Chemical Equilibrium The reversibility of reactions.

Chemical Chemical EquilibriumEquilibrium

The reversibility of reactionsThe reversibility of reactions

EquilibriumEquilibrium

Many chemical reactions do not Many chemical reactions do not go to completion. Initially, when go to completion. Initially, when reactants are present, the forward reactants are present, the forward reaction predominates. As the reaction predominates. As the concentration of products increases, concentration of products increases, the reverse reaction begins to the reverse reaction begins to become significant.become significant.


The forward reaction rate slows The forward reaction rate slows down since the concentration of down since the concentration of reactants has decreased. Since the reactants has decreased. Since the concentration of products is concentration of products is significant, the reverse reaction rate significant, the reverse reaction rate gets faster.gets faster.

Eventually, theEventually, the

forward reaction rate = reverse forward reaction rate = reverse reaction ratereaction rate


forward reaction rate = reverse forward reaction rate = reverse reaction ratereaction rate

At this point, the reaction is At this point, the reaction is in in equilibriumequilibrium. The equilibrium is . The equilibrium is dynamicdynamic. Both the forward and . Both the forward and reverse reactions occur, but there is reverse reactions occur, but there is no net change in concentration.no net change in concentration.

Equilibrium: Equilibrium: 2NO2 ↔ N2O4

The concentrations of N2O4 and NO2 level off when the system reaches equilibrium.

Equilibrium: Equilibrium: 2NO2 ↔ N2O4

The system will reach equilibrium starting with either NO2 , N2O4, or a mixture of the two.


Reactions that can reach Reactions that can reach equilibrium are indicated by double equilibrium are indicated by double arrows (↔ or ). arrows (↔ or ).

The extent to which a reaction The extent to which a reaction proceeds in a particular direction proceeds in a particular direction depends upon the reaction, depends upon the reaction, temperature, initial concentrations, temperature, initial concentrations, pressure (if gases), etc.pressure (if gases), etc.


Scientists studying many reactions Scientists studying many reactions atat equilibriumequilibrium determined that for a general determined that for a general reaction with coefficients reaction with coefficients a, b, c a, b, c and and d d ::

a a A + A + b b B ↔ B ↔ c c C + C + d d DD

K is the K is the equilibrium constant equilibrium constant for the for the reaction.reaction.

[C]c[D]d[A]a[B]b

K=


a a A + A + b b B ↔ B ↔ c c C + C + d d DD

This is the This is the equilibrium constant equilibrium constant expression expression for the reactionfor the reaction. . K is the K is the equilibrium constantequilibrium constant. The square . The square brackets indicate the concentrations of brackets indicate the concentrations of products and reactants products and reactants at equilibriumat equilibrium..

[C]c[D]d[A]a[B]b

K=

Equilibrium ConstantsEquilibrium Constants

The value of the equilibrium The value of the equilibrium constant depends upon temperature, constant depends upon temperature, but does not depend upon the but does not depend upon the initialinitial concentrations of reactants and concentrations of reactants and products. It is determined products. It is determined experimentally.experimentally.


The units of K are usually The units of K are usually omitted. The square brackets in the omitted. The square brackets in the equilibrium constant expression equilibrium constant expression indicate concentration in moles/liter.indicate concentration in moles/liter.

Some texts will use the symbol Some texts will use the symbol KKcc or K or Keqeq for equilibrium constants for equilibrium constants that use concentrations or molarity. that use concentrations or molarity.


The size of the equilibrium The size of the equilibrium constant gives an indication of constant gives an indication of whether a reaction proceeds to the whether a reaction proceeds to the right. right.

Equilibrium ConstantsEquilibrium ConstantsIf a reaction has a small value of K, If a reaction has a small value of K,

the reverse reaction will have a large the reverse reaction will have a large value of K.value of K.

At a given temperature, K = 1.3 x 10At a given temperature, K = 1.3 x 10-2-2 for: for:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

If the reaction is reversed, If the reaction is reversed,

2 NH2 NH33((gg) ↔ N) ↔ N22((gg) + 3 H) + 3 H22((gg) ) K = 1/(1.3 x 10K = 1/(1.3 x 10-2-2 ) = 7.7 x 10 ) = 7.7 x 1011


Equilibrium constants for gas phase Equilibrium constants for gas phase reactions are sometimes determined reactions are sometimes determined using pressures rather than using pressures rather than concentrations. The symbol used is Kconcentrations. The symbol used is Kpp rather than K (or Krather than K (or Kcc).).

