Dynamic Equilibrium, Keq, and the Mass Action...

1 Equilibrium January 10

Dynamic Equilibrium, Keq, and the Mass Action Expression

The Equilibrium Process

Dr. Fred Omega Garces Chemistry, Miramar College


(Dynamic) Equilibrium Chemical Equilibrium is a dynamic state in which the rates of the forward and the reverse reaction are equal.

A teeter-totter is an analogy of a system in balance. The left is in balance to the right. In a chemical reaction the Left (reactant) is changing at the same rate as the right (products).

Reactant Product


When equilibrium is establish A D B That is rate forward = rate reverse or kf[A] = kr[B] Rearranging this equation can be rearrange to kf/kb = [B] /[A] The mass action expression defines the equilibrium constant (Keq):

Keq = [B] / [A]

For any general chemical process at equilb. aA + bB D pP + qQ

A mass action expression can be written:

This is also referred to as the Law of Mass Action

Equilibrium: Mass Action Expression

K eq = A [ ]

a • B [ ] b

P [ ] p • Q [ ]

q


Substances are in different phases at equilibrium

i.e., solid and aqueous.

Which solid is more concentrated?

Concentration of a solid (and pure liquid) is always a constant; The concentration does not change ! These chemicals are not included in the mass action expression

100 g (1 cup) 200 g (2 cup)

Heterogeneous Systems (1)


Heterogeneous Systems Consider the following reaction: CaCO3(s) D CaO (s) + CO2 (g)

€

Keq =CaO[ ] • CO2[ ]

CaCO3[ ]

but, CaO[ ] = constant (solid)

and CaCO3[ ] = constant (solid)

Keq =Constant1[ ] • CO2[ ]

Constant2[ ]

Keq = Kp = CO2[ ]


Which is favored: Reactant or Product?

For Keq > 1, the ratio [Product]x > [Reactant]y

For Keq = 1, the ratio [Product]x = [Reactant]y

For Keq < 1, the ratio [Product]x < [Reactant]y

For Keq > 1, at equilibrium Product is favored.

For Keq = 1, at equilibrium Product and Reactant are equal.

For Keq < 1, at equilibrium Reactant is favored.

Meaning of Keq

K eq = A [ ] a • B [ ] b P [ ]

p • Q [ ] q

= Product [ ] x

Reactant [ ] y


Summary: Equilibrium constant

• The mass action expression is equal to the equilibrium constant.

• Equilibrium constants are generally written without

units.

For Keq >> 1, at equilibrium Product is favored.

For Keq < 1, at equilibrium Reactant is favored.


LeChatelier’s Principle Changing the concentration or pressure of a substance in a reaction at equilibrium is creating a stress. The removal of something creates a “deficit stress”, and adding something creates an “excess stress.” Reactions will always respond to stress by replacing the deficit (making more of what was removed) or by conversion of the excess (consuming some of what was added). This description works regardless of which side of the reaction is changed.

Stress/Deficit will be in the form of:

(1) Concentration,

(2) Pressure and

(3) Temperature variation


LeChatelier Principle Methods of disturbing Equilibrium and LeChatelier response. Stress on Reactant Stress on Product Relief on Reactant Relief on Product

Reactant D Product

Equilibrium Revisited: Equilibrium - Situation in which changes occur at equal rates so no net change is apparent. LeChatelier Principle: A stress on the a system at equilibrium causes changes to the system to reduce the stress until a new equilibrium is established.


Reactant Change: Stress added Consider the affect on Equilibrium as a result of increasing the

concentration of the reactant.

Stress on the reactant Which direction does the reaction shift towards?

i) A system at equilibrium.

ii) A stress is added to the reactant.

The reaction shifts to the right or left?


Reactant Change: Stress added Consider the affect on Equilibrium as a result of increasing the

concentration of the reactant.

Stress on the reactant Which direction does the reaction shift towards?



iii) The system self-adjust to reduce the stress.

iv) Until a new equilibrium is re-established

The reaction shifts to the right. The reaction favors the product.


Product Changes: Stress added Consider the affect on Equilibrium as a result of increasing the concentration of the product. Stress on the product Which direction does the reaction shift towards?


ii) A stress is added to the product.



