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Chemical Equilibrium. Chapter 14 14.1-14.5. Equilibrium. Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: 1.) the rates of the forward and reverse reactions are equal and

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chemical equilibrium

Chemical Equilibrium

Chapter 14

14.1-14.5

equilibrium
Equilibrium

Equilibrium is a state in which there are no observable changes as time goes by.

Chemical equilibrium is achieved when:

1.) the rates of the forward and reverse reactions are equal and

2.) the concentrations of the reactants and products remain constant

equilibrium3
Equilibrium
  • There are two types of equilibrium: Physical and Chemical.
    • Physical Equilibrium
      • H20 (l) ↔ H20 (g)
    • Chemical Equilibrium
      • N2O4 (g) ↔ 2NO2
chemical equilibrium6
Chemical Equilibrium

N2O4 (g) ↔ 2NO2 (g)

law of mass action
Law of Mass Action
  • Law of Mass Action- For a reversible reaction at equilibrium and constant temperature, a certain ratio of reactant and product concentrations has a constant value (K).
  • The Equilibrium Constant (K)- A number equal to the ratio of the equilibrium concentrations of products to the equilibrium concentrations of reactants each raised to the power of its stoichiometric coefficient.
law of mass action8

[C]c[D]d

K =

[A]a[B]b

aA (g) + bB (g)cC (g) + dD (g)

Law of Mass Action
  • For the general reaction:
chemical equilibrium10
Chemical Equilibrium
  • Chemical equilibrium is defined by K.
  • The magnitude of K will tell us if the equilibrium reaction favors the reactants or the products.
  • If K » 1……..favors products
  • If K « 1……..favors reactants
equilibrium constant expressions
Equilibrium Constant Expressions
  • Equilibrium constants can be expressed using Kc or Kp.
  • Kc uses the concentration of reactants and products to calculate the eq. constant.
  • Kp uses the pressure of the gaseous reactants and products to calculate the eq. constant.
equilibrium constant expressions12

Kc =

Kp =

P2

[NO2]2

NO2

[N2O4]

aA (g) + bB (g)cC (g) + dD (g)

P

N2O4

Equilibrium Constant Expressions
  • Equilibrium Constant Equations
homogeneous equilibrium

Kc =

Kp =

In most cases

Kc Kp

P2

[NO2]2

NO2

[N2O4]

P

N2O4

Homogeneous Equilibrium
  • Homogeneous Equilibrium- applies to reactions in which all reacting species are in the same phase.

N2O4 (g) ↔ 2NO2 (g)

equilibrium constant expressions18
Equilibrium Constant Expressions
  • Relationship between Kc and Kp

Kp = Kc(RT)Dn

Dn = moles of gaseous products – moles of gaseous reactants

= (c + d) – (a + b)

equilibrium constant calculations

CO (g) + Cl2(g) COCl2(g)

=

0.14

0.012 x 0.054

[COCl2]

[CO][Cl2]

Equilibrium Constant Calculations

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

= 220

Kc=

Kp = Kc(RT)Dn

Dn = 1 – 2 = -1

R = 0.0821

T = 273 + 74 = 347 K

Kp= 220 x (0.0821 x 347)-1 = 7.7

equilibrium constant calculations20

2

PNO PO

= Kp

2

PNO

2

Kp =

PO

PO

= 158 x (0.400)2/(0.270)2

2

2

2

2

PNO

PNO

2

2

Equilibrium Constant Calculations
  • The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?

= 347 atm

heterogeneous equilibrium
Heterogeneous Equilibrium
  • Heterogeneous Equilibrium- results from a reversible reaction involving reactants and products that are in different phases.
  • Can include liquids, gases and solids as either reactants or products.
  • Equilibrium expression is the same as that for a homogeneous equilibrium.
  • Omit pure liquids and solids from the equilibrium constant expressions.
heterogeneous equilibrium constant

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

Kp = PCO

2

Heterogeneous Equilibrium Constant

[CaCO3] = constant

[CaO] = constant

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

equilibrium constant calculations27

Kp = P

NH3

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

P

H2S

Equilibrium Constant Calculations

Consider the following equilibrium at 295 K:

The partial pressure of each gas is 0.265 atm. Calculate Kp and Kcfor the reaction.

= 0.265 x 0.265 = 0.0702

Kp = Kc(RT)Dn

Kc= Kp(RT)-Dn

Dn = 2 – 0 = 2

T = 295 K

Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4

multiple equilibria

A + B C + D

C + D E + F

A + B E + F

[C][D]

[E][F]

[E][F]

[C][D]

[A][B]

[A][B]

Kc =

Kc =

Kc

Kc

Kc =

x

Kc =

Kc

Kc

Multiple Equilibria
  • Multiple Equilibria- Product molecules of one equilibrium constant are involved in a second equilibrium process.