For the reaction:For the reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

KKpp = (P = (Pammoniaammonia))22

(P(PNN22)(P)(PH2H2))33


The numerical value of KThe numerical value of Kpp is usually is usually different than that for K. The values are different than that for K. The values are related, since P= related, since P= nRT nRT = MRT, where M = MRT, where M = mol/liter.= mol/liter.

For reactions involving gases:For reactions involving gases:

KKCC = K = KPP(RT)(RT)ΔnΔn

where n is the change in moles of gases where n is the change in moles of gases in the balanced chemical reaction.in the balanced chemical reaction.

V


The method for solving The method for solving equilibrium problems with Kequilibrium problems with Kpp or K is or K is the same. With Kthe same. With Kpp, you use pressure , you use pressure in atmospheres. With K or Kin atmospheres. With K or Kcc, you , you use concentration in moles/liter, or use concentration in moles/liter, or molarity.molarity.

Heterogeneous Heterogeneous EquilibriaEquilibria

Many equilibrium reactions Many equilibrium reactions involve reactants and products where involve reactants and products where more than one phase is present. more than one phase is present. These are called These are called heterogeneous heterogeneous equilibriaequilibria..

An example is the equilibrium An example is the equilibrium between solid calcium carbonate and between solid calcium carbonate and calcium oxide and carbon dioxide.calcium oxide and carbon dioxide.

CaCOCaCO33((ss) ↔ CaO() ↔ CaO(ss) + CO) + CO22((gg))


CaCOCaCO33((ss) ↔ CaO() ↔ CaO(ss) + CO) + CO22((gg))Experiments show that the Experiments show that the

position of the equilibrium does position of the equilibrium does not depend upon the amounts of not depend upon the amounts of calcium carbonate or calcium calcium carbonate or calcium oxide. That is, adding or removing oxide. That is, adding or removing some of the CaCOsome of the CaCO33((ss) or CaO() or CaO(ss) ) does not disrupt or alter the does not disrupt or alter the concentration of carbon dioxide.concentration of carbon dioxide.



This is because the This is because the concentrationsconcentrations of pure solids (or of pure solids (or liquids) cannot change. As a liquids) cannot change. As a result, the equilibrium constant result, the equilibrium constant expression for the above reaction expression for the above reaction is:is:

K=[COK=[CO22] or K] or Kpp = P = PCO2CO2


The position of a heterogeneous The position of a heterogeneous equilibrium does not depend on equilibrium does not depend on the amounts of pure solids or the amounts of pure solids or liquids present.liquids present.

The Reaction QuotientThe Reaction Quotient

If reactants and products of a If reactants and products of a reaction are mixed together, it is reaction are mixed together, it is possible to determine whether the possible to determine whether the reaction will proceed to the right or reaction will proceed to the right or left. This is accomplished by left. This is accomplished by comparing the composition of the comparing the composition of the initial mixture to that at equilibrium.initial mixture to that at equilibrium.


If the concentration of one of If the concentration of one of the reactants or products is zero, the the reactants or products is zero, the reaction will proceed so as to make reaction will proceed so as to make the missing component. the missing component.

If all components are present If all components are present initially, the initially, the reaction quotient, reaction quotient, Q, is Q, is compared to K to determine which compared to K to determine which way the reaction will go.way the reaction will go.


Q has the same form as the Q has the same form as the equilibrium constant expression, but equilibrium constant expression, but we use we use initialinitial concentrations instead concentrations instead of equilibrium concentrations. of equilibrium concentrations. Initial concentrations are usually Initial concentrations are usually indicated with a subscript zero.indicated with a subscript zero.


K = 2.2 x 10K = 2.2 x 1022 for the reaction: for the reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

Suppose 1.00 mol NSuppose 1.00 mol N22, 2.0 mol H, 2.0 mol H22 and and 1.5 mol of NH1.5 mol of NH3 3 are placed in a 1.00L are placed in a 1.00L vessel. What will happen? In which vessel. What will happen? In which direction with the reaction proceed?direction with the reaction proceed?



NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))


1. Calculate Q and compare it 1. Calculate Q and compare it to K.to K.


A comparison of Q with K will indicate the direction the reaction will go.



NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))


1. Calculate Q and compare it 1. Calculate Q and compare it to K.to K.

∆∆G for Non-Standard G for Non-Standard ConditionsConditions

The thermodynamic tables are The thermodynamic tables are for standard conditions. This for standard conditions. This includes having all reactants and includes having all reactants and products present initially at a products present initially at a temperature of 25temperature of 25ooC. All gases are C. All gases are at a pressure of 1 atm, and all at a pressure of 1 atm, and all solutions are 1 M.solutions are 1 M.

Equilibrium and Free Equilibrium and Free EnergyEnergy

Thermodynamic tables can be used to Thermodynamic tables can be used to calculate ΔGcalculate ΔGoo, the , the standard standard Gibbs free Gibbs free energy change. Since the direction in energy change. Since the direction in which a reaction will proceed depends which a reaction will proceed depends upon the reaction quotient, Q, the value of upon the reaction quotient, Q, the value of ΔG, will also depend upon the initial ΔG, will also depend upon the initial concentrations of reactants and products.concentrations of reactants and products.

ΔG = ΔGΔG = ΔGo o + RT ln Q + RT ln Q


For non-standard temperature, For non-standard temperature, concentrations or gas pressures: concentrations or gas pressures:

∆∆G = ∆GG = ∆Goo + RTlnQ + RTlnQ

Where R = 8.314 J/K-molWhere R = 8.314 J/K-mol

T is temperature in KelvinsT is temperature in Kelvins

Q is the reaction quotientQ is the reaction quotient


For non-standard temperature, For non-standard temperature, concentrations or gas pressures: concentrations or gas pressures:

∆∆G = ∆GG = ∆Goo + RT ln Q + RT ln Q

For Q, gas pressures are in For Q, gas pressures are in atmospheres, and concentrations of atmospheres, and concentrations of

solutions are in molarity, M.solutions are in molarity, M.

∆∆GGo o and Equilibrium and Equilibrium

A large negative value of ∆GA large negative value of ∆Goo indicates that the forward reaction indicates that the forward reaction or process is spontaneous. That is, or process is spontaneous. That is, there is a large there is a large driving forcedriving force for the for the forward reaction. This also means forward reaction. This also means that the equilibrium constant for the that the equilibrium constant for the reaction will be large. reaction will be large.


A large positive value of ∆GA large positive value of ∆Goo indicates that the reverse reaction indicates that the reverse reaction or process is spontaneous. That is, or process is spontaneous. That is, there is a large there is a large driving forcedriving force for the for the reverse reaction. This also means reverse reaction. This also means that the equilibrium constant for the that the equilibrium constant for the reaction will be small. reaction will be small.

When a reaction or process is at When a reaction or process is at equilibrium, ∆Gequilibrium, ∆Goo = zero. = zero.


∆∆G = ∆GG = ∆Goo + RT lnQ + RT lnQ

At equilibrium, ∆G is equal to At equilibrium, ∆G is equal to zero, and zero, and

Q = K.Q = K.

0 = ∆G0 = ∆Goo + RT lnK + RT lnK

∆∆GGoo = - RT lnK = - RT lnK


Calculate, ∆GCalculate, ∆Go o and K at 25and K at 25ooC for:C for:

HH22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) (g)



HH22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) (g)

Since hydrogen and oxygen react Since hydrogen and oxygen react explosively to form water, the reverse explosively to form water, the reverse reaction should be unfavorable, and reaction should be unfavorable, and have a positive ∆Ghave a positive ∆Go o , and a small , and a small equilibrium constant.equilibrium constant.


HH22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) (g)



HH22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) (g)

∆∆GGo o = = ΣΣ∆G∆Gffo o products –products –ΣΣ∆G∆Gff

o o reactants reactants

∆∆GGo o = [(1 mol) (∆G= [(1 mol) (∆Gffo o HH22(g)) +(½ mol) (g)) +(½ mol)

(∆G(∆Gffo o HH22(g))] –[(1 mol) (∆G(g))] –[(1 mol) (∆Gff

ooHH22O(l)]O(l)]

∆∆GGo o =0 –[( 1 mol)(-237.1kJ/mol)] = =0 –[( 1 mol)(-237.1kJ/mol)] = +237.1 kJ+237.1 kJ

