Chemistry Notes 15

Introduction to Chemical Introduction to Chemical EquilibriumEquilibrium

Chapter 15Chapter 15

CHEM 160CHEM 160

• Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2:

N2O4(g) 2NO2(g).• At some time, the color stops changing and we have a

mixture of N2O4 and NO2.• Chemical equilibrium is the point at which the

concentrations of all species are constant.• Using the collision model:

– as the amount of NO2 builds up, there is a chance that two NO2 molecules will collide to form N2O4.

– At the beginning of the reaction, there is no NO2 so the reverse reaction (2NO2(g) N2O4(g)) does not occur.

The Concept of EquilibriumThe Concept of Equilibrium


• The point at which the rate of decomposition: N2O4(g) 2NO2(g)

equals the rate of dimerization: 2NO2(g) N2O4(g).

is dynamic equilibrium.• The equilibrium is dynamic because the reaction has

not stopped: the opposing rates are equal.• Consider frozen N2O4: only white solid is present. On

the microscopic level, only N2O4 molecules are present.



• As the substance warms it begins to decompose: N2O4(g) 2NO2(g)

• A mixture of N2O4 (initially present) and NO2 (initially formed) appears light brown.

• When enough NO2 is formed, it can react to form N2O4: 2NO2(g) N2O4(g).

• At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4:

• The double arrow implies the process is dynamic.N2O4(g) 2NO2(g)


• ConsiderForward reaction: A B Rate = kf[A]Reverse reaction: B A Rate = kr[B]

• At equilibrium kf[A] = kr[B].• For an equilibrium we write:

• As the reaction progresses– [A] decreases to a constant,– [B] increases from zero to a constant.– When [A] and [B] are constant, equilibrium is achieved.

A B


• Alternatively:– kf[A] decreases to a constant,– kr[B] increases from zero to a constant.– When kf[A] = kr[B] equilibrium is achieved.


• Consider

• If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will reach equilibrium with a constant concentration of nitrogen, hydrogen and ammonia.

• However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed and N2 and H2 will be produced until equilibrium is achieved.

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

The Equilibrium ConstantThe Equilibrium Constant


• No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium.

• For a general reaction

the equilibrium constant expression is

where Kc is the equilibrium constant.

aA + bB(g) pP + qQ

ba

qpcK

BA

QP


• Kc is based on the molarities of reactants and products at equilibrium.

• We generally omit the units of the equilibrium constant.

• Note that the equilibrium constant expression has products over reactants.


The Equilibrium Constant in Terms of PressureThe Equilibrium Constant in Terms of Pressure• If KP is the equilibrium constant for reactions

involving gases, we can write:

• KP is based on partial pressures measured in atmospheres.

• We can show thatPA = [A](RT)

ba

qpP

PP

PPK

BA

QP


The Equilibrium Constant in Terms of PressureThe Equilibrium Constant in Terms of PressurePA = [A](RT)

• This means that we can relate Kc and KP:

where n is the change in number of moles of gas.• It is important to use:

n = ngas(products) - ngas(reactants)

ncP RTKK



PhosphorousProblem 1(a)

Phosphorous pentachloride dissociates on heating:

PCl5(g) PCl3(g) + Cl2(g)

If Kc = 3.26 x 10-2 at 191C, what is Kp at this temperature?

Ans.: Kp = 1.24


PhosphorousProblem 1(b)

The value of Kc for the following reaction at 900C is 0.28.

S2(g) + 4H2(g) CH4(g) + 2H2S(g)

What is Kp at this temperature?

Ans.: Kp = 3.0 x 10-5



Consider the following reaction at 1000C:

CO(g) + 3H2(g) CH4(g) + H2O(g)

At equilibrium, the following concentrations are measured: [CO] = 0.0613 M, [H2] = 0.1839 M, [CH4] = 0.0387,[H2O] = 0.0387 M. Calculate the value of Kc for this reaction.Calculate the value of Kp. ?

Ans.: Kc = 3.93, Kp = 3.60 x 10-4



A 5.00 L vessel contained 0.0185 mol of phosphorus trichloride,0.0158 mol of phosphorus pentachloride, and 0.0870 mol chlorine at503 K in an equilibrium mixture. Calculate the value of Kc at thistemperature.

PCl3(g) + Cl2(g) PCl5(g)

Ans.: Kc = 49.1

The Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• The equilibrium constant, K, is the ratio of products

to reactants.• Therefore, the larger K the more products are present

at equilibrium.• Conversely, the smaller K the more reactants are

present at equilibrium.• If K >> 1, then products dominate at equilibrium and

equilibrium lies to the right.• If K << 1, then reactants dominate at equilibrium and

the equilibrium lies to the left.


The Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium ConstantsThe Equilibrium ConstantThe Equilibrium Constant

The Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• An equilibrium can be approached from any

direction.• Example:

• hasN2O4(g) 2NO2(g)

212.0

ONNO

42

22 cK


The Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• However,

• has

• The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

2NO2(g) N2O4(g)

72.4212.01

NO

ON2

2

42 cK


Heterogeneous EquilibriaHeterogeneous Equilibria• When all reactants and products are in one phase, the

equilibrium is homogeneous.• If one or more reactants or products are in a different

phase, the equilibrium is heterogeneous.• Consider:

– experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

• The concentration of a solid or pure liquid is its density divided by molar mass.

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


Heterogeneous EquilibriaHeterogeneous EquilibriaThe Equilibrium ConstantThe Equilibrium Constant

Heterogeneous EquilibriaHeterogeneous Equilibria• Neither density nor molar mass is a variable, the

concentrations of solids and pure liquids are constant.• For the decomposition of CaCO3:

• We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions.

• The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present.

2

223

COconstant

.COconstantCOCaCO

CaO

cc

c

KK

K


• Proceed as follows:– Tabulate initial and equilibrium concentrations (or partial

pressures) given.– If an initial and equilibrium concentration is given for a

species, calculate the change in concentration.– Use stoichiometry on the change in concentration line only

to calculate the changes in concentration of all species.– Deduce the equilibrium concentrations of all species.

• Usually, the initial concentration of products is zero. (This is not always the case.)

• When in doubt, assign the change in concentration a variable.

Calculating Equilibrium ConstantsCalculating Equilibrium Constants



Carbon dioxide decomposes at elevated temperatures to carbonmonoxide and oxygen:

2CO2(g) 2CO(g) + O2(g)

At 3000 K, 2.00 mol of CO2 is placed into a 1.00 L container andallowed to come to equilibrium. At equilibrium, 0.90 mol CO2

remains. What is the value for Kc at this temperature?

Ans.: Kc = 0.82




CO(g) + 3H2(g) CH4(g) + H2O(g)

The original concentrations of CO and H2 were 0.2000 M and0.3000 M, respectively. At equilibrium, the concentration of CH4 was measured to be 0.0478 M. Calculate the values of Kc and Kp.

Ans.: Kc = 3.91, Kp = 3.60 x 10-4


PhosphorousProblem 3(c)


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

When 4.00 mol of HI was placed into the 5.0 L reaction vessel at458C, the equilibrium mixture was found to contain 0.422 mol I2.Calculate the value of Kc for the decomposition of HI.

Ans.: Kc = 1.79 x 10-2


PhosphorousProblem 3(d)

Hydrogen sulfide, a colorless gas with a foul odor, dissociates onheating:

2H2S(g) 2H2(g) + S2(g)

When 0.100 mol H2S was put into a 10.0 L vessel and heated to1132C, it gave an equilibrium mixture containing 0.0285 mol H2.Calculate the value of Kc at this temperature.

Ans.: Kc = 1.35 x 10-6 M

Predicting the Direction of ReactionPredicting the Direction of Reaction• We define Q, the reaction quotient, for a general

reaction

as

where [A], [B], [P], and [Q] are molarities at any time.• Q = K only at equilibrium.

aA + bB(g) pP + qQ

ba

qpQ

BA

QP

Applications of Equilibrium ConstantsApplications of Equilibrium Constants

Predicting the Direction of ReactionPredicting the Direction of Reaction• If Q > K then the reverse reaction must occur to reach

equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K).

• If Q < K then the forward reaction must occur to reach equilibrium.


