Equilibrium © 2009, Prentice-Hall, Inc. Chemical Equilibrium

Equilibrium

© 2009, Prentice-Hall, Inc.

Chemical Equilibrium

Equilibrium

Equilibrium


Equilibrium

Not all reactions “go to completion”


• Reactions in which limiting reagents are used up are said to “go to completion”

Equilibrium



• Reactions in which limiting reagents are used up are said to “go to completion”

• Many reactions do not “go to completion”, but approach an equilibrium state in which both reactants are forming products, and products are reforming into reactants

Equilibrium


The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

Equilibrium



• Chemical equilibrium- a dynamic state of a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction in a closed system, but no net effect is observable

• Quantites of reactants and products are measurably unchanged over time

Equilibrium

Equilibrium-Concentration versus time graph


Equilibrium


A System at Equilibrium

Once equilibrium is achieved, the amount of each reactant and product remains constant.

Equilibrium

Rate versus time graph


Equilibrium


Depicting Equilibrium

Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow.

N2O4 (g) 2 NO2 (g)

Equilibrium


The Equilibrium Constant

Equilibrium



• Forward reaction:N2O4 (g) 2 NO2 (g)

• Rate Law:Rate = kf [N2O4]

Equilibrium



• Reverse reaction:2 NO2 (g) N2O4 (g)

• Rate Law:Rate = kr [NO2]2

Equilibrium



• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Rewriting this, it becomeskf

kr [NO2]2

[N2O4]=

Equilibrium



The ratio of the rate constants is a constant at that temperature, and the expression becomes

Keq =kf

kr [NO2]2

[N2O4]=

Equilibrium



• Consider the generalized reaction

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

Equilibrium


• The products appear in the numerator and the reactants in the denominator

• Each concentration is raised to the power of its stoichiometric coefficient

• Is constant for a particular reaction at a as long as the temperature is constant


Equilibrium


• K values are written without units• Kc-

equilibrium constant for

concentration (aqueous)• Kp- Equilibrium constant for partial

pressure (gases)


Equilibrium

Not part of the equilibrium constant calculation:

• Pure solids (s)• Pure liquids (l) • Their concentration is considered to be

constant-- Both can be obtained by dividing the density of the substance by its molar mass — and both of these are constants at constant temperature.


Equilibrium

Write the equilibrium Expression for 4 NH3 (g)+ 7 O2(g)D4 NO2(g)+ 6 H2 O (g)


Equilibrium

Write the equilibrium Expression for 4 NH3 (g)+ 7 O2(g)D4 NO2(g)+ 6 H2 O (g)

• K= [NO2 ]4 [H2O]6

[NH3 ]4 [O2 ]7


Equilibrium

Write the equilibrium expression for 2KClO3 (s) ↔ 2KCl(s) + 3O2 (g)


Equilibrium

Write the equilibrium expression for 2KClO3 (s) ↔ 2KCl(s) + 3O2 (g)

• K= [O2]3


Equilibrium

Write the equilibrium expression for H2O(l) ↔ H+ (aq) + OH- (aq)


Equilibrium

Write the equilibrium expression for H2O(l) ↔ H+ (aq) + OH- (aq)

K= [H+ ] [ OH- ]


Equilibrium


The Equilibrium Constant Kp

Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

Kp =(PC

c) (PDd)

(PAa) (PB

b)

Equilibrium

Write the K and Kp of the following:

• The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas


Equilibrium


• The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas

• K= [Cl2 ]

• Kp = P Cl2


Equilibrium


• Deep blue solid copper II sulfate pentahydrate is heated to drive off water vapor to form solid copper II sulfate.


Equilibrium


• Deep blue solid copper II sulfate pentahydrate is heated to drive off water vapor to form solid copper II sulfate.

