Chemical Equilibrium - Lamar · 2021. 4. 6. · Using Equilibrium Constants •A mixture of...

Chemical Equilibrium

Chapter 15

Homogeneous equilibrium:

When all reacting species are in the same phase.

Heterogeneous equilibrium:

When all reacting species are NOT in the same phase.

Heterogeneous Equilibrium:

When reacting species are in DIFFERENT phases.

SOLID and LIQUID phases are EXCLUDED from the equilibrium expression because they do not have “concentrations”

Chemical Equilibrium

A state achieved when

the RATES of the forward and

reverse reactions are EQUAL &

the concentrations of the reactants and

products remain constant.

Equilibrium Constant

The equilibrium constant (Keq) is

independent of concentration changes, but

dependent on the temperature.

Writing Equilibrium Constant Expressions

For the general reaction

α A + β B ↔ c C + d D

Forward Rate = kF [A]α [B]ββββ

Reverse Rate = kR [C]c [D]d

At Equilibrium

Forward Rate = Reverse Rate

Forward Rate = Reverse Rate


For the general reactionα A + β B ↔ c C + d D

At Equilibrium

kF [A]α [B]ββββ = kR [C]c [D]d

Put CONSTANTS on one side of equation And

Concentrations on other side of equation


• The equilibrium expression compares reactant & product concentrations.

[ ][ ]r

p

KReactantsProducts=

Size of Equilibrium Constant

If Kc > 103, products predominate

If Kc < 10–3, reactants predominate

If Kc is in the range 10–3 to 103, appreciable concentrations of both reactants and products are present.


α A + β B ↔ c C + d DAt Equilibrium

kF [A]α [B]ββββ = kR [C]c [D]d

[ ][ ]βα BD

A

cC

kk

Kd

R

Feq

��

��

��

��

==

The Concept of EquilibriumThe Concept of Equilibrium

The point at which the rate of decomposition N2O4(g) → 2NO2(g)

equals the rate of dimerization 2NO2(g) → N2O4(g).

is a dynamic equilibrium.

The equilibrium is dynamic because the reaction has not stopped: the opposing rates are equal.

At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4

The double arrow implies the process is dynamicForward reaction: Rate = kf[N2O4]

Reverse reaction: B → A Rate = kr[NO2]

At equilibrium kf[N2O4] = kr[NO2]

N 2 O 4 ( g ) 2 N O 2 ( g )

The Direction of the Chemical Equation and Keq

An equilibrium can be approached from any direction.For Example:

has

N2O4(g) 2NO2(g)

46.642

2

ON

2NO ==

P

PKeq

In the reverse direction:

2NO2(g) N2O4(g)

46.61

155.02NO

ON

2

42 ===P

PKeq

Other Ways to Manipulate Chemical Equations and Keq Values

The reaction

Has

• which is the square of the equilibrium constant for

2N2O4(g) 4NO2(g)

2ON

4NO

42

2

P

PKeq =

N2O4(g) 2NO2(g)

Equilibrium Concentration

Amounts of components are given as molarity { moles solute / liters of solution

orpartial pressure of a gas

Kc =NO2[ ] 2

N2O4[ ] Kp =PNO2

2

PN2O4


Write the Kp expressions for:

2 N2O5(g) ↔ 4 NO2(g) + O2(g)

Kp = (P NO2 )4 (P O2) / (P N2O5)2


Write the Kc expressions for:

2 N2O5(g) ↔ 4 NO2(g) + O2(g)

KC = [NO2 ]4 [ O2] / [N2O5]2

2 N2O5(g) ↔ 4 NO2(g) + O2(g)

The Kp and Kc expressions :

Kp = (P NO2 )4 (P O2) / (P N2O5)2

KC = [NO2 ]4 [ O2] / [N2O5]2

Is Kp = KC ?

No !

Relation between Gas Pressure and Concentration

• P V = n R T Ideal Gas Equation• Concentration (M) = moles / Liter• M = n / V• Rearranging P V = n R T

RTP

Vn

M ==

When are KP and KC EQUAL ?

For which of the following is KP = KC

(a) CO2(g) + C(s) ↔ 2 CO(g)

(b) Hg(l) + Hg2+(aq) ↔ Hg22+(aq)

(c)2Fe(s) + 3H2O(g) ↔ Fe2O3(s) + 3H2(g)

(d) 2 H2O(l) ↔ 2 H2(g) + O2(g)

Homogeneous And Heterogeneous Equilibria

• Homogeneous Equilibrium When all reacting species are in the same phase.