Product Changes: Stress added Consider the affect on Equilibrium as a result of increasing the concentration of the product. Stress on the product Which direction does the reaction shift towards?



iii) The system self-adjust itself to reduce the stress.

iv) Until a new equilibrium is re-established.

The reaction shifts to the left. The reaction favor the reactant.


Reactant Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the reactant. What shift in direction occurs on the reaction ?

Deficit on the reactant


ii) A de-stress is place to the reactant.



Reactant Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the reactant. What shift in direction occurs on the reaction ?



ii) A de-stress is place to the reactant.

iii) The system self-adjust itself to replace the missing reactants.


The reaction shifts to the left. The reaction favors the reactant.


Product Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the Product. What shift in direction occurs on the reaction ?



ii) A de-stress is place to the product.



Product Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the Product. What shift in direction occurs on the reaction ?

Deficit on the product



iii) The system self-adjust itself to replace the displaced products.


The reaction shifts to the right. The reaction favors to the product.


LeChatelier’s Principle (1): Concentration Effect

If a chemical system is at equilibrium and then a substance is added (either a reactant or product), the reaction will shift so as to re-establish equilibrium by subtracting part of the added substance. Conversely, removal of a substance will result in the reaction moving in the direction that forms more of the substance.

Consider the Haber reaction at equilibrium: N2 + 3H2 → 2NH3

If some H2 is added to the reaction which was at equilibrium, the system self-adjust to remove the excess H2 by converting it to NH3 until equilibrium is re-establish; in the process some N2 is also consumed.


LeChatelier’s Principle (2): Volume / Pressure Effect

Increasing the pressure (by reducing the volume) of a gaseous mixture causes the system to shift in the direction that reduces the number of moles of gas.*

Consider the reaction : N2O4 (g) D 2NO 2 (g) Suppose the volume of the container is reduced?

The equilibrium shifts to the side that reduces the total number of moles of gas involved in the reaction. In this example, 2 moles of product (NO2) will change to 1 mol of reactant (N2O4). The total moles in the reaction mixture is reduced to compensate for the increase in pressure.

Note: Adding an inert gas will not shift the equilibrium because the inert gas is not involved in the overall reaction.

+ P



Examples: Which side is sensitive to pressure? 1) 2SO2 (g) + 2H2O (g) D 2H2S (g) + 3O 2 (g)

What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? R.S.* Reactant (Left) What happens to pressure if the volume is increase? Pressure decrease Which direction the reaction shift if the volume is increase? R.S. Product (Right)

+ P



iii) The system self-adjust itself to reduce the stress.




Examples: Which side is sensitive to pressure? 1) 2SO2 (g) + 2H2O (g) D 2H2S (g) + 3O 2 (g)

What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? R.S.* Reactant (Left) What happens to pressure if the volume is increase? Pressure decrease Which direction the reaction shift if the volume is increase? R.S. Product (Right)

+ P



iii) The system self-adjust itself to replace the displaced products.




Examples: Which side is sensitive to pressure?

2) N2 (g) + O2 (g) D 2 NO (g)

What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? No shift What happens to pressure if the volume is increase? Pressure decrease Which direction the reaction shift if the volume is increase? No shift

Add stress to reaction

System has no change Create a deficit to

system at equilibrium.




3) 2CO (g) D C (s) + CO2 (g) What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? R.S. Product (Right) What happens to pressure if the volume is increase? Pressure decrease Which direction the reaction shift if the volume is increase? R.S. Reactant (Left)

P +




iii) The system self-adjust to reduce the stress.




3) 2CO (g) D C (s) + CO2 (g) What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? R.S. Product (Right) What happens to pressure if the volume is increase? Pressure decrease Which direction the reaction shift if the volume is increase? R.S. Reactant (Left)

P +


ii) A deficient is place to the reactant.

iii) The system self-adjust itself to replace the missing reactants.



The Math Behind Pressure Effects due to Volume changes

Consider: C (s) + CO2 (g)D 2CO (g)

1 mol of CO is in equilibrium with 1 mol of CO2 in 1 L container. Keq = [1M]2 / [1 M] Now suppose the volume is reduced from 1L to 0.5 L, What happens to the concentration?