• K=[H2O]5

• Kp= PH2O 5


Equilibrium

Calculating the values of K

• The process in which ammonia is manufactured form nitrogen gas and hydrogen gas at high temperature and pressure is called the Haber process. The following equilibrium concentrations were observed for this process at 127 C

• [NH3 ]= 3.1 x 10-2 M, [N2]= 8.5 x 10-1 M

• [H2]= 3.1 x 10-3 M


Equilibrium


• [NH3 ]= 3.1 x 10-2 M, [N2]= 8.5 x 10-1 M

• [H2]= 3.1 x 10-3 M

• Calculate the value of K at 127 C for this reaction


Equilibrium


• [NH3 ]= 3.1 x 102 M, [N2]= 8.5 x 10-1 M

• [H2]= 3.1 x 10-3 M

• Calculate the value of K at 127 C for this reaction

3.8 x 104


Equilibrium


• [NH3 ]= 3.1 x 102 M, [N2]= 8.5 x 10-1 M

• [H2]= 3.1 x 10-3 M

• Calculate the value of K at 127 C for the reverse of this reaction (K’ )


Equilibrium


• [NH3 ]= 3.1 x 102 M, [N2]= 8.5 x 10-1 M

• [H2]= 3.1 x 10-3 M

• Calculate the value of K at 127 C for the reverse of this reaction (K’ )

• 2.6 x 10-5


Equilibrium


• What is the relationship between K and K’ for a reaction?


Equilibrium


• What is the relationship between K and K’ for a reaction?

•K’ = 1/K


Equilibrium

Calculating Kp

• The reaction for the formation of nitrosyl chloride is 2NO(g) + Cl2 g) D2 NOCl(g)

Was studied at 25 C. The pressures at equilibrium were found to be

PNOCl = 1.2 atm

PNO = 5.0 x 10-2 atm

P Cl2 = 3.0 x 10-1 atm

Calculate the Kp for this reaction at 25 C© 2009, Prentice-Hall, Inc.

Equilibrium

Calculating Kp

• The reaction for the formation of nitrosyl chloride is 2NO(g) + Cl2 g) D2 NOCl(g)

Was studied at 25 C. The pressures at equilibrium were found to be

PNOCl = 1.2 atm

PNO = 5.0 x 10-2 atm

P Cl2 = 3.0 x 10-1 atm

Calculate the Kp for this reaction at 25 C

1.9 x 103


Equilibrium


Relationship Between Kc and Kp

• From the Ideal Gas Law we know that

• Rearranging it, we get

PV = nRTR= .0821 L atm/mol K

P = RTnV

Equilibrium



Kc and Kp

ARE NOT INTERCHANGABLE

Equilibrium



Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes

where T is temp in kelvin , R=.0821 L atm/mol K

Kp = Kc (RT)n

n = (moles of gaseous product) - (moles of gaseous reactant)

Equilibrium



Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes

When would Kp = Kc ??

Kp = Kc (RT)n

n = (moles of gaseous product) - (moles of gaseous reactant)

Equilibrium

Calculating K from Kp

• Using the value for Kp in our previous question, calculate the value of K at 25 C for the reaction

2NO(g) + Cl2 g) D2 NOCl(g)


Equilibrium

Calculating K from Kp

• Using the value for Kp in our previous question, calculate the value of K at 25 C for the reaction

2NO(g) + Cl2 g) D2 NOCl(g)

4.6 x 104 © 2009, Prentice-Hall, Inc.

Equilibrium


What Does the Value of K Mean?

• If K>>1, the reaction is product-favored; product predominates at equilibrium.

Equilibrium


What Does the Value of K Mean?

• If K>>1, the reaction is product-favored; product predominates at equilibrium.( K greater than 10, reaction nearly goes to completion)

• If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium(K less than .1, the reaction doesn’t go very far to completion at all.)