• Heterogeneous Equilibrium: When all reacting species are NOT in the same phase.

Heterogeneous Equilibrium:

When reacting species are in DIFFERENT phases.

SOLID and LIQUID phases are EXCLUDED from the equilibrium expression because they do not have “concentrations”


The equilibrium concentrations at 1000 K

for the reaction

CO(g) + Cl2(g) ↔ COCl2(g) are

[CO] = 1.2x10–2 M, [Cl2]=0.054 M,

[COCl2] = 0.14 M

Calculate Kc and Kp.

CO(g) + Cl2(g) ↔↔↔↔ COCl2(g)

[CO] = 1.2x10–2 M, [Cl2]=0.054 M, [COCl2] = 0.14 M

Kc = [COCl2] / [CO][Cl2]

Kc = 0.14 /(1.2x10–2 )(0.054 ) = 21578.2984

and

KC = KP (RT) Why ?Kp = KC / RT = KC / (0.0821)(1000) =

Sulfur dioxide + Oxygen = Sulfur trioxide

SO2 (g) + O2 (g) = SO3 (g)2 SO2 (g) + O2 (g) = 2 SO3 (g)

[ ][ ] [ ]2

22

23

OSOSO

KC =

2 SO2 (g) + O2 (g) = 2 SO3 (g)

Initial Moles # # #

Change in Moles –2x -x + 2x

Equilibrium # - 2x # - x # + 2x

Finally Concentrations at Equilibrium

The Initial Change Equilibrium method Is used to calculate equilibrium constant

1.00 mole of SO2 and 1.00 mole of O2 are put into a 1.00 liter flask. At equilibrium, 0.0925

mole of SO3 is formed. Calculate Kc at 1000 K for the reaction

2 SO2 (g) + O2 (g) = 2 SO3 (g)


2 SO2 (g) + O2 (g) = 2 SO3 (g)

Initial Moles 1.00 1.00 0

Change in Moles -0.925 -0.925/2 +0.925

Equilibrium 0.075 0.537 0.925

Equil. Conc 0.075 M 0.537 M 0.925 M



[ ][ ] [ ]2

22

23

OSOSO

KC =

2108.2537.02075.0

2925.0xCK ==

��

��

�

��

��

��

Methane (CH4) reacts with hydrogen sulfide to yield H2 and carbon disulfide

What is the value of Kp if the partial pressures in an equilibrium mixture are 0.20 atm of CH4, 0.25 atm of H2S, 0.52 atm of CS2, and 0.10 atm of H2?

How do you solve the problem?

• 1st Write Reaction• 2nd Balance Reaction• 3rd Write Equilibrium Constant Equation• 4th Put Equilibrium Pressures in Equation• Do the Arithmetic

Using Equilibrium Constants

The equilibrium constant (K c ) for

the formation of nitrosyl chloride,

2 NO(g) + Cl2(g) ↔ 2 NOCl(g)

from nitric oxide & chlorine gas is

6.5 x 10 4 at 35 °C.

2NO(g) + Cl2(g) ↔ 2NOCl(g) Kc= 6.5x104

2.0 x 10–2 moles of NO, 8.3 x 10–3 moles of Cl2, &

6.8 moles of NOCl are mixed in a 2.0–L flask.

Is the system at equilibrium? If not, what will happen?

2NO(g) + Cl2(g) ↔ 2NOCl(g) Kc= 6.5x104

2.0 x 10–2 moles of NO, 8.3 x 10–3 moles of Cl2, 6.8 moles of NOCl in a 2.0–L flask.

][][][

105.62

2

24

ClNONOCl

xK ==

7322

2

108.2]1015.4[]100.1[

]4.3[x

xx=−−

The reaction quotient (Qc)

Predicts reaction direction.

Qc > Kc System proceeds to form reactants.

Qc = Kc System is at equilibrium.

Qc < Kc System proceeds to form products.

Using Equilibrium Constants

•A mixture of 0.500mol H2 and 0.500mol I2

was placed in a 1.00–L stainless steel flask at 700°C. The equilibrium constant Kc is 57

H2(g) + I2(g) ↔ 2HI(g)

•Calculate the equilibrium concentrations.

