Chapter 17:

Chemical EquilibriumEquilibrium: The Extent of Chemical Reactions

17.1 The Dynamic Nature of the Equilibrium State

17.2 The Reaction Quotient and the Equilibrium Constant

17.3 Expressing Equilibria with Pressure Terms: Relation between Kc and Kp

17.4 Reaction Direction: Comparing Q and K

17.5 How to Solve Equilibrium Problems

17.6 Reaction Conditions and the Equilibrium State: LeChatelier’s Principle

Chapter 17 Homework Problems

Problems have been updated for the Principles of Chemistry, Silberberg, 2nd Edition Textbook.

1,2,3,9,13,17,19,23,25,28,30,32,35,39,42,44,46,50,51,54,59,64,66,69,72,80,81,86,89,90.

Keeping the chemical themes straight

Thermodynamics tells us about the energetics of a process(enthalpy and internal energy).

Equilibrium theory tells us the extent that a chemical reaction will occur (must be experimentally determined..i.e. pre-determined by the big guy).

Kinetic theory tells us how fast or slow a chemical reaction will occurs (reaction rates).

All chemical reactions are reversible with reactions occurring simultaneously.

1) What happens to a reaction after it reacts?2) Can we predict the concentrations after the reaction “stops”?3) What does it mean to react and stop reacting?

N2O4 (g) 2NO2 (g)

“double-arrow” symbolicrepresentation of reversiblity

Chemical equilibrium is a property of all reactions, and occurs when the when the rate of the forward and reverse chemical reactions are equal!

Start with pure NO2 Start with pure N2O4 Start with a mixtureequilibrium equilibrium

equilibrium

Time Time Time

Con

cent

ratio

n

Con

cent

ratio

n

Con

cent

ratio

n

There is no change in concentration after time at equilibrium

N2O4 (g) 2NO2 (g)kf

kr

A + 2B AB2kf

kr

rateforward = kf [A][B]2

ratereverse = kr [AB2]

Equilibrium occurs when the rate of the forward reaction and the rate of the reverse reaction are equal!

kf [A][B]2 = kr [AB2]

kfkr

[AB2][A][B]2

= Kc =

ratef = rater

Assume a reaction that is also an elementary reaction

Because the reaction are elementary we can write rate laws from balanced equation

At equilibrium the reactions continue but at the same rate!

Kc is defined as the ratio of the rate constants and is tied to stoichiometry of the elementary event!

Experiment

InitialConcentrations

[N2O4] [NO2]

Inital Ratio (Q)

EquilibriumConcentration

EquilibriumRatio (K)

[N2O4]eq [NO2]eq

1 0.1000 0.0000 0.0000 3.57x10-3 0.193

2 0.10000.0000 ! 9.24x10-4 9.83x10-3

3 0.05000.0500 0.0500 2.04x10-3 0.146

4 0.02500.0750 0.00833 2.75x10-3 0.170

[NO2]2

[N2O4]

[NO2]eq2

[N2O4]eq

10.4

10.4

10.4

10.5

N2O4 (g) 2NO2 (g)

Let’s play with data and attempt to find the right equation by trial and error.

CONSTANT!N2O4 (g) 2NO2 (g) Kc =

[NO2]2

[N2O4]

1. 2.3.4.5.

CONSTANT!

NOTICE: By taking the ration of [products] to [reactants] raised to their stoichiometric factors we get a constant number.

The equilibrium constant, Kc is a unitless number that summarizes the extent of a chemical reaction. It is related to the equilibrium concentrations of reactants and products.

Unitless Equilibrium Constant

[Products] raised to stoichiometric coefficient

[Reactants] raised to stoichiometric coefficient

Kc =[NO2]

2

[N2O4]

N2O4 (g) 2NO2 (g)

Kc is temperature dependent and does not depend on the initial concentrations of either products or reactants.

