Chapter 5 Chemical Equilibrium

Chapter 5 Chemical Equilibrium

The Equilibrium State Learning objectives:  Define chemical equilibrium  State examples of chemical equilibrium  Write a balanced chemical equation for any reversible reaction.  Interpret how equilibrium state is achieved from a reaction.

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The Equilibrium State • What does equilibrium mean to you? • Chemical Equilibrium: A state achieved when the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant. – Examples are . . . • Saturated solutions • Phase equilibrium • Many reactions • Human body  equilibria involving O2 molecules and the protein Hb play a crucial role in the transport and delivery of O2 from our lungs to cells throughout our body.

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The Equilibrium State is known as •Equilibrium Chemists arebetween interested inphases these reversible reactions. One example is the following: physical equilibrium.

The decomposition of the colorless gas dinitrogen tetroxide (N2O4) to the dark brown gas nitrogen dioxide (NO2). 4

The Equilibrium State

N2O4 (s) Frozen N2O4 is nearly colorless

2NO2 (g) N2O4 (g) As N2O4 is warmed above its boiling point, it starts to dissociate into brown NO2 gas

N2O4 (g) 2NO2 (g) Eventually the color stops changing as N2O4 (g) and NO2 9g) reach concentration at which they are interconverting at the same rate. The wo gases are in equilibrium.5

The Equilibrium State N2O4(g)

2NO2(g)

Change in the concentrations of N2O4 and NO2 with time in two experiments at 25°C

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The Equilibrium State

The Equilibrium State Important lesson about equilibrium… A mixture if reactants and products is formed in which concentrations no longer change with time indicates that the reaction has reached a state of equilibrium. For equilibrium to occur, neither reactants nor products can escape from the system. At equilibrium, the particular ratio of concentration terms equals to constant. 8

The Equilibrium Constant Learning objectives:  Write equilibrium constant expression.  Calculate the equilibrium constant Kc from the equilibrium concentrations of products and reactants.

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The Equilibrium Constant Kc • Equilibrium expression for determining the equilibrium constant, Kc, for the general reversible reaction aA + bB cC +dD.

In this expression:  Product concentrations always appear in the numerator  Reactant concentrations always appear in the denominator  Each concentration is always raised to the power of its stoichiometric coefficient in the balance equation  Kc is independent of concentration changes, but dependent on the temperature. 10

The Equilibrium Constant Kc Kc for a chemical reaction is very useful because it indicates whether a reaction is product or reactant favored, and it can be used to calculate the quantity of reactant or product present at equilibrium

Experiment 1 Kc =

[NO2]2

(0.0125)2

[N2O4]

0.0337

= 4.64 x 10-3

Experiment 5 (0.0141)2 0.0429

= 4.63 x 10-3

Homogeneous & Heterogeneous Equilibria Learning objectives:  Distinguish between homogenous and heterogeneous equilibrium. 

Write the equilibrium equation for any balanced chemical equation representing a homogenous or heterogeneous equilibrium.



Calculate the equilibrium constant Kp from the equilibrium partial pressures of reactants and products.



Interconvert between Kc and Kp using a balanced equation. 12

Homogeneous Equilibria • Homogeneous Equilibrium: When all reacting species are in the same phase, all reactants and products are included in the expression. •

Eg.

Ag+ (aq) + 2NH3 (aq)

Co (g) + Cl2 (g)

Ag(NH3)2+ (aq) diamminesilver(I) cation COCl2 (g) phosgene

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Homogeneous Equilibria • Kc is obtained when equilibrium concentrations expressed in molarity are substituted into equilibrium – constant expression. • Kp is obtained equilibrium partial pressures expressed in atm are substituted into equilibrium –constant expression.

2

NO2 Kc   N2O4 

Kp

2 PNO2

PN2O4 14

Homogeneous Equilibria • We can convert between Kc and Kp using an equation derived from PV = nRT: For aA bB Kp = Kc (RT) ∆n ∆n = moles gas products – moles of gas reactants ∆n = b – a R = 0.08206 (L.atm)/K.mol)

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Heterogeneous Equilibria • Heterogeneous Equilibrium: – When reacting species are in different phases – Solution and gas phases are included, solid and liquid phases are excluded from the equilibrium equation because their concentrations “do not change.”

CaCO3(s)  CaO(s) + CO2(g) Kc = [CO2] because CaCO3 and CaO are solids.

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Heterogeneous Equilibria Thermal decomposition of calcium carbonate: CaCO3(s) <=> CaO(s) +CO2(g)

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Using Equilibrium Constants Learning objectives:  State generalizations concerning the composition of equilibrium mixtures 

Define reaction quotients. Relate reaction quotient to equilibrium constant.



Determine whether or not a system is at equilibrium for a given mixture of reactants and products. If it is not, determine the direction in which the reaction must go to achieve equilibrium.



Calculate the final concentrations of reactants and/or products from Kc and initial concentrations of reactants 18 and/or products.

Using Equilibrium Constants •

The magnitude of equilibrium constant provides us with important information about the composition of an equilibrium mixture.

