Chemical Equilibrium. Reaction Arrow Conventions 2 A single arrow indicates all the reactant...

Chemical Equilibrium

Reaction Arrow Conventions

2

• A single arrow indicates all the reactant molecules are converted to product molecules at the end

• A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products– in these notes

Reaction Dynamics• When a reaction starts, the reactants are consumed and

products are made– forward reaction = reactants products– [reactant] decrease and [product] increase– as [reactant] decreases, the forward reaction rate decreases

• Eventually, the products can react to re-form some of the reactants, the reverse reaction– reverse reaction = products reactants– assuming the products are not allowed to escape– as product concentration increases, the reverse reaction rate

increases• Processes that proceed in both the forward and reverse

direction are said to be reversible– reactants Û products

3

Hypothetical Reaction2 Red Û Blue

The reaction slows over time,

but the Red molecules never run out!

At some time between 90 and 100 sec,the concentrations of both the Red andthe Blue molecules no longer change –equilibrium has been established.

Equilibrium does not meanthat the concentrations are equal!

Once equilibrium is established, the rateof Red molecules turning into Blue is thesame as the rate of Blue molecules turning into Red

4

Hypothetical Reaction2 Red Û Blue

5

Eventually, the reaction proceedsin the reverse direction as fast asit proceeds in the forward direction.At this time equilibrium is established.

Reaction Dynamics

Time

Rate

Rate Forward

Rate Reverse

Initially, only the forwardreaction takes place

As the forward reaction proceedsit makes products and uses reactants

Because the reactant concentration decreases, the forward reaction slows.As the products accumulate, thereverse reaction speeds up.

Once equilibrium is established,the forward and reverse reactions proceed at the same rate, so theconcentrations of all materialsstay constant

6

Dynamic Equilibrium• Rate of forward reaction = Rate of reverse

reaction • At equilibrium, the concentrations of all the

chemicals remain constant– because the chemicals are being consumed and

made at the same rate

7

H2(g) + I2(g) Û 2 HI(g)At t0: reactants only only forward reaction takes place

[H2] = 8, [I2] = 8, [HI] = 0

[H2] = 6, [I2] = 6, [HI] = 4

8

At t16: both reactants & products in the mixture forward reaction & reverse reaction

H2(g) + I2(g) Û 2 HI(g)

At t32: [products] > [reactants] forward reaction rate slows, reverse reaction rate accelerates

[H2] = 4, [I2] = 4, [HI] = 8

At t48: [products] = [reactants] forward rxn rate = reverse rxn rate

Equilibrium is reached.

9

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

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

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

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

H2(g) + I2(g) Û 2 HI(g)

Equilibrium established

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

Once equilibrium is established, the concentrations no longer change

10

Equilibrium Equal

• The rates of the forward and reverse reactions are equal at equilibrium

• [reactants] [products]

• When “Equilibrium favors products”,[products]>[reactants]

• When “Equilibrium favors reactants”,[reactants]>[products]

11

Equilibrium Constant• Law of Mass Action

– The relationship between [reactants] and [products]– Must know the balanced chemical equation

aA + bB Û cC + dD

– Lowercase letters = coefficients– In determining, relationship, Productscoefficients/Reactantscoefficients

• K = equilibrium constant, a unit-less number

12

Writing Equilibrium Constant Expressions

• For the reaction aA(aq) + bB(aq) Û cC(aq) + dD(aq)

the equilibrium constant expression

•So for the reaction 2 N2O5(g) Û 4 NO2(g) + O2(g)

the equilibrium constant expression

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Implications of Keq

When Keq >> 1[product molecules] > [reactant molecules]equilibrium favors products

When Keq << 1[reactant molecules] > [product molecules]equilibrium favors reactants

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Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following

2 SO2(g) + O2(g) Û 2 SO3(g) K = 8 x 1025

N2(g) + 2 O2(g) Û 2 NO2(g) K = 3 x 10−17

favors products

favors reactants

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Relationships between K and Chemical Equations

• When the reaction is written backward, the equilibrium constant is inverted

For the reaction aA + bB Û cC + dDthe equilibrium constant expression is

For the reaction cC + dD Û aA + bB

the equilibrium constant expression is

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• When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor

For the reaction aA + bB Û cC the equilibrium constant expression is

For the reaction 2aA + 2bB Û 2cC


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• When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations

For the reactions (1) aA Û bB & (2) bB Û cC the equilibrium constant expressions are

For the reaction aA Û cC


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2 B Û A K3 = 1/Krxn 1 = 6.25 x 10−5

Z Û 2 B K4 = 1/Krxn 2 = 2Z Û A K3K4 = 1.25 x 10−4

3 Z Û 3 A (K3K4)3 = 2.0 x 10−12

Practice – When the reaction A(aq) Û 2 B(aq) reaches equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.