Balanced chemical equation

aA + bB <==> cC + dD

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

Equilibrium Constant

[Products] raised to stoichiometric coefficient

[Reactants] raised to stoichiometric coefficient

Kc gives us information on whether a reaction is product-favored or reactant-favored or both (extent)

Kc VERY LARGE

VERY SMALL

Product-favoredReactantFavored

Both reactants and products

A large Kc means the reaction favors the products; a small Kc means the reaction favors the reactants (no reaction); intermediate means both are reactants and products are present.

small K

large K

intermediate K

Kc is a experimentally derived number that characterizes the extent of a reaction under question. Here are some example observations!

CH3COOH(aq) + H2O(l) CH3COO- (aq) + H3O+ (aq)

Kc = [CH3COO-][H3O+]

[CH3COOH]

Pure solids and pure liquids are excluded from the equilibrium constant expression.

C(s) + H2O(g) <==> CO(g) + H2(g)

Kc = [H2O]

[CO][H2] =PH2O

PCO PH2 (RT)(2-1)

Same Kc

T = 650˚C

Same [CO2]

DifferentCaO and CaCO3

Solids and pure liquid solvents do not change concentration (Just as density of a material does not change whether you have a little or a lot of the substance)

2. Heterogenous equilibrium --reactants and products have different phases....solids reacting with gases or liquids. We exclude solids and pure liquids from the equilibrium expression.

1. Homogenous equilibrium---reactants and products are have the same phase. (i.e. all liquids, or all gases). All reactants and products are included in the equilibrium expression.

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

We distinguish two types of equilibrium expressions: homogeneous and hetereogeneous.

CH4 (g) + 2H2S (g) CS2 (g) + 4H2 (g) We can use molarity or partial pressures.

Pure liquids and solids are excluded from Kc and Kp

3ClO2-(aq) 2ClO3(aq) + Cl-(aq) Kc =[ClO3]

2[Cl−

]

[ClO−3 ]3

Determine whether the following reactions are homogeneous or heterogeneous, and write equilibrium equation for these reactions:

(a) CO2(g) + C(s) <==> 2 CO(g)

(b) Hg(l) + Hg2+(aq) <==> Hg22+(aq)

(c) 2 Fe(s) + 3H2O(g) <==> Fe2O3(s) + 3 H2(g)

(d) 2 H2O(l) <==> 2 H2O(g)

(e) N2(g) + 3 H2(g) <==> 2 NH3(g)

(f) 2 NH3(g) <==> N2(g) + 3 H2(g)

Suppose for reaction a Kp > 1010 what does this imply?Suppose for reaction e Kc < 10-10 what does this imply

Kc =[CO]

2

[CO2]

Kc =[Hg

2+2 ]

[Hg2+

]

Kc =[H2]3

[H2O]3

Kc =[NH3]

2

[H2]3[N2]

N2(g) + 3H2(g) <==> 2NH3(g)

Using partial pressure in atmospheres, not torr or mmHg!

Kc =[NH3]

2

[N2][[H2]3

Kp =(PNH3

)2

PN2(PH2

)3

The equilibrium constant for reaction involving only gases can be written two ways. Kc is used for solutions, Kp is used for gas phase. They are related!

Using Molarity Values

Kc = Kp RT-!n or Kp = Kc RT!n Kc = Kp RT-[(c+d)-(a+b)] = Kp RT-!n

Kc is related to Kp through the ideal gas law PV = nRT

∆n = (total moles of gaseous products) – (total moles of gaseous reactants)

aA + bB <==> cC + dD

Rearranging: PV = nRT

and substituting for [A], [B], [C], [D] in Kc

regrouping

Kc =[C]c[D]d

[A]a[B]b=

[PC/RT]c[PD/RT]d

[PA/RT]a[PB/RT]b

[C] =n

V=

PC

RT[D] =

n

V=

PD

RT

Kc =(PC)c(PD)d

(PA)a(PB)b× (RT)a+b

(RT)c+d=

(PC)c(PD)d

(PA)a(PB)b× (RT)(a+b)−(c+d)

Kc =[C]c[D]d

[A]a[B]b

Consider the following equilibrium at 295 K:

The partial pressure of each gas at equilibrium is 0.265 atm. Calculate Kp and Kc for the reaction?

