D. Changes in V and P.

1. P changes have little effect on equilibria involving only solids or liquids.

Chapter 14 Page 7

2. P changes do affect gas concentrations and thus may perturb equilibria involving gaseous species.

3. Consider same rxn: 2 NO2(g) N2O4(g) Kc = 0.25 initially 2.0 M 1.0 M given concs. in equilib.

cut volume 4.0 M 2.0 M in half Now: Q =

N2O4[ ]NO2[ ]2

=2.04.0( )2

= 0.125 < 0.25

Qc < Kc, no longer at equilib. System will respond by shifting to right, in this case.

4. Summary: Reducing volume increased the pressure shifting the gas phase rxn in the direction of decreasing the number of gas molecules.

2 NO2 → N2O4

5. P or V change has no effect on Kc, only on Qc

E. Changes in T. 1. An increase in T favors endothermic rxns.

2. A decrease in T favors exothermic rxns.

3. For an exothermic rxn, as T↑ Kc↓ .

4. For an endothermic rxn, as T↑ Kc↑ .

5. Changing T changes Kc, meaning that Qc no longer = Kc, and system must respond to

restore equilib.

(exo) 6. Example:

2 NO2(g) N2O4(g) Kc = 0.25 at T1 at equilib. 2.0 M 1.0 M at T1

Chapter 14 Page 8

Given: Raising T causes K to decrease to Kc = 0.10 at T2. Recalculate the concentration at T2. [NO2] [N2O4] at T1 equilib. 2.0 M 1.0 M at T2 equilib. 2.0 + 2x 1.0 - x

€

1.0 − x( )2.0 + 2x( )2 = 0.10← new Kc

End up with a quadratic equation to solve. x = 0.30 at T2 equilib. [NO2] = 2.0 + 2x = 2.6 M [N2O4] = 1.0 - x = 0.70 M

F. Introduction of a Catalyst. 1. Has no effect on Kc. Only increases the rates of the forward and reverse rxns. 2. Equilibrium is established more quickly in presence of catalyst, but the catalyst does

not change the amount of product or reactant produced at equilibrium.

Part Three: Other Considerations

A. Equilibrium Constants Involving Partial Pressures.

1. For gas phase rxns, convenient to express amounts of species as partial pressures in atm rather than concentrations.

2. Thus, a different form of equilibrium constant is defined, Kp. This is a pressure-based equilibrium constant instead of a concentration-based one.

3. Otherwise, its properties are the same as Kc.

4. For example, in rxn:

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

Qp =PNH3( )2

PN2 PH2( )3 Px = partial pressures, not concentrations

Qp = Kp at equilibrium.

Chapter 14 Page 9

B. Relationship Between Kp and Kc. Derive.

1. PV = nRT, so:

nV=PRT

nV

is just [ ], molar concentration.

Kc =NH3[ ]2

N2[ ] H2[ ]3=

PNH3RT

2

PN2RT

PH2RT

3

Kc =PNH3( )2

PN2( ) PH2( )3×1RT( )21RT( )4

Kc = Kp(RT)2

OR

Kp = Kc(RT)-2

In general: Kp = Kc(RT)Δn

where Δn = ngas products - ngas reactants

= change in moles of gaseous substances (coeffs in balanced equation)

C. Heterogeneous Equilibria.

1. Involve species in more than one phase. 2. Example:

2 HgO(s) 2 Hg(l) + O2(g) 3. Terms for solids and liquids do not appear in K expressions. So for this rxn: Kc = [O2(g)] OR Kp = PO2

Chapter 14 Page 10

NOTES: