Chemical Equilibrium: The Extent of Chemical Reactions

CHEMICAL EQUILIBRIUM: THE

EXTENT OF CHEMICAL REACTIONS

THE STATE OF CHEMICAL EQUILIBRIUM

Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time.

2 NO2(g)N2O4(g)

BrownColorless


2 NO2(g)N2O4(g)

THE EQUILIBRIUM CONSTANT KC

c C + d Da A + b B

For a general reversible reaction:

Kc =[A]a[B]b

[C]c[D]d

Equilibrium constant expression

Equilibrium constantReactants

Products

For the following reaction: 2 NO2(g)N2O4(g)

[N2O4]

[NO2]2

= 4.64 x 10–3 (at 25 °C)Kc =

Equilibrium equation:

Concentration is Molarity [ ]: units of mol/L

THE EQUILIBRIUM CONSTANT KC

= 4.63 x 10–3

0.0429

(0.0141)2

Experiment 5

[N2O4]

[NO2]2

Kc = = 4.64 x 10–3

0.0337

(0.0125)2

Experiment 1

THE EQUILIBRIUM CONSTANT KP

What is the relationship of mol/L to P?

P = nRT/V well isn’t n/V = M?

KP =(PA)a(PB)b

(PC)c(PD)d

Equilibrium constant expressionEquilibrium constant

Reactants

ProductsEquilibrium equation:

Pressure is like concentration of a gas: units are atm

c C + d Da A + b B

∆n

Kp = Kc(RT)∆n 0.082 06K mol

L atmR is the gas constant,

T is the absolute temperature (Kelvin).

is the number of moles of gaseous products minus the number of moles of gaseous reactants.

*If the substance is a pure solid or a liquid, put a 1 in its place.

HETEROGENEOUS EQUILIBRIA

CaO(s) + CO2(g)CaCO3(s)

LimeLimestone

(1)

(1)[CO2]= [CO2]

[CaCO3]

[CaO][CO2]=

Pure solids and pure liquids are not included.

Kc =

Kc = [CO2] Kp = PCO2

HETEROGENEOUS EQUILIBRIA

USING THE EQUILIBRIUM CONSTANT

2 H2(g) + O2(g)2 H2O(g)

Kc = 4.2 x 10–48(at 500 K)

2 H2O(g)2 H2(g) + O2(g)

Kc = 2.4 x 1047(at 500 K)

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

Kc = 57.0(at 700 K)

EXAMPLE #1

Write the equilibrium expressions for Kc for the

following:a. 2H2O(l) 2H2 (g) + O2 (g)

b.2HCl(aq) H2 (g) + Cl2 (g)

EXAMPLE PROBLEM #2

The following equilibrium concentrations were observed for the Haber process at 127oC

N2(g) + 3H2(g) 2NH3 (g)

[NH3] = 0.031M[N2] = 0.85M[H2] = 0.0031M

Calculate Kc and Kp.

EXAMPLE PROBLEM #3

The reaction below occurs at 25oC.2NO(g) + Cl2(g) 2NOCl(g)

The equilibrium partial pressures are:PNOCl = 1.2atm

PNO = 0.05atm

PCl2 = 0.30atm

Calculate Kp and Kc.

REACTION QUOTIENT

Predicting shifts in equilibrium mathematicallyQc = [C]c[D]d Qp = (PC)c(PD)d

[A]a[B]b

(PA)a(PB)b

Q<K: shifts rightQ>K: shifts leftQ=K: at equilibrium


• If Qc = Kc no net reaction occurs.

• If Qc < Kc net reaction goes from left to right (reactants to products).

• If Qc > Kc net reaction goes from right to left (products to reactants).

EXAMPLE #1

The equilibrium constant, Kp, is 2.33 for the reaction C(s) + CO2 (g) 2CO(g)

at 50oC. If 0.75g of C, 3.5atm of CO2 and 4.5 atm CO are present in a container, is the reaction at equilibrium? If not, which way will it shift?

EXAMPLE #2

The equilibrium constant, Kp, for the

reaction2H2O(l) 2H2 (g) + O2 (g)

is 5 x 10-6. If 50g of hydrogen, 25g of

oxygen and 100g of water are placed in a 5L

flask at 100oC, will the reaction be at

equilibrium? If not, which way will it shift?

EXAMPLE #3

Consider the reaction:N2(g) + 3H2(g) 2NH3 (g)

The equilibrium constant, Kc, at 25oC is 0.082. If 0.4mol of nitrogen, 0.098mol ofhydrogen and 2.32mol of ammonia arepresent in a 4L container, is the system atequilibrium? If not, which way will it shift?

DETERMINING EQUILIBRIUM

CONCENTRATIONSThe ICE method

aA + bB cC + dDI # # 0 0C -ax -bx +cx +dxE #-ax #-bx cx dx

I= initial concentrations, C= change during reaction process E = equilibrium concentrations.


At 700 K, 0.500 mol of HI is added to a 2.00 L container and allowed to come to equilibrium. Calculate the equilibrium concentrations of H2, I2, and HI. Kc is 57.0 at 700 K.