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

Consider the following equilibrium at 295 K:

The partial pressure of each gas at equilibrium is 0.265 atm. Calculate Kp and Kc for the reaction?

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

Kp = PNH3 H2SP

= 0.265 x 0.265 = 0.0702

Kc = Kp(RT)-!n !n = total mol product - total mol reactant !n = 2 – 0 = 2

T = 295 K

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

R = 0.0821

Plug and chug

Remember this equation!

Write the Kp expression.

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 74˚C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

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

Kc = [COCl2][CO][Cl2]

=0.14

0.012 x 0.054

Kp = Kc(RT)!n

!n = 1 – 2 = -1

R = 0.0821T = 273 + 74 = 347 K

Kp = 216 x (0.0821 x 347)-1 = 7.6

= 216

The reaction quotient Q is used to predict the direction of a chemical reaction.

N2O4 (g) 2NO2 (g) Qc =[NO2]

2

[N2O4]

At equilibrium

• If Qc > Kc reaction proceeds from right to left • If Qc = Kc the system is at equilibrium• If Qc < Kc reaction proceeds from left to right

A reaction will start out at non-equilibrium conditions and move towards equilibrium until the reaction concentrations satisfy the number Kc (the God number).

N2O4 (g) 2NO2 (g)

Not at equilibrium

At equilibrium

Qc =[NO2]

2

[N2O4]

Qc uses non-equilibrium concentrationsKc uses equilibrium concentrations

Qc Kc

equilibrium

At equilibrium

• If Qc > Kc reaction proceeds from right to left • If Qc = Kc the system is at equilibrium• If Qc < Kc reaction proceeds from left to right

Qc is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression and comparing it to the value of Kc.

N2O4 (g) 2NO2 (g) Qc =[NO2]

2

[N2O4]

Write the reaction quotient, Qc, for each of the following reactions:

(a) The decomposition of dinitrogen pentoxide, N2O5(g) NO2(g) + O2(g)

(b) The combustion of propane gas, C3H8(g) + O2(g) CO2(g) + H2O(g)

Write the reaction quotient, Qc, for each of the following reactions:

(a) The decomposition of dinitrogen pentoxide, N2O5(g) NO2(g) + O2(g)

(b) The combustion of propane gas, C3H8(g) + O2(g) CO2(g) + H2O(g)

SOLUTION:

42(a) N2O5(g) NO2(g) + O2(g) Qc = [NO2]4[O2]

[N2O5]2

3 45(b) C3H8(g) + O2(g) CO2(g) + H2O(g) Qc = [CO2]3[H2O]4

[C3H8][O2]5

Qc is the same as Kc except Q is not at equilibrium

You are given a mixture containing 1.57 mol N2, 1.92 mol H2, and 8.13 mol NH3 in a 20.0 L vessel at 500 K. The equilibrium constant Kc for the production of NH3 from N2 and H2 = 170 for the reaction. Is reaction mixture at equilibrium? If not, what direction will the reaction occur?

N2 + 3 H2 2NH3

Given:Kc = 170 Initial [N2] = 1.57 mol/20.0 L = 0.0785 M; Initial [H2]t = 1.92 mol/20.0 L = 0.0960 M Initial [NH3]t = 8.13mol/20.0 L = 0.407 M

Using Qc to see if a reaction is at equilibrium1. Balance the chemical equation

2. Write the equilibrium expression for the reaction

3. Calculate Qc by substituting initial concentrations

4. Compare Qc/p to Kc. Is Q > K or is Q < K?

5. Determine which way the reaction must go to attain equilibrium.

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

Kc = [NH3]2/[N2][H2]3

Qc = [NH3]t2/[N2]t[H2]t3 = (0.46)2/(0.0785)(0.096)3 = 2370

• Qc > Kc (2370 >> 170)

• therefore the reaction mixture is not at equilibrium and the reaction will proceed from right to left to reduce Q to 170.