[CO] = 1 mol / .5 L = 2 M, [CO2] = 1 mol / .5 L = 2 M Q (not at equilb) = [2M]2 / [2 M] = 2 2 > 1, but reaction needs to go back to equilb. Q reduces to become Keq again. Reaction shifts to the left.

Q = 2 Keq = 1


Energy of Reaction: Review

(b) Heat reactant

(+) Endothermic Reactant + E → Product

The energy of a reaction depends on the energies of the reactant relative that of the product. If the reaction is downhill (a) then energy is a product (Exothermic Reaction). If the reaction is uphill (b), then energy is a reactant (Endothermic)

(a)  Heat product

(-) Exothermic Reactant → Product + E


Summary of Thermicity

(+) Endothermic Reactant + E → Product

Energy - Heat is Absorb Heat is a Reactant ΔH is (+)

(-) Exothermic Reactant → Product + E Energy - Heat is Release Heat is a Product ΔH is (-)


Sign Convention ... monetary analogy

$ign convention and the flow of energy/heat between system and surrounding$.

System point of View Exothermic(-) ____________ Surrounding View Endothermic (+)

System point of View Endothermic (+) ___________ Surrounding View Exothermic (-)

Cash Flow Analogy

Wages (+) cash flow (pay in) Debt (-) cash flow (pay out)

System

Surrounding

Heat System

Surrounding

Heat

World

$ (+) (-)

Wages Debt $


Temperature Effect (3): Endothermic Consider the affect on Equilibrium as a result increasing the temperature of an endothermic reaction. What shift in direction occurs on the reaction ?

Recall that an endothermic reaction treats heat (energy) as a reactant. Increase in temperature, increase in stress of reactant.


ii) An increase in temperature is a stress to the reactant.

iii) The system self-adjust itself to reduce the stress..


The reaction shifts to the right.

E E E


LeChatelier’s Principle (3): Temperature Effect

Examples: Which side is sensitive to pressure? Question 1) 2SO2 (g) + 2H2O (g) D 2H2S (g) + 3O 2 (g) + E

Which direction the reaction shift if the temperature decreases? Which direction the reaction shift if the temperature is increases?

2) N2 (g) + O2 (g) D 2NO (g) , ( ΔH = + 120 kJ )


3) 2CO (g) D C (s) + CO2 (g) (Exothermic reaction )



LeChatelier’s Principle (3): Temperature Effect

Examples: Which side contains Energy? 1) 2SO2 (g) + 2H2O (g) D 2H2S (g) + 3O 2 (g) + E

Which direction the reaction shift if the temperature decreases? R.S. Product (Right) Which direction the reaction shift if the temperature is increases? R.S. Reactant (Left)

2) N2 (g) + O2 (g) D 2NO (g) , ( ΔH = + 120 kJ )

Which direction the reaction shift if the temperature decreases? R.S. Reactant (Left) Which direction the reaction shift if the temperature is increases? R.S. Product (Right)

3) 2CO (g) D C (s) + CO2 (g) (Exothermic reaction )

Which direction the reaction shift if the temperature decreases? R.S. Product (Right) Which direction the reaction shift if the temperature is increases? R.S. Reactant (Left)


LeChatelier’s Principle (3): Temperature increase An increase in the temperature of a system at equilibrium will shift the

reaction so that it will absorb the heat. The equilibrium constant changes with temperature. It will either

increase or decrease depending on the exothermicity or endothermicity of the reaction.

Endothermic: Reactant + E D products Energy = Reactant Temp increase Shift rxn to Right , (Keq increases) Temp increase Shift rxn to Right, (Keq increases)

Exothermic: Reactant D products + E Energy = Product Temp increase Shift rxn to Left, (Keq decrease) Temp increase Shift rxn to Left, (Keq decreases)

Keq(old) Keq(new)

Keq(new) Keq(old)

T increase

T increase


LeChatelier’s Principle (3): Temperature decrease An decrease in the temperature of a system at equilibrium will shift the

reaction so that it will compensate for the energy deficit. The equilibrium constant changes with temperature. It will either

decrease or increase depending on the exothermicity or endothermicity of the reaction.

