The Equilibrium Condition, the Equilibrium Constant and Equilibrium

in Terms of Pressures

Chemistry 142 BAutumn Quarter 2004J. B. Callis, Instructor

Lecture #17

Chemical Equilibrium

Until Now we have assumed that a chemical reaction goes to completion as written. For example, we might suppose that when H2O is introduced into a flask with CO and sealed, it will all convert to H2 and CO2:

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

However, we note that the reaction does not go to completion, but rather forms a certain, predictable amount of products and does not proceed further. This new stable state of the system which includes both reactants and products is called the equilibrium state.

Molecular Picture of Equilibrium2 NO2(g) = N2O4(g)

Concentration vs. TimeCO(g) + H2O(g) = CO2(g) + H2(g)

Molecular Picture of Establishment of Equilibrium

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

Kinetics of Approach to Equilibrium

The Haber ProcessN2(g) + 3 H2(g) 2 NH3(g)

Characteristics of True Chemical Equilibria

• They show no macroscopic evidence of change.

• They are reached through spontaneous processes.

• A dynamic balance of forward and reverse processes exists within them.

• They are the same regardless of the direction from which they are approached.

The Equilibrium Constant - DefinitionFor a reaction of the type

jA + kB = lC+mD

Where A, B, C and D represent chemical species and j, k, l, and m are coefficients of the balanced chemical equation, the law of mass action is represented in the following equilibrium expression:

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

kj

ml

BADCK

Characteristics of the Equilibrium Expression

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

• If the original reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power n.

• The apparent units for K are determined by the powers of the concentration terms.

Problem 17-1 Expressing the Equilibrium Constant

At a certain temperature the value of the equilibrium constant for the reaction is

CS2(g) + 3 O2(g) = CO2(g) + 2 SO2(g)

is K1.

(a) Write an expression for the equilibrium constant, K1.

(b) Write an expression for the equilibrium constant, K2, of the reverse reaction.

(c) Write an expression for the related equilibrium constant, K3

(1/3) CS2(g) + O2(g) = (1/3) CO2(g) + (2/3) SO2(g)

Problem 17-1 Calculation of the Equilibrium Constant

Answers:

(a)

(b)

(c)

Problem 17-2 Calculation of the Equilibrium Constant

At 454 K, the following reaction takes place:

3 Al2Cl6(g) = 2 Al3Cl9(g)

At this temperature, the equilibrium concentration of Al2Cl6(g) is 1.00 M and the equilibrium concentration of Al3Cl9(g) is 1.02 x 10-2 M. Compute the equilibrium constant at 454 K.

Answer:

Equilibrium Expressions Involving PressuresFor a reaction of the type

jA + kB = lC+mD

It is sometimes convenient to write the equilibrium expression in terms of partial pressures, e.g.

The the Ps indicate the partial pressures of the species in equilibrium and KP is a constant called the equilibrium constant in terms of partial pressures.

kB

jA

mD

lC

PPPPPK

How is KP related to K?Answer: Through the use of the ideal gas law.

RTCRTRTnVP AAA

Vnor P A

A

nkjmlkj

ml

kB

jA

mD

lC

kB

jA

mD

lC

kB

jA

mD

lC

P

RTKRTKRTRT

CCCC

RTCRTCRTCRTC

PPPPK

)()( )(

Where n = l+m-(j+k)

(Apparent) Units for K and KP

• At first sight, it would seem that the units for K would be in concentrations raised to a reaction-specific power, and those for KP in pressure units raised to a reaction-specific power. But these are ‘apparent’ units.

• For theoretical reasons, we will refer each concentration or pressure to a reference state, which always causes the units of concentration or pressure to cancel. Thus, K and KP are expressed without units.

Problem 17-3: Converting between K and KP

Calculate Kp for the following equilibrium:

N2(g) + 3 H2(g) = 2 NH3(g), K = 2.4 x 10-3 at 1000 K

Answers to Problems in Lecture 17

1. (a) (b) (c)

2.

3.

322

222

1 OCSSOCO

K

12

22

322

21

SOCOOCS

KK

311

231

2

322

312

3 OCSSOCO KK

141004.1 MK

7106.3 PK