Using Qc to see if a reaction is at equilibrium

Comparing Q and K To Determine Rxt Direction

For the reaction N2O4(g) 2NO2(g), Kc = 0.21 at 100 oC. At a point during the reaction, [N2O4] = 0.12 M and [NO2] = 0.55 M. Has the reaction reached equilibrium? If not, in which direction is it progressing?

PLAN: Write an expression for Qc, substitute with the values given, and compare the Qc with the given Kc.

Comparing Q and K To Determine Reaction Direction For the reaction N2O4(g) 2NO2(g), Kc = 0.21 at 100 oC. At a point during the reaction, [N2O4] = 0.12 M and [NO2] = 0.55 M. Has the reaction reached equilibrium? If not, in which direction is it progressing?

Qc =[NO2]2

[N2O4]=

(0.55)2

(0.12) = 2.5 > 0.21

Qc > Kc, therefore the reaction is not at equilibrium and will proceed from right to left, from products to reactants, until Qc = Kc.

The equilibrium constant Kp for the reaction

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?2

2NO2 (g) 2NO (g) + O2 (g)

Kp = 2PNO PO2

PNO2

2

PO2 = Kp

PNO2

2

PNO2

PO2 = 158 x (0.400)2/(0.270)2 = 347 atm

Sample Integrative ProblemConverting Between Kc and Kp

A chemical engineer injects limestone (CaCO3) into the hot flue gas of a coal-burning power plant for form lime (CaO), which scrubs SO2 from the gas and forms gypsum. Find Kc for the following reaction, if CO2 pressure is in atmospheres.

CaCO3(s) CaO(s) + CO2(g) Kp = 2.1x10-4 (at 1000K)

Sample Integrative ProblemConverting Between Kc and Kp

A chemical engineer injects limestone (CaCO3) into the hot flue gas of a coal-burning power plant for form lime (CaO), which scrubs SO2 from the gas and forms gypsum. Find Kc for the following reaction, if CO2 pressure is in atmospheres.

CaCO3(s) CaO(s) + CO2(g) Kp = 2.1x10-4 (at 1000K)

PLAN: We know Kp and can calculate Kc after finding !ngas. R = 0.0821 L*atm/mol*K.

SOLUTION: !ngas = 1 - 0 since there is only a gaseous product and no gaseous reactants.

Kp = Kc(RT)!n

Kc = Kp/(RT)!n = (2.1x10-4)(0.0821 x 1000)-1 = 2.6x10-6

Sample Integrative ProblemMethane (CH4) reacts with hydrogen sulfide to yield H2 and carbon disulfide, a solvent used in manufacturing. What is the value of Kp at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of CH4, 0.25 atm of H2S, 0.52 atm of CS2, and 0.10 atm of H2?

Computing K From Equilibrium Data

Sample Integrative ProblemMethane (CH4) reacts with hydrogen sulfide to yield H2 and carbon disulfide, a solvent used in manufacturing. What is the value of Kp at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of CH4, 0.25 atm of H2S, 0.52 atm of CS2, and 0.10 atm of H2?

Kp = (0.52)(0.10)4/(0.20)(0.25)2 = 4.2 ! 10-3

Note that partial pressure must be in units of atmosphere!

Balance equation

Products raised to stoichiometric coeff divided by reactants raised to stoich coeff!

Plug in the “equilibrium numbers in ATM!

Computing K From Equilibrium Data

K�eq =

[N2O4][NO2]2

=1

Keq

=1

0.212

1. Kc of a reaction in the reverse direction is the reciprocal of the Kc in the forward reaction.

N2O4(g)←−→ 2 NO2(g)

2 NO2(g)←−→ N2O4(g)

2. Kc for a reaction multiplied by a constant is the equilibrium constant raised to the power of the constant.

Now multiply the equation by 2

N2O4(g)←−→ 2 NO2(g)

2 N2O4(g)←−→ 4 NO2(g)

Kc is squared!