Equilibrium

If K<<1, the reaction is

• There are substantial amounts of reactant and product present at equilibrium


Equilibrium


Manipulating Equilibrium Constants

The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

Kc = = 0.212 at 100 C[NO2]2

[N2O4]N2O4 (g) D2 NO2 (g)

Kc = = 4.72 at 100 C[N2O4][NO2]2

N2O4 (g)2 NO2 (g) D

Equilibrium


Manipulating Equilibrium ConstantsThe equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

Kc = = 0.212 at 100 C[NO2]2

[N2O4]N2O4(g) D 2 NO2(g)

Kc = = (0.212)2 at 100 C[NO2]4

[N2O4]22 N2O4(g) D 4 NO2(g)

Equilibrium


Manipulating Equilibrium Constants

The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.(multiple equilibrium rule)

Equilibrium

Multiple equilibria rule

I) 2A (aq) + B(aq)D4C(aq) K1= [C]4

[A]2 [B]

II) 4 C(aq) + E(aq)D2F(aq) KII [F]2

[C]4

[E]

III) 2A(aq) + B(aq) +E(aq) D2F(aq)

KIII= KI x KII


Equilibrium

Can you…..• Write an equilibrium constant expression?• Calculate K if given equilibrium

concentrations of reactants and products• Tell how K changes if the stoichiometric

coefficients are changed in an equation?• Tell how to find K for a summary equation?• Explain what the magnitude of K is telling you

about a reaction?• Convert Kc to Kp ?


Equilibrium


Equilibrium Calculations

Equilibrium

THE REACTION QUOTIENT

Q


Equilibrium

The Reaction Quotient

• Used when a system is not in equilibrium

• Calculated in the same way as K, however the concentrations are not necessarily at equilibrium


Equilibrium


The Reaction Quotient

• Consider the generalized reaction

• The reaction quotient Q for this reaction would be

Qc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

Equilibrium

What is Q good for???

• If Q is less than K, the reaction is not at equilibrium

• Reactants products for the reaction to be at equilibrium


Equilibrium


• If Q is greater than K, the reaction is not at equilibrium

• Products reactants for the reaction to be at equilibrium


Equilibrium


• If Q equals K, the reaction is at equilibrium!!!


Equilibrium


If Q < K,there is too much reactant, and the

equilibrium shifts to the right.

Equilibrium


If Q = K,

the system is at equilibrium.

Equilibrium


If Q > K,there is too much product, and the

equilibrium shifts to the left.

Equilibrium

Using the Reaction Quotient

• For the synthesis of ammonia at 500 C, the equilibrium constant = 6.0 x 10-2. Predict the direction in which equilibrium will shift in each of the following cases:

• [NH3]= 1.0 x 10-3 M; [N2]=1.0 x 10-5 M; [H2]= 2.0 x 10-3 M

• (toward the reactants)


Equilibrium

Some Calculations with the

Equilibrium Constant


Equilibrium


An Equilibrium Problem

A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448 C for the reaction taking place, which is

H2 (g) + I2 (s) D 2 HI (g)

Equilibrium


What Do We Know?

[H2], M [I2], M [HI], M

Initia 1.000 x 10-3 2.000 x 10-3 0

Change

Equilibrium 1.87 x 10-3

Equilibrium


[HI] Increases by 1.87 x 10-3 M

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3


Equilibrium


Stoichiometry tells us [H2] and [I2] decrease by half as much.

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3


Equilibrium


We can now calculate the equilibrium concentrations of all three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

Equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

Equilibrium


…and, therefore, the equilibrium constant.

Kc =[HI]2

[H2] [I2]

Equilibrium


…and, therefore, the equilibrium constant.

Kc =[HI]2

[H2] [I2]

= 51

=(1.87 x 10-3)2

(6.5 x 10-5)(1.065 x 10-3)

Equilibrium


The Reaction Quotient (Q)

• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

• To calculate Q, one substitutes the initial concentrations of reactants and products into the equilibrium expression.

Equilibrium


Le Châtelier’s Principle

Equilibrium



“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.”

Equilibrium



• When a stress is added, shifts in the reactant and product concentrations occur to reestablish equilibrium position

• Think about K :[product]/[reactants]

Equilibrium

Things that “Stress a System”

• Adding or removing a reactant or product

• Increasing or decreasing the pressure or volume in systems with gaseous reactants/products (sometimes)

• Increasing or decreasing the temperature (which will change K)


Equilibrium

Change in Reactant or Product Concentrations

• Adding a reactant or product shifts the equilibrium away from the increase.

• Removing a reactant or product shifts the equilibrium towards the decrease.

