Chemical Equilibrium

CHEMICAL EQUILIBRIUM

Basic Concepts

Reversible reactions do not go to completion. They can occur in either direction Symbolically, this is represented as:

dD cC bB+aA

Basic Concepts

Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate. A chemical equilibrium is a reversible

reaction that the forward reaction rate is equal to the reverse reaction rate.

Chemical equilibria are dynamic equilibria. Molecules are continually reacting, even

though the overall composition of the reaction mixture does not change.

Basic Concepts

4

One example of a dynamic equilibrium can be shown using radioactive 131I as a tracer in a saturated PbI2 solution.

solution. into go williodine eradioactiv theof Some

solution. filter the then minutes, few afor Stir 2

I 2 Pb PbI

solution. PbI saturated ain PbI solid Place 1-(aq)

2(aq)

OH

2(s)

2*22

Basic Concepts

5

Graphically, this is a representation of the rates for the forward and reverse reactions for this general reaction.

aA(g) + bB(g) ⇄ cC(g) + dD(g)

Basic Concepts

6

One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction.

2SO2(g) + O2(g) ⇄ 2SO3(g)

7

The Equilibrium Constant

rate. reverse therepresents which DCkRate

rate. forward therepresents which BAkRate

rr

ff

For a simple one-step mechanism reversible reaction such as:

The rates of the forward and reverse reactions can be represented as:

(g)(g)(g)(g) D C B A


10

When system is at equilibrium:Ratef = Rater

BA

DC

k

k

torearrangeswhich

DCkBAk

:give toiprelationsh rate for the Substitute

r

f

rf


11

Because the ratio of two constants is a constant we can define a new constant as follows :

kk

K and

KC DA B

f

rc

c


Similarly, for the general reaction:

we can define a constant

12

reactions. allfor validis expression This

BA

DCK

products

reactants ba

dc

c

D d C c B b A a (g)(g)(g)(g)


13

Kc - Equilibrium constant

Kc - product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

Kc values are dimensionless because they actually involve a thermodynamic quantity called activity. Activities are directly related to molarity


14

Write equilibrium constant expressions for the following reactions at 500oC. All reactants and products are gases at 500oC.

5

23c

235

PCl

ClPClK

ClPClPCl


15

HI 2 I + H 22

22

2

c IH

HIK

5

24

3

62

4

c

223

ONH

OHNO=K

OH 6+NO 4O 5 + NH 4


Example: One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for the reaction.

Equil [ ]’s 0.028 M 0.172 M 0.086 M

16

235 ClPClPCl


17

Kc PCl3 Cl2

PCl5 Kc

0.172 0.086 0.028

Kc 0.53

KPCl

PCl Cl

KK

or K K

c' 5

3 2

cc' c

'

c

0 0280172 0 086

19

1 1 10 53 19

.. .

.

. .

235 ClPClPCl


18

Example: The decomposition of PCl5 was studied at another temperature. One mole of PCl5 was introduced into an evacuated 1.00 liter container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl3 was present in the container. Calculate the equilibrium constant at this temperature.

0 0 1.00 Initial

ClPClPCl g2g3g5

M


Tanother at 90.0

40.0

60.060.0K

0.60 0.60 0.40 mEquilibriu

0.60+ 0.60+ 0.60- Change

0 0 1.00 Initial

ClPClPCl

'c

g2g3g5

MMM

MMM

M


Example : At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH3 was present. Calculate Kc for the reaction.

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

Initial 0.80 M 0.90 M 0

Change - 0.10 M - 0.30 M +0.20 M

Equilibrium 0.70 M 0.60 M 0.20 M

Kc NH3 2

N2 H2 3 0.20 2

0.70 0.60 3 0.26

Variation of Kc with the Form of the Balanced Equation

Large equilibrium constants indicate that Large equilibrium constants indicate that most of the reactants are converted to most of the reactants are converted to products.products.

Small equilibrium constants indicate that Small equilibrium constants indicate that only small amounts of products are only small amounts of products are formed.formed.2 main calculations:

Equilibrium Concentrations Kc

Kc equilibrium concentrations

ICE table

Predicting the direction of reactions: The REACTION COEFFICIENT, Qc

Tank of water in Equilibrium

The Reaction Quotient

For this general reaction :

aA +bB cC +dD

Q C c

D d

A aB b

The mass action expression or reaction quotient has the symbol Q. Q has the same form as Kc

The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values.


Why do we need another “equilibrium constant” that does not use equilibrium concentrations?

Q will help us predict how the equilibrium will respond to an applied stress - compare Q with Kc.

When

Q = Kc : the system is in equilibrium

Q > Kc : the system goes to the left (), towards reactants

Q < Kc : the system goes to the right (), towards products


The equilibrium constant for the following reaction is 49 at 450oC. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

Calculate Qc, compare with Kc

Use ICE table to calculate equilibrium [ ]s

H2 + I2 ⇄ 2HI

Equilibrium calculations and reaction quotient The equilibrium constant Kc, is 3.00 for the

following reaction at a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?