•

We can make the following generalizations concerning the composition of equilibrium mixtures: If Kc > 103, products predominate over reactants. If Kc is very large, the reaction is said to proceed to completion. If Kc is in the range 10–3 to 103, appreciable concentrations of both reactants and products are present. If Kc < 10–3, reactants predominate over products. If Kc is very small, the reaction proceeds hardly at all. 19

Using the Equilibrium Constant 2H2O(g) (at 500 K)

2H2(g) + O2(g) Kc = 4.2 x 10-48

H2(g) + I2(g) 2HI(g) (at 500 K) Kc = 57.0

2H2(g) + O2(g)

2H2O(g)

(at 500 K) Kc = 2.4 x 1047 20

Using the Equilibrium Constant aA + bB

Reaction quotient:

cC + dD

Qc =

[C]tc[D]td [A]ta[B]tb

The reaction quotient, Qc, is defined in the same way as the equilibrium constant, Kc, except that the concentrations in Qc are not necessarily equilibrium values.

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Using the Equilibrium Constant • To determine the direction in which the reaction will proceed to achieve equilibrium, we compare the values of Qc and Kc: net reaction goes from left to right (reactants to products).

•

If Qc < Kc

•

If Qc > Kc net reaction goes from right to left (products to reactants).

•

If Qc = Kc no net reaction occurs. 22

Using Equilibrium Constants Using the reaction quotient

• Predicting the direction of reaction. The direction of reaction depends on the relative values of Qc and Kc. • If Qc < Kc, the reaction goes from left to right If Qc = Kc, the reaction is at equilibrium If Qc > Kc, the reaction goes from right to left 23

Steps to follow in calculating equilibrium concentrations from initial concentrations.

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Using the Equilibrium Constant Set up a table: H2(g) + I2(g)

2HI(g)

I

0

0

0.250

C

+x

+x

-2x

E

x

x

0.250 - 2x

Substitute values into the equilibrium expression: [HI]2 Kc =

[H2][I2]

(0.250 - 2x)2 57.0 = x2 25

Using the Equilibrium Constant Solve for “x”: (0.250 - 2x)2 57.0 =

x = 0.0262 x2

Determine the equilibrium concentrations: H2: 0.0262 M I2: 0.0262 M HI: 0.250 - 2(0.0262) = 0.198 M 26

Le Châtelier’s Principle Learning objectives:  State Le Chateliers’s Principle. 

State the factors that alter the composition of an equilibrium mixture based on Le Chateliers’s Principle.



Explain the effect of the changes on reactant or product when a stress is applied to a system.



Determine the reaction direction when a system at equilibrium reacts to a stress applied to the system, including changes in concentrations, pressure and volume, or temperature.



Describe the effect of adding a catalyst to a system at 27 equilibrium.

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

The concentration of reactants or products can be changed.

•

The pressure and volume can be changed.

•

The temperature can be changed. 28

Le Châtelier’s Principle Changes in Concentration Haber process for synthesis of ammonia. N2(g) + 3 H2(g)

2 NH3(g)

Kc = 0.291 at 700 K

Given an equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 at 700 K, what happens when the concentration of N2 is increased to 1.50 M?

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Altering an Equilibrium Mixture: Concentration N2(g) + 3H2(g)

2NH3(g)

Le Châtelier’s Principle In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Châtelier’s principle predicts that

– The concentration stress of an added reactant or product is relieved by reaction in the direction that consumes the added substance. – The concentration stress of a removed reactant or product is relieved by reaction in the direction that replenishes the removed substance. 31

Le Châtelier’s Principle Changes in Pressure and Volume In general Le Châtelier’s principle predicts that An increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas. A decrease in pressure by enlarging the volume will bring about net reaction in the direction that increases the number of moles of gas. 32

Effect of pressure and volume N2(g) + 3 H2(g)

2 NH3(g) Kc = 0.291 at 700 K

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Le Châtelier’s Principle Changes in Temperature In general, when an equilibrium is disturbed by a change in temperature, Le Châtelier’s principle predicts that •

the equilibrium constant for an exothermic reaction (negative H°) decreases as the temperature increases. Equilibrium shifts to form more reactants (reverse reaction).

•

the equilibrium constant for an endothermic reaction (positive H°) increases as the temperature increases. Equilibrium shifts to form more product (forward reaction).

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Altering an Equilibrium Mixture: Temperature N2(g) + 3H2(g)

2NH3(g) H° = -2043 kJ

As the temperature increases, the equilibrium shifts from products to reactants.

Effect of a catalyst on equilibrium Potential energy profiles for a reaction whose activation energy is lowered by the presence of a catalyst. The activation energy for the catalyzed pathway (red curve) is lower than that for the uncatalyzed pathway (blue curve) by an amount ΔEa. The catalyst lowers the activation energy barrier for the forward and reverse reactions by exactly the same amount.

The catalyst therefore accelerates the forward and reverse reactions by the same factor, and the composition of the equilibrium mixture is unchanged.

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