When the reaction 2 B(aq) Û Z(aq) reaches equilibrium [B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.

Calculate the equilibrium constant for each of these reactions and the equilibrium constant for the reaction 3

Z(aq) Û 3 A(aq)

Krxn 1 = 1.6 x 104

Krxn 2 = 5.0 x 10−1

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K for Reactions Involving Gases• In a mixture, [gas] α to its Pp (partial pressure)• Therefore, K can be expressed as the ratio of

the Pp of the gases• For aA(g) + bB(g) Û cC(g) + dD(g) the

equilibrium constant expressions are

or

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Kc and Kp

• In calculating Kp, the units of Pp = atm (always)• Therefore, values of Kp ≠ Kc , necessarily

– difference in units– Kp = Kc when Dn = 0

• The relationship between them is

Dn is the difference between the number of moles of reactants and moles of products

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Deriving the Relationshipbetween Kp and Kc

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Deriving the RelationshipBetween Kp and Kc

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

substituting

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Calculate the value of Kp or Kc for each of the following at 27 °C:

2 SO2(g) + O2(g) Û 2 SO3(g) Kc = 8 x 1025

N2(g) + 2 O2(g) Û 2 NO2(g) Kp = 3 x 10−17

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Heterogeneous Equilibria• [Pure solids] and [pure liquids] do not change during

the course of a reaction– Because it is not in solution, the amount changes, but the

amount of it in solution doesn’t – Because their concentration doesn’t change, solids and

liquids are not included in the equilibrium constant expression

• For the reaction aA(s) + bB(aq) Û cC(l) + dD(aq) the equilibrium constant expression is

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Heterogeneous Equilibria

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

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HNO2(aq) + H2O(l) Û H3O+(aq) + NO2

−(aq) K = 4.6 x 10−4

Ca(NO3)2(aq) + H2SO4(aq) Û CaSO4(s) + 2 HNO3(aq) K = 1 x 104

Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following

27Tro: Chemistry: A Molecular Approach, 2/e

Calculating Equilibrium Constants from Measured Equilibrium Concentrations

• Most direct way of finding K:– measure the amounts of reactants and products in a mixture

at equilibrium– 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• the value of the equilibrium constant is independent of

the initial amounts of reactants and products

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

Con

Initialcentra tion

EqCon

uilibricentr

umation

EquilibriumConstant

[H2] [I2] [HI] [H2] [I2] [HI]

0.50 0.50 0.0 0.11 0.11 0.78

0.0 0.0 0.50 0.055 0.055 0.39

0.50 0.50 0.50 0.165 0.165 1.17

1.0 0.5 0.0 0.53 0.033 0.93429

Tro: Chemistry: A Molecular Approach, 2/e

Calculating Equilibrium Concentrations

• To use stoichiometry to determine equilibrium concentrations of reactants and products – You must know initial concentrations and one

equilibrium concentration• Use the change in the concentration of the

material that you know to determine the change in the other chemicals in the reaction

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Because Product C increased by 0.50 mol/L, you must have used ½ that amount of Reactant A, according to the stoichiometry.

Calculating Equilibrium Concentrations:An Example

• Suppose you have a reaction 2 A(aq) + B(aq) Û 4 C(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.

+0.50-¼(0.50) -½(0.50)

0.75 0.8831

Because Product C increased by 0.50 mol/L, you must have used ¼ that amount of Reactant B, according to the stoichiometry.

2 A(aq) + B(aq) Û 4 C(aq) Product C increased by 0.50 mol/L

Practice –The following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and Kc.

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Practice – A second experiment was done and the following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine

values of Kp and Kc. Compare them to the first experiment.