(g)3(g)2(g)2(g) NO SO NO SO

The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?

H2 + I2 ⇄ 2HI

An equilibrium mixture from the following reaction was found to contain 0.20 mol/L of A, 0.30 mol/L of B, and 0.30 mol/L of C.

(a) What is the value of Kc for this reaction?

(b) If the volume of the reaction vessel were suddenly doubled while the temperature remained constant, what would be the new equilibrium concentrations?

ggg C B A

A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 moles of COCl2, 0.60 moles of CO and 0.20 mole of Cl2.

a) Calculate the equilibrium constant.

b) An additional 0.80 mole of Cl2 is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl2, and COCl2 when the new equilibrium is established

g2g2g COCl Cl CO

Comparing Q and Kc

K

Q

K

Q

KQ

Equilibrium Formation of reactants

Formation of products

Q = Kc Q > KcQ < Kc

∴ Equilibrium is aSTABLE condition

Disturbing a System at Equilibrium: Le Chatelier’s Principle

Le Chatelier’s Principle - If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium.

Some possible stresses to a system at equilibrium are:

1. Changes in concentration of reactants or products.

2. Changes in pressure or volume (for gaseous reactions)

3. Changes in temperature.

Le Chatelier’s Principle

Changes in Concentration of Reactants and/or Products

Look at the following system at equilibrium at 450oC.H2 + I2 ⇄ 2HI Kc = 49

]I][[H

HI][K

22

2

c

- Removing HI: forward – toward product-side

- Removing I2 : backward – toward reactant-side

- Adding H2 : forward – product-side

- Adding HI and removing I2 : backward – reactant-side

- Adding HI and adding H2 : ? Calculate Q


Changes in Volume • (and pressure for reactions involving gases) Consider the following system

2NO2(g) ⇄

N2O4(g)

Move the piston downward: ⇩ volume, ⇧ pressure

Shift toward products (lower #molecules)

Move the piston upward: ⇧ volume, ⇩ pressureShift toward reactants (greater #molecules)


Effect of Temperature on Equilibrium Consider the following reaction at equilibrium:

reaction? in thisproduct or reactant aheat Is

kJ/mol 198H SO 2 O SO 2 orxn3(g)2(g)2(g)

2SO2(g) + O2(g) ⇄ 2SO3(g) + 198 kJAdd heat to the system: increase O2 and SO2

Remove heat from the system: increase SO3

Place the system in an ice bath: decrease SO2 and O2, increase SO3

Predictions are only true for EXOTHERMIC reactions


Effect of Temperature on Equilibrium

Add heat to the system: clear solution

Remove heat from the system: increase precipitate

Place the system in an ice bath: increase precipitate

NH4NO3 (s) + 98.9 kJ ⇄ NH4+

(aq) + NO3-

(aq)

The effect of temperature on reactions in equilibrium depends on the sign of ∆Hrxn

Introduction of a Catalyst Catalysts decrease the activation energy of both the forward

and reverse reaction equally. Catalysts do not affect the position of equilibrium.

The concentrations of the products and reactants will be the same whether a catalyst is introduced or not.

Equilibrium will be established faster with a catalyst.


Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the following?

a) Increasing the temperature

b) Decreasing the external temperature c) Increasing the volume

d) Increasing the pressure

e) Adding He(g) in the system

f) Adding N2(g) in the system

g) Placing solid Pt in the system as catalyst

Backward, decrease productsForward, increase products

Forward, increase products

Backward, decrease products

Forward, increase products

No effect

No effect

A certain indicator was extracted from mayana leaves, and its dissociation in water can be shown in the following thermochemical equation:

At 25oC the predominant color of the neutral extract is light green. Determine the expected color when a solution of the extract is applied with the following stress:

a) Placed in an ice bath

c) Placing sodium carbonate in the solution

b) Placing sodium sulfate in the system

d) Compressing the solution

e) Adding 3 drops of nitric acid f) Adding hot water to the systemg) Adding small amounts of red dye in c)

Yellow

Yellow

Light Green

Light Green

Dark Green

Dark Green

Orange

Equilibrium Constant Expression inGas-Phase Reactions For gas phase reactions the equilibrium

constants can be expressed in partial pressures rather than concentrations.

For gases, the pressure is proportional to the concentration. PV = nRT P = nRT/V n/V = M P= MRT and M = P/RT

For systems involving gases, we can use Kp instead of Kc

Equilibrium Constant Expression inGas-Phase Reactions

Consider this system at equilibrium at 5000C.