Endothermic: Reactant + E D products Energy = Reactant Temp decrease Shift rxn to Left , (Keq decreases) Temp decrease Shift rxn to left, (Keq decreases)

Exothermic: Reactant D products + E Energy = Product Temp decrease Shift rxn to Right, (Keq increase) Temp decrease Shift rxn to Right, (Keq increases)

Keq(old) Keq(new)

Keq(new) Keq(old) T decrease

T decrease


Catalyst added to a system at equilibrium lowers the activation energy of a reaction and therefore accelerates the rate of the reaction in both direction. A catalyst does not change the value of the equilibrium constant.

LeChatelier’s Principle (4): Catalyst

€

Keq =kf

kr=

2kf

2kr=

100kf

100kr

A catalyst decrease the activation barrier between the reactant and product. This result in increasing the rate of the forward and the reverse reaction to the same extent. kf = kr, the Keq is unaffected:


Summary of Temperature & Pressure Effects on Solubility

ΔH (s, l or g) Temp Direction Solubility

(+) Endothermic h g Prod h increase (+) Endothermic i f React i decrease

(-) Exothermic h f React i decrease (-) Exothermic i g Prod h increase

*Pressure Direction Solubility Gas solute h g Prod h increase Gas solute i f React i decrease

* An inert gas will not change solubility.

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Activity 6: Chemical Equilibrium and Application of LeChâtelier’s Principle

Objective: The purpose of this exercise is to use the concept of LeChâtelier's Principle in order to predict the direction of a reaction

system, originally at equilibrium, due to a stress that is imposed to the reaction system.

Discussion: At the start of a reaction, only reactants are present. As the reaction proceeds, reactants are transformed into products.

Reactant concentrations decrease while product concentrations increase.

A + B g C + D (forward reaction)

When the products come into contact with each other, they too can react, in this case to reform some or the reactants.

A + B f C + D (reverse reaction)

The forward reaction and the reverse reaction occur simultaneously. As the reaction progresses forward, reactant concentrations

decrease and product concentrations increase to a point at which the rates of the forward and reverse reactions are equal.

A + B D C + D

This is referred to as equilibrium. Reactant and product continue to be in equilibrium until some kind of stress is applied to the reaction

system. When stress is applied, the equilibrium is disturbed and the system reacts to relieve this stress. Eventually the system is

restored to a new point of equilibrium. This effect of stress on a system at equilibrium is referred to as LeChâtelier's Principle.

Le Châtelier's Principle states that if a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of

one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance. Changing the

concentration or pressure of a substance in a reaction at equilibrium is creating a stress. The removal of something creates a “deficit

stress,” and adding something creates an “excess stress.” Reactions will always respond to stress by adding to the deficit (making more

of what was removed) or by reducing the excess (consuming some of what was added). This description works regardless which side of

the reaction is changing, as long as the chemicals undergoing the change are either in the aqueous or gas phase.

Le Châtelier's Principle also applies if the temperature changes for a system at equilibrium. An increase in the temperature of a system

at equilibrium will shift the reaction so that it will absorb the heat. A catalyst does not change the value of the equilibrium constant,

however, and therefore does not affect the direction of the reaction.

In this exercise, you will state what occurs when the concentrations of reactants and of products in a system at equilibrium are

changed. One possible change is to increase a concentration by adding more of one substance. If, for example, more B is added, some A

reacts to remove the excess B. In the process, the reaction shifts to the right as more C and D are produced. The products C and D

increase concentration as the concentrations of A and B decreases from its concentration after B was added. This process continues

until a new equilibrium is reached at which time the concentration remains constant for all chemicals.

Another possibility is to decrease a concentration by removing one of the chemicals involved in the equilibrium. If for example, some B

is removed, some C and D react to restore some of the B that was removed. In the process, the reaction shifts to the left as the

concentrations of A and B increase from its concentration after B was removed. At the same time the concentrations of C and D

decrease as they are used up to form A and B. This occurs until equilibrium is reestablished.