Free Energy and Free Energy and EquilibriumEquilibrium

∆∆GGoo = - RT lnK = - RT lnK

where R = 8.314 J/K-molwhere R = 8.314 J/K-mol

lnK = –∆GlnK = –∆Goo /RT /RT

For HFor H22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) :(g) :

lnK = –237.1 kJ(10lnK = –237.1 kJ(1033J/kJ)/(8.314J/mol-K) J/kJ)/(8.314J/mol-K) (298K)(298K)

ln K = –95.70ln K = –95.70

K = eK = e–95.70–95.70 = 2.7 x 10 = 2.7 x 10-42-42

Free Energy and Free Energy and EquilibriumEquilibrium

In summary, an unfavorable In summary, an unfavorable reaction has a positive free energy reaction has a positive free energy change, and a small equilibrium change, and a small equilibrium constant.constant.

Problem: Calculation of Problem: Calculation of KK

At a certain temperature, 3.00 moles At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles vessel. At equilibrium, 2.50 moles remain. Calculate K for the remain. Calculate K for the reaction:reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))



NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

1. Write the equilibrium constant 1. Write the equilibrium constant expression.expression.


At a certain temperature, 3.00 moles At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:remain. Calculate K for the reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))


K = [NHK = [NH33]]22/([N/([N22][H][H22]]33))



NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

2. Make a table of concentrations.2. Make a table of concentrations.


NN22((gg) + 3 H) + 3 H22((gg) ↔ ) ↔ 2NH2NH33((gg)) [N[N22]]

[H[H22

]][NH[NH33

]]

initialinitial

3.003.00molmol 2.00L2.00L

changchangee

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL


NN22((gg) + 3 H) + 3 H22((gg) ↔ ) ↔ 2NH2NH33((gg))

3. Complete the 3. Complete the table.table.

[N[N22]][H[H22

]][NH[NH33

]]

initialinitial

3.003.00molmol 2.00L2.00L

changchangee

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL


NN22((gg) + 3 H) + 3 H22((gg) ↔ ) ↔ 2NH2NH33((gg))


[N[N22]][H[H22

]][NH[NH33

]]

initialinitial

3.003.00molmol 2.00L2.00L

changchangee

-.50m-.50molol2.00 2.00 LL

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL


NN22((gg) + 3 H) + 3 H22((gg) ↔ ) ↔ 2NH2NH33((gg))


[N[N22]][H[H22

]][NH[NH33

]]

initialinitial 00 00

3.003.00molmol 2.00L2.00L

changchangee

-.50m-.50molol2.00 2.00 LL

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL

[N[N22]] [H[H22]][NH[NH33

]]

initiainitiall

00 00

3.003.00molmol 2.00L2.00L

chanchangege

+.25m+.25molol2.00 L2.00 L

+.75m+.75molol2.00L2.00L

-.50m-.50molol2.00 2.00 LL

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL

NN22((gg) + 3H) + 3H22((gg) ) ↔2NH↔2NH33((gg))

[N[N22]] [H[H22]][NH[NH33

]]

initiainitiall

00 00

3.003.00molmol 2.00L2.00L

chanchangege

+.25m+.25molol2.00 L2.00 L

+.75m+.75molol2.00L2.00L

-.50m-.50molol2.00 2.00 LL

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL

NN22((gg) + 3H) + 3H22((gg) ) ↔2NH↔2NH33((gg))

The equilibrium values are the sum of the initial concentrations plus any changes that occur.

[N[N22]] [H[H22]] [NH[NH33]]

initiainitiall

00 003.00m3.00molol 2.00L2.00L

chanchangege

+.25m+.25molol2.00 L2.00 L

+.75m+.75molol2.00L2.00L

-.50mo-.50moll2.00 L2.00 L

equiliequili--

libriulibriumm

+.13+.13MM

+.38+.38MM

+1.25+1.25MM

NN22((gg) + 3H) + 3H22((gg) ) ↔2NH↔2NH33((gg))

The equilibrium values are the sum of the initial concentrations plus any changes that occur.

4. Substitute and 4. Substitute and solve for K.solve for K.

K K == [NH [NH33]]22/([N/([N22]][H[H22]]33))

[N[N22]] [H[H22]] [NH[NH33]]

equiliequili--

libriulibriumm

+.13+.13MM

+.38+.38MM

+1.25+1.25MM

NN22((gg) + 3H) + 3H22((gg) ) ↔2NH↔2NH33((gg))

K=(1.25)2/[(.13)(.38)3] = 2.2 x 102

ProblemProblem

HH22((gg) + I) + I22((gg) ↔ 2HI() ↔ 2HI(gg) K) Kpp = 1.00 = 1.00 x 10x 1022

Initially, PInitially, Phydrogenhydrogen = P = Piodineiodine = 0.500 atm. = 0.500 atm. Calculate the equilibrium partial Calculate the equilibrium partial pressures of all three gases.pressures of all three gases.