Equilibrium Constant Expression inGas-Phase Reactions Using the expression [ ] = P/RT

From the previous slide we can see that the relationship between Kp and Kc is:

reactants) gaseous of moles of (#-products) gaseous of moles of (#=n

RTKKor RTKK npc

ncp

Relationship Between Kp and Kc

Relationship Between Kp and Kc

Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of Kp?

g2gg Br + NO 2 NOBr 2

Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and 1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter vessel,

What is the total pressure inside the reaction vessel after equilibrium is reached?

gg2g2 HI 2I H

Heterogeneous Equilibria Heterogeneous equilibria have more

than one phase present. For example, a gas and a solid or a liquid and a

gas.

How does the equilibrium constant differ for heterogeneous equilibria? Pure solids and liquids have activities of unity. Solvents in very dilute solutions have activities that

are essentially unity. The Kc and Kp for the reaction shown above are:

2(g)(s)s3 O32KCl 2KClO

3Op

32c 2

P=K ][O=K

2(g)(l)2(aq)32 COOHCOH

C)25at (F 2CaCaF o-aq

2aqs2

Gibbs’ Free Energy and Equilibrium

G0 is the standard free energy change. G0 is defined for the complete conversion of all

reactants to all products. With superscript 0

G is the free energy change at nonstandard conditions• For example, concentrations other than 1 M or

pressures other than 1 atm.

• Temperature, however, is same as in G0rxn


G is related to Go by the following relationship -

quotientreaction =Q

re temperatuabsolute = T

constant gas universal =R

Q log RT 303.2G=G

or lnQ RTG=Go

o


At equilibrium, G=0 and Q=Kc. Then we can derive this relationship:

K log RT 2.303 -=G

orK ln RT -=G

: torearrangeswhich

K log RT 303.2G0

orK ln RTG0

0

0

0

0


D ofactivity theis C ofactivity theis

B ofactivity theis A ofactivity theis

where=K

dD + cC bB +aA

DC

BA

bB

aA

dD

cC

aa

aa

aa

aa

For the following generalized reaction, the thermodynamic equilibrium constant is defined as follows:


General equation; THERMODYNAMIC FORM

All gaseous reactants and products

All solutions of reactants and products

Used when both gaseous and solution forms appear in the chemical equation


Calculate the equilibrium constant, Kp, for the following reaction at 25oC.

g2g42 NO 2ON

g2g42 NO 2ON

NON SPONTANEOUSFORWARD

[N2O4] > [NO2]

Kp for the reverse reaction at 25oC can be calculated easily - it is the reciprocal of the above reaction.

2 NO2(g) N2O4(g)

G rxno 4.78 kJ/mol

Kp'

1

Kp

1

0.1456.90

PN2O4 PNO2 2


It is difficult to analyze the relationship between free energy and equilibrium constant at non-standard conditions

The basic relationship between ∆G0rxn

and Kc appears only for the standard conditions form

Gorxn K Spontaneity at unit concentration

< 0 > 1Forward reaction spontaneous,

More products than reactants at equilibrium

= 0 = 1 IDEAL system, very RARE

> 0 < 1Reverse reaction spontaneous,

More reactants than products at equilibrium



Gibbs’ Free Energy and Equilibrium Nitrosyl bromide, NOBr, is 34% dissociated

by the following reaction at 25oC, in a vessel in which the total equilibrium pressure is 0.25 atmosphere.

Calculate the ∆G0rxn.

What is the ∆G when 5.00% of 2.00 atm NOBr has dissociated?

EVALUATION OF EQUILIBRIUM CONSTANTS AT DIFFERENT TEMPERATURES From the value of Ho and K at one

temperature, T1, we can use the van’t Hoff equation to estimate the value of K at another temperature, T2.

OR

The equilibrium constant Kc of the reaction

H2(g) + Br2(g) ⇄ 2HBr(g)

is 1.6 x 105 at 1297 K and 3.5 x 104 at 1495 K.

(a)Is the reaction exothermic or endothermic?(b)Find Kc for the reaction: ½H2(g) + ½Br2(g) ⇄

HBr(g)

(c)Pure HBr is placed in a container of constant volume and heated to 1297 K. What percentage of the HBr is decomposed to Br2 and H2 at equilbrium?

At its normal boiling point of 100oC, the heat of vaporization is 40.66 kJ/mole. What is the equilibrium vapor pressure of water at 25oC?

In the distant future, when hydrogen may be cheaper than coal, steel mills make iron by the reaction

Fe2O3(s) + 3H2(g) ⇄ 2Fe(s) + 3H2O(g)

For this reaction, ∆H0 = 96 kJ/mole and Kc = 8.11 at 1000 K. (a)What percentage of the H2 remains unreacted after the reaction has come to equilibrium at 1000 K?(b) Is this percentage greater or less if the temperature is decreased below 1000 K?

Chemical Equilibrium

Documents

le chateliers

decrease products

equilibrium

balanced equation

adding h2

equilibrium

phase reactions

o2