Equilibrium laws can be reversed, scaled and added.

3. Equilibrium in Multiple Reactions3. If a reaction can be expressed as the sum of two or more reactions, Kc for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

A + B C + D

C + D E + F

A + B E + F

Kc =`[C][D][A][B]

Kc ‘Kc ‘‘Kc

Kc =[E][F][C][D]``

Kc = K �c ×K”c =

[E][F ][A][B]

=[E][F ][C][D]

× [C][D][A][B]

1. Change the Direction of the Reaction2. Multiply Stoichiometric Coefficients by a Constant

3. Sum Chemical Reactions (Hess’s Law)

Summary What happens to Kc or Kp when we:

aA + bB cC + dD Kc = [C]c[D]d

[A]a[B]b

cC + dD aA + bB K’ = 1/Kc

aA + bB cC + dDn Kc’ = (Kc)n

For a sequence of equilibria, Koverall = K1 x K2 x K3 x !

1.

2.

3.

Sample Integrative ProblemAt 1565 K suppose we have the following elementary reactions giving a net reation (3). What is the equilibrium constant for the overall reaction?

2CO2(g) <==> 2CO(g) + O2(g)2H2O(g) <==> 2H2(g) + O2(g)

CO2(g) + H2 <==> CO(g) + H2O (g)

K2 = 1.3 X 10-13

Sample Integrative ProblemAt 1565 K suppose we have the following elementary reactions giving a net reation (3). What is the equilibrium constant for the overall reaction?

2CO2(g) <==> 2CO(g) + O2(g)2H2O(g) <==> 2H2(g) + O2(g)

CO2(g) + H2 <==> CO(g) + H2O (g)

2H2(g) + O2(g) <==> 2H2O(g) K � =1

1.3× 10−13

2CO2(g) <==> 2CO(g) + O2(g)

2CO2(g) + 2H2 <==> 2CO(g) + 2H2O (g) K’3 = K1 X K’ = 123.07

K3 =√

123.07 = 11.1

Divide equation by 2 ============> Square root of K3

Reverse ==> Take Inverse

Understanding reactions involving N2 and O2, the most abundant gases in air, is essential for solving problems dealing with atmospheric pollution. Here is a reaction sequence between N2 and O2 to form nitrogen dioxide, a toxic pollutant that contributes to photochemical smog.

(a) Show that the Qc for the overall reaction sequence is the same as the product of the Qcs of the individual reactions.

(1) N2(g) + O2(g) 2NO(g) Kc1 = 4.3 x 10-25

(2) 2NO(g) + O2(g) 2NO2(g) Kc2 = 6.4 x 109

(b) Calculate the Kc for the overall reaction.

SOLUTION:

the equilibrium constant, Kc, is 2.4x10-3 at 1000K. If we change the coefficients of the equation, which we’ll call the reference (ref) equation, what are the values of Kc for the following balanced equations?

For the ammonia formation reactionN2(g) + 3H2(g) 2NH3(g)

(a) 1/3N2(g) + H2(g) 2/3NH3(g)(b) NH3(g) 1/2N2(g) + 3/2H2(g)

(a) The reference equation is multiplied by 1/3, so Kc(ref) will be to the 1/3 power. Kc = [Kc(ref)]1/3 = (2.4x10-3)1/3 = 0.13

(b) The reference equation is reversed and halved, so Kc(ref) is to the -1/2 power. Kc = [Kc(ref)]-1/2 = (2.4x10-3)-1/2 = 20.

Why is K Unitless?The real scoop: units of equilibrium constants

Equilibrium constants are really defined in terms of activity, not concentration.

becomes:

Activity is unitless, so K is unitless.

