Chapter 14 chemical equilibrium
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Chemistry II. Chapter 14 Chemical Equilibrium. Speed of a chemical reaction is determined by kinetics How fast a reaction goes Extent of a chemical reaction is determined by thermodynamics How far a reaction goes. Reaction Dynamics. forward reaction: reactants  products

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Chapter 14 Chemical Equilibrium

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Chapter 14 chemical equilibrium

Chemistry II

Chapter 14Chemical Equilibrium


Chapter 14 chemical equilibrium

Speed of a chemical reaction is determined by kinetics

  • How fast a reaction goes

    Extent of a chemical reaction is determined by thermodynamics

  • How far a reaction goes


Reaction dynamics

Reaction Dynamics

  • forward reaction:

    reactants  products

    • therefore the [reactant] decreases and the [product] increases

    • as [reactant] decreases, the forward reaction rate decreases

  • reverse reaction:

    products  reactants

    • assuming the products are not allowed to escape

    • as [product] increases, the reverse reaction rate increases

  • processes that proceed in both the forward and reverse direction are said to be reversible

    • reactants ⇌ products


Reaction dynamics1

Rate Forward

Rate Reverse

Reaction Dynamics

Initially, only the forward

reaction takes place.

As the forward reaction proceeds

it makes products and uses reactants.

Because the reactant concentration

decreases, the forward reaction slows.

As the products accumulate, the

reverse reaction speeds up.

Once equilibrium is established,

the forward and reverse reactions

proceed at the same rate, so the

concentrations of all materials

stay constant.

Eventually, the reaction proceeds

in the reverse direction as fast as

it proceeds in the forward direction.

At this time equilibrium is established.

Rate

Time


H 2 g i 2 g 2 hi g

Concentration 

Equilibrium Established

Time 

H2(g) + I2(g) ⇌ 2 HI(g)

Since the reactant concentrations are decreasing, the forward reaction rate slows down

Since the [HI] at equilibrium is larger than the [H2] or [I2], we say the position of equilibrium favors products

As the reaction proceeds, the [H2] and [I2] decrease and the [HI] increases

At equilibrium, the forward reaction rate is the same as the reverse reaction rate

Once equilibrium is established, the concentrations no longer change

And since the product concentration is increasing, the reverse reaction rate speeds up


Dynamic equilibrium

Dynamic Equilibrium

  • some reactions reach equilibrium only after almost all the reactant molecules are consumed –

    equilibrium favors the products

  • other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed –

    equilibrium favors the reactants


An analogy pwad left pwad right

An Analogy: Pwad(left) ⇌ Pwad(right)

Rules:

  • Each student wads up two paper wads.

  • You must start and stop as the timekeeper says.

  • Throw only one paper wad at a time.

  • If a paper wad lands next to you, you must throw it back.


Equal number of students on each side of the classroom

Equal Number of Students on Each Side of the Classroom

Pwad (left) Pwad (right)


Most students on the left side 2 students on the right side

Most Students on the Left Side– 2 Students on the Right Side

Pwad (left) ⇌ Pwad (right)


Most students on the left side 2 students on the right side1

Most Students on the Left Side– 2 Students on the Right Side

Pwad (left) ⇌ Pwad (right)


Common misconceptions

Common Misconceptions

  • Equilibrium means equal amounts of reactant and product.

    • No - A reaction can be at equilibrium and have more paper wads on the product side of the room

  • A reaction at equilibrium has stopped.

    • No - The paper wads keep flying in both directions even after equilibrium is achieved

  • Equilibrium can only be achieved by starting with reactants.

    • No - Equilibrium can also be achieved when starting with all of the paper wads on the product side of the room.