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The Reaction Quotient• In a mixture containing both

reactants and products not at equilibrium how can we determine in which direction it will proceed?– compare the current

concentration ratios to the equilibrium constant

– The reaction quotient, Q is the concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients)

for the gas phase reaction aA + bB Û cC + dDthe reaction quotient is:

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The Reaction Quotient:Predicting the Direction of Change

• If Q > K, the reaction will proceed fastest in the reverse direction: [products] and [reactants]

• If Q < K, the reaction will proceed fastest in the forward direction: [products] and [reactants]

• If Q = K, the reaction is at equilibrium there will be no Δ in [products] and [reactants]

• If Q = 0, the reaction mixture contains only reactants, the reaction will go forward

• If Q = ∞, a reaction mixture contains only products, the reaction will reverse

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Q, K, and the Direction of Reaction

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Practice – For the reaction CH4(g) + 2 H2S(g) Û CS2(g) + 4 H2(g) Kc = 3.59 at 900 °C. For each of the following measured concentrations determine whether the

reaction is at equilibrium. If not at equilibrium, in which direction will the reaction proceed to get to equilibrium?

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Practice – A sample of PCl5(g) is placed in a 0.500 L container and heated to 160 °C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635

PCl5 Û PCl3 + Cl2

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

Step 1: decide which direction the reaction will proceed– compare Q to K

Step 2: define the changes of all materials in terms of x– use the coefficient from the chemical equation as 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 fromStep 3: 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

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Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?

(Hint: you will need to use the quadratic formula to solve for x)

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Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the

equilibrium concentrations of [I2] and [I]?

because [I]initial = 0, Q = 0 and the reaction must proceed forward

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Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the


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Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the


x = 0.00216

0.500 0.00216 = 0.498[I2] = 0.498 M

2(0.00216) = 0.00432[I] = 0.00432 M

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Approximations to Simplify the Math

• When the K is very small, the position of equilibrium favors the reactants

• For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium– the [X]equilibrium = ([X]initial ax) [X]initial

• we are approximating the equilibrium concentration of reactant to be the same as the initial concentration

– assuming the reaction is proceeding forward

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Checking the Approximation and Refining as Necessary

• Check approximation by comparing the approximate value of x to the initial concentration

• If the approximate value of x is less than 5% of the initial concentration, the approximation is valid

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Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?

(use the simplifying assumption to solve for x)

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because [I]initial = 0, Q = 0 and the reaction must proceed forward

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the approximation is valid!!

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x = 0.00217

0.500 0.00217 = 0.498[I2] = 0.498 M

2(0.00217) = 0.00434[I] = 0.00434 M

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Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is 0.0125 M, what will be the equilibrium

concentrations of [N2O4] and [NO2]?

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because [NO2]initial = 0, Q = 0 and the reaction must proceed forward

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without approximation with approximation

the approximation is valid!!

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x = 1.82 x 10−4



x = 1.83 x 10−4

[N2O4]= 0.0125 1.83 x 10−4

[N2O4] = 0.0123 M

[NO2]=2(1.83 x 10−4)[NO2] = 3.66 x 10−4 M

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Disturbing and RestoringEquilibrium

• Once a reaction is at equilibrium, the [reactants] and [products] remain the same

• However, if conditions change, the concentrations of all the chemicals will change until equilibrium is restored

• The new concentrations will be different, but the equilibrium constant will be the same– unless you change the temperature

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Le Châtelier’s Principle

• Le Châtelier's Principle helps us in predict the effect various changes in conditions have on the position of equilibrium – if a system at equilibrium is disturbed, the position of

equilibrium will shift to minimize the disturbance – disturbances all involve making the system open

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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; the new populations will have different numbers of people than the old ones.

An Analogy: Population Changes

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When an influx of population enters Country Bfrom somewhere outside Country A, it disturbs the

equilibrium established between Country A and Country B.

When the populations of Country A and Country Bare in equilibrium, the emigration rates between thetwo countries are equal so the populations stay constant.