Calculation of Equilibrium ConcentrationsCalculation of Equilibrium Concentrations• The same steps used to calculate equilibrium

constants are used.• Generally, we do not have a number for the change in

concentration line.• Therefore, we need to assume that x mol/L of a species

is produced (or used).• The equilibrium concentrations are given as algebraic

expressions.



Phosphorous

Problem 4

The following reaction has an equilibrium constant Kc of 3.07 x 10-4 at 24C:

2NOBr(g) 2NO(g) + Br2(g)

For each of the following compositions, decide whether the reactionis at equilibrium. If not, decide which direction the reaction should go.

(a) [NOBr] = 0.0610 M,[NO] = 0.0151 M, [Br2] = 0.0108 M(b) [NOBr] = 0.115 M,[NO] = 0.0169 M, [Br2] = 0.0142 M(c) NOBr] = 0.181 M,[NO] = 0.0123 M, [Br2] = 0.0201 M

Ans.: a.) goes left; b.) equilibrium; c.) goes right


Phosphorous

Problem 5(a)

Nitrogen and oxygen form nitric oxide:

N2(g) + O2(g) 2NO(g) If an equilibrium mixture at 25C contains 0.040 mol/L N2 and0.010 mol/L O2, what is the concentration of NO in this mixture?Kc = 1 x 10-30 at this temperature.

Ans.: 2 x 10-17 M


Phosphorous

Problem 5(b)

An equilibrium mixture at 1200 K contains 0.30 mol CO, 0.10 mol H2,and 0.020 mol H2O, plus an unknown amount of CH4 in each liter.What is the concentration of CH4 in this mixture if Kc = 3.92?The reaction is:

CO(g) + 3H2(g) CH4(g) + H2O(g)

Ans.: 0.059 M


Phosphorous

Problem 6(a)

The reaction:

CO(g) + H2O(g) CO2(g) + H2(g)

is used to increase the ratio of hydrogen in synthesis gas (mixtures ofCO and H2). Suppose you start with 1.00 mol each of carbonmonoxide and water in a 50.0 L vessel. What is the concentrationof each substance in the equilibrium mixture at 1000C given thatKc = 0.58 at this temperature?

Ans.: [CO] = [H2O] = 0.0114 M, [CO2] = [H2] = 0.0086 M


Phosphorous

Problem 6(b)

Hydrogen iodide decomposes to hydrogen gas and iodine gas:

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

At 800 K, Kc for this reaction is 0.016. If 0.50 mol HI is placed intoa 5.0 L flask, what will be the composition of the mixture atequilibrium?

Ans.: [HI] = 0.080 M, [H2] = [I2] = 0.010 M


Phosphorous

Problem 7(a)

N2O4 decomposes to NO2 according to the reaction:

N2O4(g) 2NO2(g)

At 100C, Kc = 0.36. If a 1.00 L flask initially contains 0.100 molN2O4, what will be the concentrations of N2O4 and NO2 at equilibrium?

Ans.: [N2O4] =0.040 M, [NO2] = 0.12 M


Phosphorous

Problem 7(b)

Hydrogen and iodine react according to the equation:

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

Suppose 1.00 mol H2 and 2.00 mol I2 are placed into a 1.00 L vessel.How many moles of each substance are in the equilibrium mixture at458C if Kc = 49.7 at this temperature?

Ans.: 1.86 mol HI, 0.07 mol H2, 1.07 mol I2


Phosphorous

Problem 7(c)

Iodine and bromine react to give iodine monobromide:

I2(g) + Br2(g) 2IBr(g)

What is the equilibrium composition of a mixture at 150C thatinitially contained 0.0015 mol each of iodine and bromine in a 5.0 Lvessel if Kc = 1.2 x 102 at this temperature?

Ans.: [IBr] = 5.1 x 10-4 M, [Br2] = [I2] = 4.7 x 10-5 M

• Consider the production of ammonia

• As the pressure increases, the amount of ammonia present at equilibrium increases.

• As the temperature decreases, the amount of ammonia at equilibrium increases.

• Can this be predicted?• Le Châtelier’s Principle: if a system at equilibrium is

disturbed, the system will move in such a way as to counteract the disturbance.

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

Le Châtelier’s PrincipleLe Châtelier’s Principle


Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• Consider the Haber process

• If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier).