Le Chatelier’s principle:

If an external stress is applied to a system at equilibrium, the system adjusts in such a

way that the stress is partially offset.

Le Chatelier’s principle:

Stress may be changes in

• concentration,

• pressure, volume, or

• temperature

that removes the system from equilibrium.

Concentration Changes:

The concentration stress of an added

reactant or product is relieved by

reaction in the direction that consumes

the added substance.

Concentration Changes:

The concentration stress of a removed

reactant or product is relieved by

reaction in the direction that

replenishes the removed substance.

Use the synthesis of ammonia as an example

N2(g) + 3H2(g) ↔↔↔↔ 2NH3(g)

At equilibrium [N2 ] = 0.50M [H2 ] = 3.00M

and [NH3 ] = 1.98 M at 700K

N2(g) + 3H2(g) ↔↔↔↔ 2NH3(g)

At equilibrium [N2 ] = 0.50M [H2 ] = 3.00M

and [NH3 ] = 1.98 M at 700K

what happens when the concentration

of N2 is increased to 1.50M?

N2(g) + 3H2(g) ↔↔↔↔ 2NH3(g)

when the concentration of N2 is increased

Le Châtelier’s principle says the

reaction will relieve the stress by

converting the N2 to NH3

N2(g) + 3H2(g) ↔↔↔↔ 2NH3(g) Kc = 0.291 at 700K

If H2 is increased

Le Châtelier’s principle tells us the


converting the extra H2 to NH3

N2(g) + 3H2(g) ↔↔↔↔ 2NH3(g) Kc = 0.291 at 700K

If NH3 is removed

Le Châtelier’s principle tells us the


producing more NH3

Volume and Pressure ChangesP V = n R T

Pressure is inversely proportional to volume

Increasing pressure = Decreasing volume

From P V = n R T

P = M (RT)

increasing pressure or decreasing volume increases concentration.

N2(g) + 3H2(g) ↔↔↔↔ 2NH3(g)

Use the synthesis of ammonia as an example

There are Four (4) units of gas reactants but

only Two (2) units of gas products

Therefore, an increase in pressure would

shift the equilibrium to the right

Le Châtelier’s Principle

Volume and Pressure Changes

Only reactions containing gases are

affected by changes in volume and

pressure.

The reaction of iron (III) oxide with carbon monoxide occurs in a

blast furnace when iron ore is reduced to iron metal:

Fe2O3(s) + 3 CO(g) ↔↔↔↔ 2 Fe(l) + 3 CO2(g)

Fe2O3(s) + 3 CO(g) ↔↔↔↔ 2 Fe(l) + 3 CO2(g)

• Use Le Châtelier’s principle to predict the direction of reaction when an equilibrium mixture is disturbed by:

(a) Adding CO

SHIFT TO RIGHT

Fe2O3(s) + 3 CO(g) ↔↔↔↔ 2 Fe(l) + 3 CO2(g)


(b) Removing CO2

SHIFT TO RIGHT

Fe2O3(s) + 3 CO(g) ↔↔↔↔ 2 Fe(l) + 3 CO2(g)


(c) Increasing Pressure

NO EFFECT

Why ????

Fe2O3(s) + 3 CO(g) ↔↔↔↔ 2 Fe(l) + 3 CO2(g)


(d) Adding Fe2O3

NO EFFECT

Why ????

Temperature Changes:

1. Changes the equilibrium constant

2. Endothermic processes are favored when

temperature increases.

3. Exothermic processes are favored when

temperature decreases.

Le Châtelier’s Principle & Temperature

In the first step of the Ostwald process

for synthesis of nitric acid, ammonia

is oxidized to nitrogen monoxide

4NH3(g) + 5O2(g) ↔↔↔↔ 4NO(g) + 6H2O(g)

�H° = –905.6 kJ

4NH3(g) + 5O2(g) ↔↔↔↔ 4NO(g) + 6H2O(g)�H° = –905.6 kJ

What happens as temperature is increased ?

Reaction is EXO thermic therefore an increase in temperature will shift the

equilibrium toward the reactants

Le Châtelier’s Principle

Catalysis: No effect.

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Chemical Equilibrium - Lamar · 2021. 4. 6. · Using Equilibrium Constants •A mixture of...

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