Dilute regimewhere

concentration works fine in

Kc

Must use activities and

activity coefficients as

Kc is non-linear

How Do We Use The Equilibrium Constant?

1. Write the Equilibrium Expression, Kc or Kp from equation.

2. Hess’s Law with K

3. Predict The Direction of the Reaction: Qc vs Kc

3. Using Equilibrium Values, calculate Kc or one unknown

4. Given K, determining the equilibrium concentrations

ORKc = [C]c[D]d[A]a[B]b Kp =

(Pc)c (Pd)d

(Pa)a (Pb)b

Review and Perspective in Equilibrium The Van’t Hoff equation relates the equilibrium constant K1 at T1 to the equilbrium constant K2 at T2.

R = 8.314 J/mol K!H = enthalpy (watch units)

K is a function of temperature! Any K needs a temperature specified.

lnK2

K1=−∆H

◦rxn

R

�1T2− 1

T1

�

The Arrhenius, Clausius-Claperyon and the Van Hoff’t equations take similar mathematical forms.

lnK2

K1=−∆H

◦rxn

R

�1T2− 1

T1

�

lnP2

P1=−∆H

◦vap

R

�1T2− 1

T1

�

lnk2

k1=−Ea

R

�1T2− 1

T1

�

Clausius-Claperyon: vapor pressure and temperature

Arrhenius: rate constant and temperature

Van Hoff’t: equilibrium constant and temperature

R = universal gas constant = 8.314 J/mol K

K1 is the equilibrium constant at T1 and K2 is the equilibrium constant at T2

Stearic acid, nature's most common fatty acid, dimerizes when dissolved in hexane:

2C17H35COOH <=> (C17H35COOH)2; !H°rxn = -172 kJ

The equilibrium constant for this reaction at 28°C is 2900. Estimate the equilibrium constant at 38°C.

lnK2

K1=−∆H

◦rxn

R

�1T2− 1

T1

�

17.5 Solving Equilibrium Problems• Two basic kinds of chemical equilibrium problems

1. Equilibrium quantities are given (concentrations or partial pressures) and we solve for Kc. This is easy--plug and chug.

2. More than one unknown. We use ICE table and algebra to calculate either Kc from initial quantities or we calculate equilibrium concentrations given Kc.

In either case we can use a book-keeping techniqueInitial, Change Equilibrium Method

Let’s try it in an example problem!

RULES FOR ICE BOOK-KEEPING1. Alway write a balanced chemical equation.

2. Make an ICE (initial, change, equilibrium) Table

3. Use as the unknown variable x for one substance and use the stoichiometry to determine relative concentrations of other reactants and products. Think, before choosing the x-variable.

6. Substitute the equilibrium concentrations into the equilibrium expression from the balanced equation.

7. Solve the equation for x using any math methods at your disposal (1. simple squares, 2. quadratic equation, 3. Simplifying assumption.

8. Answer the questions, remember significant figures.

Determining equilibrium concentrations from Kc

A chemical engineer mixes gaseous CH4 and H2O in a 0.32 L flask at 1200 K. At equilibrium, the flask contains 0.26 mol of CO, 0.091 mol of H2, and 0.041 mol of CH4. What is the [H2O] at equilibrium? A Handbook states that Kc = 0.26 for this reaction is:

CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = 0.26

EASIEST TYPE PLUG AND CHUGPLAN: Set up a reaction table, write Kc

[H2O]eq = [CO]eq[H2]eq3

[CH4]eq Kc= 0.53 M= (0.81)(0.28]3

(0.13)(0.26]

= 0.13 M

0.26mol0.32 L

= 0.81 M

0.091 mol0.32 L

= 0.28 M

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

initialchange

equilibrium

?

? ??

? ?

0.041 mol/0.32 L ?

?

?