ProblemProblem



1. Write the K1. Write the Kpp expression.expression.

ProblemProblem



1. Write the K1. Write the Kpp expression.expression.

KKpp = (P = (PHIHI))22

(P(PH2H2)(P)(PI2I2))

ProblemProblem



2. Make a table of initial, change 2. Make a table of initial, change and and equilibrium pressures.equilibrium pressures.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) )

HH22 II22 HIHI

initialinitial .500 .500 atmatm

.500 .500 atmatm 00

changchangee

equili-equili-

briumbrium


HH22 II22 HIHI


.500 .500 atmatm 00

changchangee

-x-x -x-x +2x+2x

equili-equili-

briumbrium


HH22 II22 HIHI


.500 .500 atmatm 00

changchangee

-x-x -x-x +2x+2x

equili-equili-

briumbrium.500-x.500-x .500-x.500-x 2x2x

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) ) HH22 II22 HIHI

equili-equili-


3. Substitute and solve.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) ) HH22 II22 HIHI

equili-equili-


3.Substitute and solve.

Kp= (PHI)2/(PH2)(PI2) = 1.00 x 10= 1.00 x 1022

(2x)2 = 1.00 x = 1.00 x 101022

(.500-x) (.500-x)


3. Substitute and solve.Kp= (PHI)2/(PH2)(PI2) = 1.00 x 10= 1.00 x 1022

(2x)2 = 1.00 x 10= 1.00 x 1022

(.500-x) (.500-x)

Take the square root of both sides: (2x) = 10.0= 10.0

(.500-x)


3. Substitute and solve. (2x) = 10.0= 10.0

(.500-x)

2x = 10.0 (.500-x) = 5.00 -10.0x

12.0 x = 5.00

x = .417


4. Answer the question: Calculate the Calculate the

equilibrium partial pressures of all three equilibrium partial pressures of all three

gases.gases.

x = .417

HH22 II22 HIHI

equili-equili-

briumbrium .500-x.500-x .500-x.500-x 2x2x


4. Answer the question: Calculate the Calculate the

equilibrium partial pressures of all three equilibrium partial pressures of all three

gases.gases.

x = .417

HH22 II22 HIHI

equili-equili-

briumbrium.500-x.500-x

0.083 atm0.083 atm.500-x.500-x

0.083 atm0.083 atm2x2x

.834 atm.834 atm


5. Check your answer, if possible. Does (PHI)2/(PH2)(PI2) = 1.00 x 10= 1.00 x 102 2 ??

(.834)(.834)22/(.083)(.083) = (.696)/(.0069) = 1.0 x 10/(.083)(.083) = (.696)/(.0069) = 1.0 x 1022

HH22 II22 HIHI

equili-equili-

briumbrium.500-x.500-x

0.083 atm0.083 atm.500-x.500-x

0.083 atm0.083 atm2x2x

.834 atm.834 atm

Problem: NProblem: N22((gg) + O) + O22((gg) ↔ ) ↔ 2NO(2NO(gg))

At 2200 At 2200 ooC, K = 0.050 for the above C, K = 0.050 for the above reaction. Initially, 1.60 mol of reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. are sealed in a 2.00 liter vessel. Calculate the concentration of all Calculate the concentration of all species at equilibrium.species at equilibrium.


At 2200 At 2200 ooC, K = 0.050 for the above C, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the 2.00 liter vessel. Calculate the concentration of all species at equilibrium.concentration of all species at equilibrium.


K = 0.050 = [NO]K = 0.050 = [NO]22/[N/[N22][O][O22]]


At 2200 At 2200 ooC, K = 0.050 for the above C, K = 0.050 for the above reaction. Initially, 1.60 mol of reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. are sealed in a 2.00 liter vessel. Calculate the concentration of all Calculate the concentration of all species at equilibrium.species at equilibrium.

2. Make a table of initial, change 2. Make a table of initial, change and and equilibrium concentrations.equilibrium concentrations.