Kc

slide31

Writing Equilibrium Constant Expressions

  • The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.
  • The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.
  • The equilibrium constant is a dimensionless quantity.
  • In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.
  • If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

14.2

what does the equilibrium constant tell us
What does the Equilibrium Constant tell us?
  • We can:
    • Predict the direction in which a reaction mixture will proceed to reach equilibrium
    • Calculate the concentration of reactants and products once equilibrium has been reached
predicting the direction of a reaction

[1.98]2

[HI]02

[0.243][0.146]

[H2]0 [I2]0

Kc =

Kc =

Predicting the Direction of a Reaction
  • The Kc for hydrogen iodide in the following equation is 53.4 at 430ºC. Suppose we add 0.243 mol H2, 0.146 mol I2 and 1.98 mol HI to a 1.00L container at 430ºC. Will there be a net reaction to form more H2 and I2 or HI?

H2 (g) + I2 (g) → 2HI (g)

Kc= 111

reaction quotient
Reaction Quotient

The reaction quotient (Qc)is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.

IF

  • Qc > Kcsystem proceeds from right to left to reach equilibrium
  • Qc = Kcthe system is at equilibrium
  • Qc < Kc system proceeds from left to right to reach equilibrium
calculating equilibrium concentrations
Calculating Equilibrium Concentrations
  • If we know the equilibrium constant for a reaction and the initial concentrations, we can calculate the reactant concentrations at equilibrium.
  • ICE method

Reactants Products

Initial (M):

Change (M):

Equilibrium (M):

calculating equilibrium concentrations39

Br2 (g) 2Br (g)

Br2 (g) 2Br (g)

Calculating Equilibrium Concentrations
  • At 1280ºC the equilibrium constant (Kc) for the reaction is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Let x be the change in concentration of Br2

0.012

0.063

Initial (M)

-x

+2x

Change (M)

0.063 - x

0.012 + 2x

Equilibrium (M)

calculating equilibrium concentrations40
Calculating Equilibrium Concentrations

-b ± b2 – 4ac

x =

2a

[Br]2

[Br2]

Kc =

= 1.1 x 10-3

Kc =

(0.012 + 2x)2

0.063 - x

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x

4x2 + 0.0491x + 0.0000747 = 0

x = -0.0105

ax2 + bx + c = 0

x = -0.00178

calculating equilibrium concentrations41
Calculating Equilibrium Concentrations

Br2 (g) 2Br (g)

Initial (M)

0.063

0.012

Change (M)

-x

+2x

Equilibrium (M)

0.063 - x

0.012 + 2x

At equilibrium, [Br] = 0.012 + 2x = -0.009 M

or 0.00844 M

At equilibrium, [Br2] = 0.063 – x = 0.0648 M

calculating equilibrium concentrations42
Calculating Equilibrium Concentrations
  • Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
  • Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
  • Having solved for x, calculate the equilibrium concentrations of all species.
factors that affect chemical equilibrium
Factors that Affect Chemical Equilibrium
  • Chemical Equilibrium represents a balance between forward and reverse reactions.
  • Changes in the following will alter the direction of a reaction:
    • Concentration
    • Pressure
    • Volume
    • Temperature
le ch tlier s principle
Le Châtlier’s Principle
  • Le Châtlier’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 as the system reaches a new equilibrium position.
  • Stress???
changes in concentration
Changes in Concentration
  • Increase in concentration of reactants causes the equilibrium to shift to the ________.
  • Increase in concentration of products causes the equilibrium to shift to the ________.
changes in concentration50
Changes in Concentration

ChangeShift in Equilibrium

Increase in [Products] left

Decrease in [Products] right

Increase in [Reactants] right

Decrease in [Reactants] left

changes in concentration51
Changes in Concentration

FeSCN2+(aq) ↔ Fe3+(aq) + SCN-(aq)

a.) Solution at equilibrium

b.) Increase in SCN-(aq)

c.) Increase in Fe3+(aq)

d.) Increase in FeSCN2+(aq)

changes in volume and pressure
Changes in Volume and Pressure
  • Changes in pressure primarily only concern gases.
  • Concentration of gases are greatly affected by pressure changes and volume changes according to the ideal gas law.

PV = nRT

P = (n/V)RT

changes in pressure and volume
Changes in Pressure and Volume

ChangeShift in Equilibrium

Increase in Pressure Side with fewest moles

Decrease in Pressure Side with most moles

Increase in Volume Side with most moles

Decrease in Volume Side with fewest moles

changes in temperature
Changes in Temperature
  • Equilibrium position vs. Equilibrium constant
  • A temperature increase favors an endothermic reaction and a temperature decrease favors and exothermic reaction.

ChangeEndo. RxExo. Rx

Increase T K decreases K increases

Decrease T K increases K decreases

changes in temperature59
Changes in Temperature

Consider: N2O4(g) ↔ 2NO2(g)

The forward reaction absorbs heat; endothermic

heat + N2O4(g)↔ 2NO2(g)

So the reverse reaction releases heat; exothermic

2NO2(g) ↔N2O4(g) + heat

Changes in temperature??

effect of a catalyst
Effect of a catalyst
  • How would the presence of a catalyst affect the equilibrium position of a reaction?