Kc =[CO][H2]3

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

Determining equilibrium concentrations from initial concentrations and Kc (Perfect Square)Fuel engineers use the extent of the change from CO and H2O to CO2 and H2 to regulate the proportions of synthetic fuel mixtures. If 0.250 mol of CO and 0.250 mol of H2O are placed in a 125 mL flask at 900K, what is the composition of the mixture at equilibrium? At 900K, Kc is 1.56 for this reaction.

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

PLAN:

1) Balance equation and write equilibrium expression2) Set up ICE table, 3) find the concentrations of all species at initial conditions or equilibrium in the problem 4) use algebra to determine equilibrium concentrations and then substitute into a Kc expression.

Kc = 1.56

[CO] = [H2O] = 2.00 - x = 0.89 Mx = 1.11M = [CO2] = [H2]

Initial concentrations must be calculated as M, we have from the data given [CO] = [H2O] = 0.250/0.125L = 2.00M.

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

changeequilibrium

2.00 2.00 0 0-x -x +x +x

2.00 -x 2.00 -x x x

_________________________________________________

+

Kc =[CO2][H2][CO][H2O]

=x2

(2.00− x)(2.00− x)=

x2

(2.00− x)2= 1.56

x

2.00− x=√

1.56 = ±1.25

x = 1.25(2.00− x) = 2.50− 1.25x

2.25x = 2.50

It’s a perfect square and easy to solve!We toss out the negative result!

Same problem....but not a perfect square!Fuel engineers use the extent of the change from CO and H2O to CO2 and H2 to regulate the proportions of synthetic fuel mixtures. If 0.250 mol of CO and 0.125 mol of H2O are placed in a 125 mL flask at 900K, what is the composition of the mixture at equilibrium? At 900K, Kc is 1.56 for this reaction.

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

PLAN: 1)Balance equation, 2)write equilibrium expression, 3)set up ICE table, 4)find the concentrations of all species at initial conditions or equilibrium in the problem 5)use algebra to determine equilibrium concentrations and then substitute into a Kc expression.

Kc = 1.56

Initial concentrations must be calculated as M, we have from the data given [CO] = [H2O] = 0.250/0.125L = 2.00M.

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

changeequilibrium

2.00 1.00 0 0-x -x +x +x

2.00 -x 1.00 -x x x

_________________________________________________

Kc =[CO2][H2][CO][H2O]

=x2

(2.00− x)(1.00− x)=

x2

x2 − 3.00x + 2.00= 1.56

x2 = 1.56(x2 − 3.00x + 2.00)x2 = 1.56x2 − 4.68x + 3.12)

0.56x2 - 4.68x + 3.12 = 0

ax2 + bx + c =0

-b ± b2 – 4ac "2ax =

Kc =[CO2][H2][CO][H2O]

=x2

(2.00− x)(1.00− x)=

x2

x2 − 3.00x + 2.00= 1.56

x2 = 1.56(x2 − 3.00x + 2.00)x2 = 1.56x2 − 4.68x + 3.12)

0.56x2 - 4.68x + 3.12 = 0

ax2 + bx + c =0

-b ± b2 – 4ac "2ax =

- (-4.68 ± (-4.68)2 – 4(0.56)(3.12) "2(0.56)

x =

x = 7.6M and x = 0.73M

x = 7.6M makes 2.00 - x < 0 and makes no sense-toss it

x = 0.73 = [CO2] = [H2] and [CO] = [H2O] = 2.00 - 0.73

PLAN:

Calculating Kc from concentration dataIn a study of hydrogen halide decomposition, a researcher fills an evacuated 2.00 L flask with 0.200 mol of HI gas and allows the reaction to proceed at 453˚C. At equilibrium, it is found that the concentration of [HI] = 0.078 M. What is the equilibrium constant Kc?

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

Find the molar concentration of the starting material (in this case [HI] and then use algebra and the balanced chemical equation to determine the amount of reactants and products at equilibrium.