NN22 OO22 NONO

initialinitial1.60mo1.60mo

ll2.00 L2.00 L

.400 .400 molmol

2.00 L2.00 L00

changechange -x-x -x-x +2x+2x

equili-equili-

briumbrium.800 - x.800 - x .200 - x.200 - x 2x2x


NN22 OO22 NONO

equili-equili-

briumbrium.800 - x.800 - x .200 - x.200 - x 2x2x


K = 0.050 = [NO]K = 0.050 = [NO]22/[N/[N22][O][O22]]

Approaches to Solving Approaches to Solving ProblemsProblems

Some equilibrium problems Some equilibrium problems require use of the quadratic require use of the quadratic equation to obtain an accurate equation to obtain an accurate solution. In cases where the solution. In cases where the equilibrium constant is quite small equilibrium constant is quite small (roughly 10(roughly 10-5-5 or smaller), it is often or smaller), it is often possible to make assumptions that possible to make assumptions that will simplify the mathematics.will simplify the mathematics.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

At 35At 35ooC, K = 1.6 x 10C, K = 1.6 x 10-5-5 for the above for the above reaction. Calculate the equilibrium reaction. Calculate the equilibrium concentrations of all species present concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of if 2.0 moles of NOCl and 1.0 mole of ClCl22 are placed in a 1.0 liter flask. are placed in a 1.0 liter flask.

1. K = 1.6 x 101. K = 1.6 x 10-5-5 =[NO]=[NO]22[Cl[Cl22]/[NOCl]]/[NOCl]22


At 35At 35ooC, K = 1.6 x 10C, K = 1.6 x 10-5-5 for the above for the above reaction. Calculate the equilibrium reaction. Calculate the equilibrium concentrations of all species present concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of if 2.0 moles of NOCl and 1.0 mole of ClCl22 are placed in a 1.0 liter flask. are placed in a 1.0 liter flask.

2. Make a table of concentrations.2. Make a table of concentrations.


NOClNOCl NONO ClCl22

initiainitiall

2.0mo2.0moll

1.00 1.00 LL

001.0 1.0 molmol

1.00 L1.00 L

chanchangege -2x-2x +2x+2x +x+x

equiliequili--

briubriumm

2.0 - 2.0 - 2x2x 2x2x 1.0+x1.0+x



equiliequili--

briubriumm

2.0 - 2.0 - 2x2x 2x2x 1.0+x1.0+x


K = 1.6 x 10K = 1.6 x 10-5-5 =[NO] =[NO]22[Cl[Cl22]/[NOCl]]/[NOCl]22

1.6 x 101.6 x 10-5-5 =[2x] =[2x]22[1.0 + x]/[2.0-2x][1.0 + x]/[2.0-2x]22



1.6 x 101.6 x 10-5-5 = [2x] = [2x]2 2 [1.0 + x] [1.0 + x] [2.0-2x][2.0-2x]22

Since K is small, x, the amount of Since K is small, x, the amount of product formed, will be a small product formed, will be a small number. As a result, 1.0 + x ≈1.0, number. As a result, 1.0 + x ≈1.0, and 2.0-2x≈2.0and 2.0-2x≈2.0



1.6 x 101.6 x 10-5-5 = [2x] = [2x]2 2 [1.0 + x] [1.0 + x] [2.0-2x][2.0-2x]22

The expression simplifies to:The expression simplifies to:1.6 x 101.6 x 10-5-5 = [2x] = [2x]22 [1.0] [1.0]

[2.0][2.0]22

0

0


3.Solve for x.

1.6 x 101.6 x 10-5-5 = [2x] = [2x]22 [1.0] [1.0] [2.0][2.0]22

4x4x22 = 1.6 x 10 = 1.6 x 10-5-5 (4)/1 = 6.4 x 10-5

xx22 = 1.6 x 10 = 1.6 x 10-5 -5 ; x =0.0040 ; x =0.0040


4. Check your assumption.

x =0.0040x =0.0040

Does 1.0 + x ≈1.0, and 2.0-Does 1.0 + x ≈1.0, and 2.0-2x≈2.0?2x≈2.0?

1.0 + .004 = 1.01.0 + .004 = 1.0

2.0 – 2(.004) = 2.0 - .008=2.02.0 – 2(.004) = 2.0 - .008=2.0

Validity of the AssumptionValidity of the Assumption

These assumptions are generally considered valid if they cause an insignificant error. In this case, due to significant figures, no error is introduced. Usually, if the difference is within 5%, the error is considered acceptable.



equiliequili--

briubriumm

2.0 - 2.0 - 2x2x 2x2x 1.0+x1.0+x

5.Answer the question.

