Chapter 13 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there.

Chapter 13Chemical Equilibrium

The state where the concentrations of all reactants and products remain constant with time.

On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

Figure 13.1 A Molecular Representation of the Reaction 2NO2(g) g) Over Time in a Closed Vessel

Figure 13.2 Changes in ConcentrationsH2O(g) + CO(g) H2(g) + CO2(g)

Figure 13.4 The Changes with Time in the Rates of Forward and Reverse ReactionsH2O(g) + CO(g) H2(g) + CO2(g)

Figure 13.5 The Ammonia Synthesis Equilibrium

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

The Law of Mass Action

jA + kB lC + mDwhere A, B, C, and D represents chemical species and j, k, l, and m are their coefficient in the balanced equation.The law of mass action is represented by the equilibrium expression:

The square brackets indicate the concentrations of the chemical species at equilibrium, and K is a constant called the equilibrium constant.

Kl m

j k C DA B

Equilibrium Expression

Write the equilibrium expression for the following reaction:

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

Applying the law of mass action gives,

Superscript 4, 6, 4, and 7 are the coefficients of NO2, H2O, NH3, and O2 respectively. The value of the equilibrium constant at a given temperature can be calculated if we know the equilibrium concentrations of the reaction components.

K NO H O

NH O2

2

24 6

34 7

Notes on Equilibrium Expressions (EE)

The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse.

When the equation for a reaction is multiplied by n, EEnew = (EEoriginal)n

The units for K depend on the reaction being considered. K values are customarily written without units.

Equilibrium Expressions Involving Pressures:

Equilibria involving gases can be described in terms of pressures.

Ideal gas equation:

where, C equals n/v or number of moles n per unit volume V. Thus C represents the molar concentration of the gas.

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

In terms of the equilibrium partial pressures of the gasses

P V = n R T P =n

vR T = C R T

K =[N H ]

[N ][H ]

C

(C )(C )K

3

2 2

N H

N Hc

3

2 2

K =P

(P )(P )p

N H

N H

3

2 2

22

3

2

3

3

K v. Kp

For

jA + kB lC + mD

Kp = K(RT)n

n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.

KP P

P P

C R T C R T

C R T C R T

C C

C C

R T

R TK R T K R T

pC D

A B

c D

A B

c D

A B

( )( )

( )( )

( ) ( )

( ) ( )

)( )

( )( )

( )

( )( ) ( )

(

l l

l

m m

m

j j

j

k k

k

l m

j k

( ) ( )l m j k n

Heterogeneous Equilibria

. . . are equilibria that involve more than one phase.

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

K = [CO2]

The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

Figure 13.6 CaCO3(s) CaO(s) + CO2(g)

ExamplesThe decomposition of liquid water to gaseous hydrogen and oxygen,

2H2O(l) 2H2(g) + O2(g)

K = [H2]2[O2] and Kp=(P2H2)(PO2)

Water is not included in either equilibrium expression because it is a pure liquid. However, if water is a gas rather than a liquid,

2H2O(g) 2H2(g) + O2(g)

because the concentration or pressure of water vapor can change.

KH O

H OK p

P P

P

H O

H O

[ ] [ ]

[ ]

( )( )2 2

2

2 2

2

an d

2

2

2

2

Reaction Quotient

. . . helps to determine the direction of the move toward equilibrium.

The reaction quotient is obtained by applying the law of mass action using initial concentrations instead of equilibrium concentrations.

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

where the subscript zeros indicate initial concentrations.

Q HF

H F2 2

02

0 0

Direction of reaction

To determine in which direction a system will shift to reach equilibrium, we compare the values of Q and K. There are three possible cases:

1.Q is equal to K. The system is at equilibrium; no shift will occur.

2.Q is greater than K. The system shifts to the left, consuming products and forming reactants, until equilibrium is achieved.

3.Q is less than K. The system shifts to the right, consuming reactants and forming products, to attain equilibrium.

Solving Equilibrium Problems

1. Balance the equation.

2. Write the equilibrium expression.

3. List the initial concentrations.

4. Calculate Q and determine the shift to equilibrium.

5. Define equilibrium concentrations.

6. Substitute equilibrium concentrations into equilibrium expression and solve.

7. Check calculated concentrations by calculating K.

Le Châtelier’s Principle

. . . if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

Effects of Changes on the System

1. Concentration: The system will shift away from the added component.

2. Temperature: K will change depending upon the temperature (treat the energy change as a reactant).

3. Pressure:

a. Addition of inert gas does not affect the equilibrium position.

b. Decreasing the volume shifts the equilibrium toward the side with fewer moles.

Figure 13.8 A Mixture of N2, H2, and NH3

Figure 13.9 The Effect of Decreased Volume on the Ammonia Synthesis Equilibrium