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


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

Kc =[H2][I2]

[HI]2

I 0 0 0.250

C +x +x –2x

E x x 0.250 – 2x

57.0 =x2

(0.250 – 2x)2

Substitute values into the equilibrium expression:

Set up a table:


Determine the equilibrium concentrations:

x = 0.0262x2

(0.250 – 2x)2

57.0 =

Solve for “x”:

HI: 0.250 – 2(0.0262) = 0.198 M

H2: 0.0262 M

I2: 0.0262 M

EXAMPLE PROBLEM #1

For the reactionN2O4 (g) 2NO2 (g)

the equilibrium constant, Kp, is 0.133. If2.71 atm of N2O4 are introduced into acontainer, what will the equilibrium partialpressure be for all the species in thereaction?

EXAMPLE PROBLEM #2

For the reaction:CO(g) +H2O(g) CO2(g) + H2 (g)

the equilibrium constant, Kc, is 5.10 at 700K.If 3 moles of CO and 3 moles of H2O areintroduced into a 5L flask, what will theequilibrium concentrations be for all thespecies present in the reaction?

EXAMPLE #3

For the reaction2NOCl(g) 2NO(g) + Cl2(g)

the equilibrium constant, Kc, is 1.6 x 10-5 at35oC. If 3 moles of NOCl are placed in a 6L container, what will the concentrations of all the species reach at equilibrium?

EXAMPLE #4

In a 2L flask 2 moles of HF are introduced. The reaction is allowed to come to equilibrium. At equilibrium the concentration of the hydrogen gas reaches 0.064M. Determine the equilibrium constant.

2HF (g) H2 (g) + F2 (g)

EXAMPLE #5

Initially the system contains 0.4M A and 1.2M B. After the system reaches equilibrium, 26% of the A has reacted.Determine the equilibrium constant for this reaction.

4A + 2B 3C

REVERSE ICE

EXAMPLE #1

The equilibrium constant for the reaction below at 448oC is 50.5.

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

Determine the equilibrium concentrations if 3 moles of HI are introduced into a 4 L reaction vessel.

EXAMPLE #2

For the reaction below at 500oC the equilibrium constant, Kp, is 68956.52atm2.

2NH3(g) N2(g) + 3H2 (g)

If 2.5 atm of N2 and 5.5 atm of H2 are placed in a 10 L reaction vessel, what are the equilibrium partial pressures of all the gases?

EXAMPLE #3

Fifteen grams of solid NH4HS(s) is placed in an evacuated flask at 24oC. At equilibrium the total pressure in the flask is 0.614atm. What is the equilibrium constant, Kp, for the reaction below?

NH4HS(s) NH3(g) + H2S(g)

EXAMPLE #4

For the reaction2SO2(g) + O2(g) 2SO3 (g)

The equilibrium constant, Kc, is 3 x 104. If 240 grams of SO3 are introduced into a 5L reaction vessel, what will the equilibrium concentrations be for all the species present?

EXAMPLE #5

For the reaction:2H2O(l) 2H2 (g) + O2 (g)

If 5000g of H2O are introduced into a 5L flask, the total pressure at equilibrium is 4.5atm. Calculate the Kp value for this reaction.

LECHATELIER’S PRINCIPLE

If a stress is placed on a system at

equilibrium, that system will shift to relieve the stress.

CHANGE CONCENTRATION

• Add a substance, the arrow points away from what you added.

• Remove a substance, the arrow points toward what you removed.

You cannot change the concentration of solids or liquids (adding solid or liquid has no effect on equilibrium)

THE EFFECT OF CONCENTRATION CHANGES ON AN EQUILIBRIUM MIXTURE

2 NH3(g)N2(g) + 3 H2(g)

EXAMPLE OF CONCENTRATION

CHANGES

2H2(g) + O2(g) 2H2O

(l)

a. Add H2

b.Remove O2

c. Add H2O

d.Remove H2

CHANGING TEMPERATURE

DH: positive, energy is a reactant (endothermic)DH: negative, energy is a product (exothermic)

Heat adds energyCooling removes energy

EXAMPLE OF TEMPERATURE CHANGE

C(s) + O2(g) CO2(g) DH = -394kJ/mol

a. Heatb. Cool

CHANGING PRESSURE

Changing pressure by changing volume:-increase pressure, shifts to the side with less gas particles.-decrease pressure, shifts to the side with more gas particles.

*Change pressure by adding a gas not present in the reaction, no effect.

THE EFFECT OF PRESSURE AND VOLUME CHANGES ON AN EQUILIBRIUM MIXTURE

2 NH3(g)N2(g) + 3 H2(g)

EXAMPLE FOR PRESSURE CHANGES

CaCO3 (s) CaO(s) + CO2(g)

a. Increase volumeb.Decrease volumec. Increase pressure by

adding neon.

ADD A CATALYST

A catalyst will increase the rate of both the forward and reverse reactions at the same time, thus, no effect on equilibrium.

THE EFFECT OF A CATALYST ON EQUILIBRIUM

EXAMPLE PROBLEM

Na2CO3(s) Na2O(s) + CO2 (g) DH >0

a. Add Na2Ob. Heatc. Add CO2

d. Increase volumee. Coolf. Remove Na2CO3

Chemical Equilibrium: The Extent of Chemical Reactions

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