The equilibrium constant

The Equilibrium Constant

  • [reactants] and [products] are not equal at equilibrium, there is a relationship between them

    Law of Mass Action:

    aA + bB ⇌cC + dD,

    Keq (or Kc) is called the equilibrium constant

    • Units vary from reaction to reaction, unitless in this case


The equilibrium constant1

The Equilibrium Constant

so for the reaction:

2 N2O5⇌ 4 NO2 + O2

Units: mol3/L3


The equilibrium constant significance of k eq

The Equilibrium ConstantSignificance of Keq

  • Keq >> 1, when the reaction reaches equilibrium there will be more product than reactant molecules

    • the position of equilibrium favors products

  • Keq << 1, when the reaction reaches equilibrium there will be more reactant molecules than product molecules

    • the position of equilibrium favors reactants


A large equilibrium constant

A Large Equilibrium Constant

Remember: does not tell us about how fast, only how far


A small equilibrium constant

A Small Equilibrium Constant


Relationships between k and chemical equations

Relationships between Kand Chemical Equations

  • reaction is written backwards, the equilibrium constant is inverted

for the reaction aA + bB ⇌cC + dD

the equilibrium constant expression is:

for the reaction cC + dD ⇌aA + bB

the equilibrium constant expression is:


Relationships between k and chemical equations1

Relationships between Kand Chemical Equations

  • coefficients of an equation are multiplied by a factor, K is raised to that factor

for the reaction aA + bB ⇌cC

equilibrium constant expression is:

for the reaction 2aA + 2bB ⇌ 2cC

equilibrium constant expression is:


Relationships between k and chemical equations2

Relationships between Kand Chemical Equations

  • when you add equations to get a new equation, Keq for the new equation is the product of the equilibrium constants of the old equations

for the reactions (1) aA ⇌ bB and (2) bB ⇌cC the K expressions are:

for the overall reaction aA ⇌cC

the K expression is:


Ex 14 2 compute the equilibrium constant at 25 c for the reaction nh 3 g 0 5 n 2 g 1 5 h 2 g

K

K’

Ex 14.2 – Compute the equilibrium constant at 25°C for the reaction NH3(g) ⇌0.5 N2(g) + 1.5 H2(g)

Given:

Find:

for N2(g) + 3 H2(g) ⇌ 2 NH3(g), Keq = 3.7 x 108 at 25°C

Keq for NH3(g) ⇌ 0.5N2(g) + 1.5H2(g), at 25°C

Concept Plan:

Relationships:

Kbackward = 1/Kforward, Knew = Koldn

Solution:

N2(g) + 3 H2(g)  2 NH3(g)K1 = 3.7 x 108

2 NH3(g)⇌ N2(g) + 3 H2(g)

NH3(g)⇌ 0.5 N2(g) + 1.5 H2(g)


Equilibrium constant in terms of pressure reactions involving gases

Equilibrium Constant in Terms of PressureReactions Involving Gases

  • the concentration of a gas in a mixture is proportional to its partial pressure

    aA(g) + bB(g) ⇌cC(g) + dD(g)

or


K c and k p

Kc and Kp

  • when calculating Kp, partial pressures are always in atm

  • the values of Kp and Kc are not necessarily the same

    • because of the difference in units

  • the relationship between them is:

Δn = c + d – (a + b) or no. moles of reactants minus no. Moles products

When n = 0, Kp = Kc


Ex 14 3 find k c for the reaction 2 no g o 2 g 2 no 2 g given k p 2 2 x 10 12 @ 25 c

Kp

Kc

Ex 14.3 – Find Kc for the reaction 2 NO(g) + O2(g) ⇌ 2 NO2(g), given Kp = 2.2 x 1012 @ 25°C

Given:

Find:

Kp = 2.2 x 1012 atm-1

Kc

Concept Plan:

Relationships:

Solution:

2 NO(g) + O2(g) ⇌ 2 NO2(g)

Dn = 2  3 = -1

Check:

K has units L mol-1

since there are more moles of reactant than product, Kc should be larger than Kp, and it is


Heterogeneous equilibria reactions involving solids and liquids

Heterogeneous Equilibria: Reactions Involving Solids and Liquids

aA(s) + bB(aq) ⇌cC(l) + dD(aq)

  • Concentrations of solids and liquids doesn’t change

    • its amount can change, but the amount of it in solution doesn’t because it isn’t in solution

  • solids and liquids are not included in the Keq expression


Heterogeneous equilibria

Heterogeneous Equilibria

The amount of C is different, amounts of CO and CO2 remain the same. C has no effect on the position of equilibrium.