Disturbing Equilibrium:Adding or Removing Reactants

• After equilibrium is established, a reactant is added – as long as the added reactant is included in the equilibrium constant expression

(not a solid or liquid)…

• How will this affect the rate of the forward reaction? Initially,…– Adding a reactant increases the forward reaction rate

• How will it affect the rate of the reverse reaction?– No effect on the reverse reaction rate – The reaction proceeds to the right until equilibrium is restored

• How will it affect the value of K?– At the new equilibrium position, you will have

• more of the products than before • less of the non-added reactants than before• Slightly less of the added reactant

– At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same 57

Disturbing Equilibrium:Adding or Removing Reactants

• After equilibrium is established, a reactant is removed – as long as the added reactant is included in the equilibrium constant expression

• i.e., not a solid or liquid

• How will this affect the rate of the forward reaction? – Initial decrease in forward reaction rate, no effect on the reverse reaction rate

• How will it affect the rate of the reverse reaction? – Reverse reaction goes faster– Reaction proceeds to the left until equilibrium is restored

• How will it affect the value of K?– At the new equilibrium position, you will have

• less of the products than before• more of the non-removed reactants than before• Slightly more of the removed reactant

– At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

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The Effect of Concentration Changes on Equilibrium

• Adding a reactant will – decrease the amounts of the other reactants– increase the amount of the products – Re-establish a new position of equilibrium is found – K stays the same

• Removing a product will – increase the amounts of the other products– decrease the amounts of the reactants.– Drive the reaction to completion

• Equilibrium shifts away from side with added chemicals or toward side with removed chemicals– Remember, adding more of a solid or liquid does not change its

concentration – and therefore has no effect on the equilibrium 59


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

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When N2O4 is added, some of itdecomposes to make more NO2

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

• Adding a gaseous reactant increases its partial pressure equilibrium shifts to the right– increase in partial pressure increase in concentration– No 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 affect the partial pressures of the gases in the

reaction

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Effect of Volume Change on Equilibrium• Decrease container volume increase in all [gas]

– Increase in gas Pp

– It does not change the concentrations of solutions!!

• As Pp increase, the Ptotal increases• According to Le Châtelier’s Principle

– equilibrium should shift to remove pressure– system reduces the pressure by reducing the number of gas

molecules in the container

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

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Disturbing Equilibrium:Changing the Volume

• After equilibrium is established, the container volume is decreased

• How will it affect the concentration of solids, liquid, solutions, and gases?

• How will this affect the total pressure of solids, liquid, and gases?

• How will it affect the value of K?

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Disturbing Equilibrium: Reducing the Volume

• Decreasing the container volume increases the concentration of all gases, but not solids, liquid, or solutions– same number of moles, but different number of liters, resulting in a

different molarity• 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• Because 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

• K doesn’t change: at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that K is the same

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The Effect of Volume Changes on Equilibrium

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Because there are more gas molecules on the reactants side of the

reaction, when the pressure is increased the position of

equilibrium shifts toward the side with fewer molecules to

decrease the pressure

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

The Effect of Temperature Changes on Equilibrium Position

• Exothermic reactions release energy and endothermic reactions absorb energy

• To useLe Châtelier’s Principle to predict the effect of temperature changes, write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction

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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 concentrations of products – increase the concentrations of reactants– decrease the value of K

• How will decreasing the temperature affect the system?

aA + bB cC + dD + Heat68

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 concentrations of reactants– increase the concentrations of products– increase the value of K

• How will decreasing the temperature affect the system?

Heat + aA + bB cC + dD69

The Effect of Temperature Changes on Equilibrium

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Not Changing the Position of Equilibrium: The Effect of Catalysts

• Catalysts provide an alternative, more efficient mechanism

• Catalysts work for both forward and reverse reactions

• Catalysts affect the rate of the forward and reverse reactions by the same factor

• Therefore, catalysts do not affect the position of equilibrium

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Practice – Le Châtelier’s Principle2 SO2(g) + O2(g) Û 2 SO3(g)

DH° = −198 kJ at equilibriumHow will each of the following changes affect the

equilibrium concentrations of each gas once equilibrium is restored?

– 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 72

Practice – Le Châtelier’s Principle2 SO2(g) + O2(g) Û 2 SO3(g)

DH° = −198 kJ at equilibriumHow will each of the following changes affect the equilibrium

concentrations of each gas once equilibrium is restored?– adding more O2 to the container shift to SO3

– condensing and removing SO3 shift to SO3

– compressing the gases shift to SO3

– cooling the container shift to SO3

– doubling the volume of the container shift to SO2

– warming the mixture shift to SO2

– adding helium to the container no effect– adding a catalyst to the mixture no effect

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Chemical Equilibrium. Reaction Arrow Conventions 2 A single arrow indicates all the reactant...

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