• To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier).


Equilibrium


• Effects of Volume and Pressure• The system shifts to remove gases and

decrease pressure.• An increase in pressure favors the

direction that has fewer moles of gas.• In a reaction with the same number of

product and reactant moles of gas, pressure has no effect.

• Consider N2O4 (g D


Equilibrium


N2O4(g) 2NO2(g)

• An increase in pressure (by decreasing the volume) favors the formation of colorless N2O4.

• The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased.

• The system moves to reduce the number moles of gas (i.e. the forward reaction is favored).

• A new equilibrium is established in which the mixture is lighter because colorless N2O4 is favored.

Equilibrium



Effect of Temperature Changes• The equilibrium constant is temperature dependent.• For an endothermic reaction, H > 0 and heat can be

considered as a reactant.• For an exothermic reaction, H < 0 and heat can be

considered as a product.• Adding heat (i.e. heating the vessel) favors away

from the increase:– if H > 0, adding heat favors the forward reaction,– if H < 0, adding heat favors the reverse reaction.

Equilibrium



Effect of Temperature Changes• The equilibrium constant is temperature dependent.• For an endothermic reaction, H > 0 and heat can be

considered as a reactant.• For an exothermic reaction, H < 0 and heat can be

considered as a product.• Adding heat (i.e. heating the vessel) favors away

from the increase:– if H > 0, adding heat favors the forward reaction,– if H < 0, adding heat favors the reverse reaction.

Equilibrium


The Haber Process

The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance.

Equilibrium


The Haber Process

If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3.

Equilibrium


The Haber Process

This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid.

Equilibrium

• Consider Co(H2O)6

2+(aq) + 4 Cl(aq

D CoCl4 (aq) + 6 H2O (l)

• for which DH > 0.– Co(H2O)6

2+ is pale pink and CoCl42- is blue.


Equilibrium


The Effect of Changes in Temperature

Co(H2O)62+

(aq) + 4 Cl(aq) CoCl4 (aq) + 6 H2O (l)

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.

Equilibrium


Equilibrium

• ex CaCO3 (s) ↔ CaO(s) + CO2 (g)

• What effect will adding CaCO3 to the system have on the equilibrium?

• What will happen to the concentration of CO and CO2?

• If the volume of the reaction vessel is decreased, what will happen to the pressure?


Equilibrium


If the pressure is increased, the equilibrium will shift (left/right)____________ . The reverse reaction is favored because(more/less) CO2 is produced. What effect will this volume change have on the equilibrium constant K?

Equilibrium


Gibbs fee energy and equilibrium

Equilibrium

Do you remember…

• Calculating Gibbs free energy enables us to determine whether a reaction is spontaneous at a particular temperature

• When DG is negative, a reaction is spontaneous

• DG= DH-TDS


Equilibrium

• K is an expression of theconcentations (or partial presures) of reactants and products at equ’m

• DG indicates whether a reaction will have a higher forward or reverse rate and “move” in the forward or reverse reaction to achieve equilibrium


Equilibrium


Equilibrium

Relationship between DG and K


Equilibrium

• The relationship between Gibbs free energy and the position of equilibrium in a reaction are brought together by the expression:

DG f = -RT lnK

DG= Gproducts- Greactants

DGf- Gibbs free energy when all reactants and products are in their stadard states

1M concentrtion 1 atm presure ( not 298 K!)


Equilibrium

K, lnK, and DG

K ln K DG Situation

Greater than 1 positive negative Products favored at eq’m(forward reaction is spontaneous)

Less than 1 negative positive Reactants favored at eq’m (reverse reaction is spontaneous)

1 0 0 Products and reactants equally favored at eqq’m


Equilibrium

Calculating Gibbs free energy from Q

• DG = DG f + RTlnQ• Can be used to calculate DG in

nonstandard conditions• At equilibrium DG=0


Equilibrium


Catalysts

Catalysts increase the rate of both the forward and reverse reactions.

Equilibrium


Catalysts

When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.

Documents

Equilibrium © 2009, Prentice-Hall, Inc. Chemical Equilibrium