Calculating equilibrium constants from measured equilibrium concentrations

Calculating Equilibrium Constants from Measured Equilibrium Concentrations

  • To find K measure the amounts of reactants and products in a mixture at equilibrium

  • may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same

    • as long as the temperature is kept constant


Initial and equilibrium concentrations for h 2 g i 2 g 2hi g @ 445 c

Initial and Equilibrium Concentrations forH2(g) + I2(g) ⇌ 2HI(g) @ 445°C


Calculating equilibrium constants from measured equilibrium concentrations1

Calculating Equilibrium Constants from Measured Equilibrium Concentrations

  • Stoichiometry can be used to determine the equilibrium [reactants] and [products] if you know initial concentrations and one equilibrium concentration

    e.g.2A(aq) + B(aq) ⇌ 4C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M.

-½(0.50)

-¼(0.50)

+0.50

0.88

0.75


Calculating equilibrium concentrations

Calculating Equilibrium Concentrations

e.g.2A(aq) + B(aq) ⇌ 4C(aq)

-½(0.50)

-¼(0.50)

+0.50

0.88

0.75

Referred to as ICE table (initial, change, equilibrium)

Keq = [C]4/[A]2[B] = 0.13


Chapter 14 chemical equilibrium

Ex 14.6  Find the value of Kc for the reaction2 CH4(g) ⇌ C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M

+0.035


Chapter 14 chemical equilibrium

Ex 14.6  Find the value of Kc for the reaction2 CH4(g) ⇌ C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M

+0.035

-2(0.035)

+3(0.035)

0.105

0.045


The reaction quotient

The Reaction Quotient

for the gas phase reaction

aA + bB ⇌cC + dD

the reaction quotient is:

  • reaction mixture not at equilibrium; how can we determine which direction it will proceed?

  • the answer is to compare the current concentration ratios to the equilibrium constant

    reaction quotient, Q


The reaction quotient predicting the direction of change

The Reaction Quotient:Predicting the Direction of Change

  • if Q > K, the reaction will proceed fastest in the reverse direction

    • Q must decrease, the [products] will decrease and [reactants] will increase

  • if Q < K, the reaction will proceed fastest in the forward direction

    • Q must increase, the [products] will increase and [reactants] will decrease

  • if Q = K, the reaction is at equilibrium

    • the [products] and [reactants] will not change

  • if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction

  • if a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction


Q k and the direction of reaction

Q, K, and the Direction of Reaction


Chapter 14 chemical equilibrium

PI2, PCl2, PICl

Q

Ex 14.7 – For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm & PICl = 0.355 atm

Given:

Find:

for I2(g) + Cl2(g) ⇌ 2 ICl(g), Kp = 81.9

direction reaction will proceed

Concept Plan:

Relationships:

If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse

Solution:

I2(g) + Cl2(g) ⇌ 2 ICl(g)Kp = 81.9

since Q (10.8) < K (81.9), the reaction will proceed to the right


Chapter 14 chemical equilibrium

K, [COF2], [CF4]

[CO2]

Ex 14.8 If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and Kc = 2.00 @ 1000°C, find the [CO2]eq for the reaction given.