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Activity 6a: Chemical Equilibrium and Application of LeChâtelier’s Principle; Part 1 ____ / ____ Score Name (last)____________________(first)____________________

Lab Section: Day _________ Time _______

For numerical calculations, show your complete work on a separate piece of paper, write units and express answers to the correct number of significant figures, box your answer, and then write the answer on this worksheet. i. Consider the following system at equilibrium: __CH4(g) + __C(s) + __O2(g) D __H2CO (g) DH = - 135.2 Kcal Complete the following table. Indicate changes in moles and concentrations by entering I, D, N, or ? in the table.

(I = increase, D = decrease, N = no change, ? = insufficient information to determine)

Change or stress imposed on Direction of shift, left, right or no change to re-establish, equilibrium

Change in number of moles

Change in molar concentration

Chg

the system at equilibrium (After stress has been imposed) CH4 C O2 H2CO CH4 C O2 H2CO Temp

a) Remove C

b) Remove CH4

c) Add H2CO

d) Increase O2 pressure

e) Decrease volume of reaction vessel

f) Decrease temperature

g) Add catalyst

h) 2.0 mol of O2 & H2CO are

added at the same time

ii. Consider the reaction: PCl5(g) D PCl3(g) + Cl2(g)

At 250. °C, 45.0% of the original PCl5 is decomposed according to the reaction above.

a) If 1.100 mol of PCl5(g) is introduced into a 1.00 L container at 250. °C, what is the equilibrium concentrations of PCl5, PCl3 and Cl2?

__________________ PCl5 _________________PCl3 ___________________Cl2 (Answer)

b) What is the value of Keq at 250. °C ? ______________________ Keq (Answer)

iii. For the reaction: H2 (g) + l2 (g) D 2 HI (g), Keq = 0.170, at 500.0 K.

a) What concentration of l2(g) will be in equilibrium with H2 at 0.040 M and Hl at 0.150 M? ___________(Answer)

iv. The value of Kc = 2.20•10-10 for the equilibrium: COCl2 (g) D CO(g) + Cl2 (g)

If the concentration of Cl2 at equilibrium is 4.80•10-6 M,

a) What are the equilibrium concentrations of the other two substances? [CO] = ___________(Answer)

[COCl2] = ___________(Answer)

b) What is the initial concentration of COCl2? (Express in scientific notation) ___________(Answer)

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Activity 6b: Chemical Equilibrium and Application of Le Châtelier’s Principle; Part2

____ / ____ Score Name (last)____________________(first)____________________

Lab Section: Day _________ Time _______

v. The solubility of CaCO3 at 25 °C is 6.90 •10-5 M. The reaction is: CaCO3 (s) D Ca2+

(aq) + CO3

2-(aq)

a) Calculate the concentration of Ca2+ and CO32- at equilibrium.

_____________________ Ca2+ ___________________ CO3

2- (Answer)

b) Calculate the equilibrium constant for the dissolution reaction. (This equilibrium constant is also known as the solubility product.)

___________(Answer)

vi. A 0.400 M HClO solution was found to have an H+ concentration of 1.10•10-4 M. The reaction is: HClO(aq) D H+(aq) + ClO-

(aq)

a) Calculate the equilibrium constant for the ionization reaction (also known as the acid dissociation constant, Ka).

________(Answer)

b) Calculate the ratio ( D[HClO] / [HClO] initial) * 100 . This value is also known as the percent ionization (a).

___________(Answer)

vii. At 1285 °C the equilibrium constant for the reaction: Br2(g) D 2Br(g) is Kc = 1.04•10-3.

A 0.200-L vessel containing an equilibrium mixture of the gases has 0.245 g Br2(g). What is the mass of Br(g) in the vessel?

___________(Answer)

viii. For the reaction: CO2(g) + H2(g) D CO(g) + H2O(g), the equilibrium concentrations are [CO2] = 0.0190 M, [H2] = 0.0340 M, [CO] = [H2O] = 0.0220 M at a temperature of 298 K when contained in a volume of 1.00 L.

a) What is the value of Kc? ___________(Answer)

b) What is the value of Kp? ___________(Answer)

b) What are the concentrations of the four substances in mole fraction?

_______________ cCO2 _______________cH2 _______________c H2 O ________________cCO (Answer)

c) How does the value of Kc compare if the Kc (c) is determined using mole fraction instead of molarity, Kc (M) ?

(Answer in words and numbers)

Dynamic Equilibrium, Keq, and the Mass Action...

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