• That is, the system must consume the H2 and produce products until a new equilibrium is established.

• Therefore, [H2] and [N2] will decrease and [NH3] increases.

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


Change in Reactant or Product ConcentrationsChange in Reactant or Product ConcentrationsLe Châtelier’s PrincipleLe Châtelier’s Principle

Change in Reactant or Product ConcentrationsChange 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).

• We illustrate the concept with the industrial preparation of ammonia

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


Change in Reactant or Product ConcentrationsChange in Reactant or Product ConcentrationsLe Châtelier’s PrincipleLe Châtelier’s Principle

Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• N2 and H2 are pumped into a chamber.• The pre-heated gases are passed through a heating

coil to the catalyst bed.• The catalyst bed is kept at 460 - 550 C under high

pressure.• The product gas stream (containing N2, H2 and NH3)

is passed over a cooler to a refrigeration unit.• In the refrigeration unit, ammonia liquefies but not N2

or H2.


Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• The unreacted nitrogen and hydrogen are recycled

with the new N2 and H2 feed gas.• The equilibrium amount of ammonia is optimized

because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added.

Effects of Volume and PressureEffects of Volume and Pressure• As volume is decreased pressure increases.• Le Châtelier’s Principle: if pressure is increased the

system will shift to counteract the increase.



Phosphorous

Problem 8

Predict the effect of the following concentration changes on thereaction below.

CH4(g) + 2S2(g) CS2(g) + 2H2S(g)

a) (a) Some CH4(g) is removed.(b) Some S2(g) is added.(c) Some CS2(g) is added.(d) Some H2S(g) is removed.(e) Some argon gas is added.

Ans.: a.) goes left; b.) goes right; c.) goes left; d.) goes right ; e.) no effect

Effects of Volume and PressureEffects of Volume and Pressure• That is, 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) 2NO2(g)


Effects of Volume and PressureEffects of Volume and Pressure• 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.



Phosphorous

Problem 9

Predict the effect of increasing pressure (decreasing volume) oneach of the following reactions.

a) (a) CH4(g) + 2S2(g) CS2(g) + 2H2S(g) (b) H2(g) + Br2(g) 2HBr(g)(c) CO2(g) + C(s) 2CO(g)(d) PCl5(g) PCl3(g) + Cl2(g)(e) N2O4(g) 2NO2(g)

Ans.: a.) no effect; b.) no effect; c.) goes left; d.) goes right ; e.) goes left

Effect of Temperature ChangesEffect 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.


Effect of Temperature ChangesEffect of Temperature Changes• Removing heat (i.e. cooling the vessel), favors towards

the decrease:– if H > 0, cooling favors the reverse reaction,– if H < 0, cooling favors the forward reaction.

• Consider

for which H > 0.– Co(H2O)6

2+ is pale pink and CoCl42- is blue.

Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)


Effect of Temperature ChangesEffect of Temperature ChangesLe Châtelier’s PrincipleLe Châtelier’s Principle

Effect of Temperature ChangesEffect of Temperature Changes

– If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue.

– Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl4

2-. – If the room temperature equilibrium mixture is placed in a

beaker of ice water, the mixture turns bright pink.– Since H > 0, removing heat favors the reverse reaction

which is the formation of pink Co(H2O)62+.

Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)



Phosphorous

Problem 10

Predict the effect of increasing temperature on each of the followingreactions. What effect does this change have on Kc?

a) (a) CO(g) + 3H2(g) CH4(g) + H2O(g) H < 0 (b) CO2(g) + C(s) 2CO(g) H > 0(c) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) H < 0 (d) 2H2O(g) 2H2(g) + O2(g) H > 0

Ans.: a.) goes left, Kc decreases; b.) goes right, Kc increases; c.) goes left, Kc decreases; d.) goes right, Kc increases

The Effect of CatalystsThe Effect of Catalysts• A catalyst lowers the activation energy barrier for the

reaction.• Therefore, a catalyst will decrease the time taken to

reach equilibrium.• A catalyst does not effect the composition of the

equilibrium mixture.


Date post:	03-Jan-2017
Category:	Documents
Upload:	doanquynh
View:	236 times
Download:	0 times

Chemistry Notes 15

Documents