Given:

Find:

Sort: You’re given the reaction and Kc. You’re also given the [X]eq of all but one of the chemicals

2 COF2 ⇌ CO2 + CF4

[COF2]eq = 0.255 M, [CF4]eq = 0.118 M

[CO2]eq

Concept Plan:

Relationships:

Strategize: You can calculate the missing concentration by using the equilibrium constant expression

Solution:

Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts

Check:

Check: Round to 1 sig fig and substitute back in

Units & Magnitude OK


Chapter 14 chemical equilibrium

Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures

  • first decide which direction the reaction will proceed

    • compare Q to K

  • define the changes of all materials in terms of x

    • use the coefficient from the chemical equation for the coefficient of x

    • the x change is + for materials on the side the reaction is proceeding toward

    • the x change is  for materials on the side the reaction is proceeding away from

  • solve for x

    • for 2nd order equations, take square roots of both sides or use the quadratic formula

    • may be able to simplify and approximate answer for very large or small equilibrium constants


Chapter 14 chemical equilibrium

Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations

since Qp(1) < Kp(81.9), the reaction is proceeding forward


Chapter 14 chemical equilibrium

Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations

x

x

+2x

0.100+2x

0.100x

0.100x


Chapter 14 chemical equilibrium

Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations

x

x

+2x

0.100+2x

0.100x

0.100x


Chapter 14 chemical equilibrium

Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations

x

x

+2x

-0.0729

-0.0729

2(-0.0729)

0.100+2x

0.100x

0.100x

0.027

0.027

0.246


Disturbing and re establishing equilibrium

Disturbing and Re-establishingEquilibrium

  • once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same

  • however if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established

  • the new concentrations will be different, but the equilibrium constant will be the same

    • unless you change the temperature


Le ch telier s principle

Le Châtelier’s Principle

  • Le Châtelier's Principle:

    if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance

    • disturbances all involve making the system open

    • Concentration, temperature, volume, pressure


An analogy population changes

The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established, however the new populations will have different numbers of people than the old ones.

When an influx of population enters Country B

from somewhere outside Country A, it disturbs the

equilibrium established between Country A and Country B.

When the populations of Country A and Country B

are in equilibrium, the emigration rates between the

two states are equal so the populations stay constant.

An Analogy: Population Changes


The effect of concentration changes on equilibrium

The Effect of Concentration Changes on Equilibrium

  • Adding reactant :

    aA + bB ⇌ cC + dD

    System shifts in a direction to minimize the disturbance

  • increases the amount of products until a new equilibrium is found

    • that has the same K


The effect of concentration changes on equilibrium1

The Effect of Concentration Changes on Equilibrium

  • Removing product:

    aA + bB ⇌ cC + dD

    System shifts in a direction to minimize the disturbance

  • will increase the amounts of products and decrease the amounts of reactants

    • you can use this to drive a reaction to completion!


  • The effect of adding a gas to a gas phase reaction at equilibrium

    The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium

    • adding a gaseous reactant increases its partial pressure, causing the equilibrium to the right

      • increasing its partial pressure increases its concentration

      • does not increase the partial pressure of the other gases in the mixture

    • adding an inert gas to the mixture has no effect on the position of equilibrium

      • does not effect the partial pressures of the gases in the reaction

    Tro, Chemistry: A Molecular Approach


    The effect of concentration changes on equilibrium2

    The Effect of Concentration Changes on Equilibrium

    Describe the effect of adding more NO2(g) to the following reaction:

    N2O4(g) ⇌ NO2(g)


    The effect of concentration changes on equilibrium3

    The Effect of Concentration Changes on Equilibrium

    Q = K

    Q = K

    Q > K

    When NO2 is added, some of it combines to make more N2O4


    The effect of concentration changes on equilibrium4

    The Effect of Concentration Changes on Equilibrium

    Describe the effect of adding more N2O4(g) to the following reaction:

    N2O4(g) ⇌ NO2(g)


    The effect of concentration changes on equilibrium5

    The Effect of Concentration Changes on Equilibrium

    N2O4(g) ⇌ NO2(g)

    When N2O4 is added, reaction shifts to make more NO2


    Summarizing

    Summarizing

    If a chemical system is in equilibrium:

    • Inc. [reactant] (Q < K) reaction shifts to right

    • Inc. [product] (Q > K) reaction shifts to left

    • Dec. [reactant] (Q > K) reaction shifts to left

    • Dec. [product] (Q < K) reaction shifts to right


    Effect of volume change on equilibrium

    Effect of Volume Changeon Equilibrium

    • decreasing the size of the container increases the concentration of all the gases in the container

      • increases their partial pressures

    • if their partial pressures increase, then the total pressure in the container will increase

    • according to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure

    • the way the system reduces the pressure is to reduce the number of gas molecules in the container

    • when the volume decreases, the equilibrium shifts to the side with fewer gas molecules


    Chapter 14 chemical equilibrium

    When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side.

    Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products.

    The Effect of Volume Changes on Equilibrium


    Disturbing equilibrium reducing the volume

    Disturbing Equilibrium: Reducing the Volume

    • for solids, liquids, or solutions, changing the size of the container has no effect on the concentration, therefore no effect on the position of equilibrium

    • decreasing the container volume will increase the total pressure

      • Boyle’s Law

      • if the total pressure increases, the partial pressures of all the gases will increase

        • Dalton’s Law of Partial Pressures

    • decreasing the container volume increases the concentration of all gases

      • same number of moles, but different number of liters, resulting in a different molarity

    • since the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules

      • shift toward the side with fewer gas molecules

    • at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same


    Summarizing1

    Summarizing

    If a chemical system is in equilibrium:

    • Dec. volume causes reaction to shift in direction that has fewer moles of gas particles

    • Inc. volume causes reaction to shift in direction that has the greater number of moles of gas particles

    • If equal moles of gas on both sides, no effect

    • Adding an inert gas has no effect


    The effect of temperature changes on equilibrium position

    The Effect of Temperature Changes on Equilibrium Position

    • exothermic reactions release energy and endothermic reactions absorb energy

    • if we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes

      • even though heat is not matter and not written in a proper equation


    The effect of temperature changes on equilibrium for exothermic reactions

    The Effect of Temperature Changes on Equilibrium for Exothermic Reactions

    • for an exothermic reaction, heat is a product

    • increasing the temperature is like adding heat

    • according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat

      adding heat to an exothermic reaction will decrease the [products] and increase the [reactants]

      adding heat to an exothermic reaction will decrease the value of K

    • how will decreasing the temperature affect the system?

    aA + bB ⇌cC + dD + Heat


    The effect of temperature changes on equilibrium for endothermic reactions

    The Effect of Temperature Changes on Equilibrium for Endothermic Reactions

    • for an endothermic reaction, heat is a reactant

    • increasing the temperature is like adding heat

    • according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat

      adding heat to an endothermic reaction will decrease the [reactants] and increase the [products]

      adding heat to an endothermic reaction will increase the value of K

    • how will decreasing the temperature affect the system?

    Heat+ aA + bB ⇌cC + dD


    The effect of temperature changes on equilibrium

    The Effect of Temperature Changes on Equilibrium


    Not changing the position of equilibrium catalysts

    Not Changing the Position of Equilibrium - Catalysts

    • catalysts provide an alternative, more efficient mechanism

    • works for both forward and reverse reactions

    • affects the rate of the forward and reverse reactions by the same factor

    • therefore catalysts do not affect the position of equilibrium


    Practice le ch telier s principle

    Practice - Le Châtelier’s Principle

    2 SO2(g) + O2(g) ⇌ 2 SO3(g) DH° = -198 kJ

    How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established?

    • adding more O2 to the container

    • condensing and removing SO3

    • compressing the gases

    • cooling the container

    • doubling the volume of the container

    • warming the mixture

    • adding the inert gas helium to the container

    • adding a catalyst to the mixture


    Practice le ch telier s principle1

    Practice - Le Châtelier’s Principle

    2 SO2(g) + O2(g) ⇌ 2 SO3(g)

    • adding more O2 to the container

    • condensing and removing SO3

    • compressing the gases

    • cooling the container

    • doubling the volume of the container

    • warming the mixture

    • adding helium to the container

    • adding a catalyst to the mixture

    shift to SO3

    shift to SO3

    shift to SO3

    shift to SO3

    shift to SO2

    shift to SO2

    no effect

    no effect


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