[NOCl] = 2.0M; [NO]= 0.0080M; [NOCl] = 2.0M; [NO]= 0.0080M;

[Cl[Cl22] = 1.0M] = 1.0M


5.Check your answer.

[NOCl] = 2.0M; [NO]= 0.0080M; [NOCl] = 2.0M; [NO]= 0.0080M;

[Cl[Cl22] = 1.0M] = 1.0M

Does [NO]Does [NO]22[Cl[Cl22]/[NOCl]]/[NOCl]2 2 = 1.6 x 10= 1.6 x 10--

55 ? ?

(.0080)(.0080)22(1.0)/(2.0)(1.0)/(2.0)22 = 1.6 x 10 = 1.6 x 10-5-5

Le ChatelierLe Chatelier’’s Principles Principle

If a stress is applied to a system at If a stress is applied to a system at equilibrium, the position of the equilibrium, the position of the equilibrium will shift so as to equilibrium will shift so as to counteract the stress.counteract the stress.


If a stress is applied to a system at If a stress is applied to a system at equilibrium, the position of the equilibrium, the position of the equilibrium will shift so as to counteract equilibrium will shift so as to counteract the stress.the stress.

Types of stresses: Types of stresses: Addition or removal of reactants or Addition or removal of reactants or

productsproducts Changes in pressure or volume (for gases)Changes in pressure or volume (for gases) Changes in temperatureChanges in temperature

FeFe3+3+(aq)(aq) + SCN + SCN-- ↔ ↔ FeSCNFeSCN2+2+

Heat + NHeat + N22OO44(g) ↔ 2 (g) ↔ 2 NONO22(g)(g)

CoClCoCl442- 2- + 6 H+ 6 H22OOCo(HCo(H22O)O)66

2+2++ + 4Cl4Cl1-1- + +

heatheat


For reactions involving gaseous For reactions involving gaseous products or reactants, changes in products or reactants, changes in pressure of volume pressure of volume maymay shift the shift the equilibrium. The equilibrium will shift equilibrium. The equilibrium will shift only if there is an unequal number of only if there is an unequal number of moles of gas on either side of the moles of gas on either side of the reaction. Reactions involving liquids reaction. Reactions involving liquids or solids are not significantly affected or solids are not significantly affected by changes in pressure or volume.by changes in pressure or volume.

Le ChatelieLe Chatelierr ’’s Principles Principle

Pressure ChangesPressure Changes


For the reaction For the reaction

C(C(ss) + 2 H) + 2 H22(g) ↔ CH(g) ↔ CH44(g) ∆H(g) ∆Hoo=-75kJ=-75kJ

Assuming the reaction is initially at Assuming the reaction is initially at equilibrium, predict the shift in equilibrium, predict the shift in equilibrium (if any) for each of the equilibrium (if any) for each of the following changes. following changes.

a) add carbona) add carbon

b) remove methaneb) remove methane

c) increase the temperaturec) increase the temperature


The value of K will change with The value of K will change with temperature. The effect of a change in temperature. The effect of a change in temperature on K depends on whether temperature on K depends on whether the reaction is exothermic or the reaction is exothermic or endothermic.endothermic.

The easiest way to predict the result The easiest way to predict the result when temperature is changed is to write when temperature is changed is to write in heat as a product (for exothermic in heat as a product (for exothermic reactions) or a reactant (for reactions) or a reactant (for endothermic reactions).endothermic reactions).

Le ChatelierLe Chatelier’’s Principles Principle For the reaction For the reaction

C(C(ss) + 2 H) + 2 H22(g) ↔ CH(g) ↔ CH44(g) ∆H(g) ∆Hoo=-75kJ=-75kJ Assuming the reaction is initially at Assuming the reaction is initially at equilibrium, predict the shift in equilibrium, predict the shift in equilibrium (if any) for each of the equilibrium (if any) for each of the following changes. following changes.

d) add a catalystd) add a catalyste) decrease the volumee) decrease the volumef) remove some carbonf) remove some carbong) add hydrogeng) add hydrogen

Chemical Equilibrium The reversibility of reactions.

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Transcript of Chemical Equilibrium The reversibility of reactions.