The problem said that: [HI]eq = 0.078 M therefore:

0.100 - 2x = [HI]eq = 0.078 M Therefore: x = 0.011 M

Kc = [H2] [I2][HI]2

=[0.011][0.011]

[0.078]2= 0.020

2HI(g) H2(g) + I2(g) Kc = [H2] [I2][HI]2

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

initialchange

equilibrium

.200 mol/2.00 L-2x xx

0 0

0.100 - 2x x x+

1. Write down balanced equation and Kc

2. Set up the ICE table (initial, change, equilibrium)

Determining equilibrium concentrations from Kc

If 1.0 mol H2 is allowed to react with 1.0 mol of Cl2 to form hydrogen chloride in a 10.0 L vessel at 700 K, what are concentrations of H2, Br2 and HCl at equilibrium? What is the composition of equilibrium mixture if Kc for the reaction of H2 and Cl2 at 700 K is 57.0?

H2 + Cl2 2 HCl

Initial concentrations [H2] = [Cl2] = 0.10 M

1. Write a balanced equation:

0. What does the problem ask for (equilibrium values)

2. What information is provided?

Determining equilibrium concentrations from Kc, cont’t

H2 + I2 2 HI

Initial [H2] = [I2] = 0.10 M

1. Write a balanced equation:

0. What does the problem ask for (equilibrium values)

2. What information is provided?

Initial (M) 0.10 0.10 0Change (M) -x -x +2xEquilibrium (M) (0.10 – x) (0.10 – x) 2x

H2 + I2 2 HI

Kc =[HI]2

[H2][I2]= 57.0

• Solving for x: two solutions +7.55(0.10-x) = 2x x = 0.755/9.55 = 0.0791 M -7.55(0.10-x) = 2x x = -0.755/5.55 = 0.136 M• Initial concentrations are 0.10 M, so x can’t exceed 0.10 M

• Kc = 57.0 = [HI]2/[H2][I2] = (2x)2/(0.10-x)(0.10-x) = (2x/(0.10-x))2

"57.0 = ± 7.55 = 2x/(0.10-x)

Kc =[HI]2

[H2][I2]= 57.0

Kc =[2x]2

[0.10− x][0.10− x]=

�[2x]

[0.10− x]

�2

= 57.0

Kc =[2x]

[0.10− x]=√

57.0 = ±7.55

Text

H2 + I2 2 HI

• Solving for x: two solutions +7.55(0.10-x) = 2x x = 0.755/9.55 = 0.0791 M -7.55(0.10-x) = 2x x = -0.755/5.55 = 0.136 M• Initial concentrations are 0.10 M, so x can’t exceed 0.10 M

• Equilibrium concentrations from calculated value of x: [H2] = [I2] = 0.10 – x = 0.021 M [HI] = 2x = (2)(0.0791) = 0.158 M

• Check result: Kc = [HI]2/[H2][I2] = (0.158)2/(0.021)(0.021) = 57

• Number of moles in equilibrium mixture: moles of H2 = moles I2 = (0.021 mol/L)(10.0 L) = 0.21 mol moles of HI = (0.158 mol/L)(10.0 L) = 1.58 mol

PLAN:

Calculating equilibrium concentration using a simplfying assumption (no quadratic equation)

The reaction of nitrogen with oxygen giving nitrogen oxide contributes to air pollution whenever a fuel is burnt at high temperature. At 1500K the K = 1.0 X 10-5. Suppose a sample of air has [N2] = 0.80 M, and [O2] = 0.20M before a reaction occurs. Calculate the equilibrium concentrations at equilibrium.

Set up ICE table and begin filling it in.

N2(g) + O2(g) <==> 2NO(g) Kc = 1.0 X 10-5

= 0.020

1. Write down what we solve: Kc and what we know.

2. Set up the ICE table

Kc = [N2] [O2][NO]2 = 1.0 X 10-5

Molarity

initialchange

equilibrium

0.80-x +2x-x

0.20 0

0.80 - x 2x+

N2(g) + O2(g) <=> 2NO(g)

0.20 - x

=[.80 -x]

[2x]2Kc = [N2] [O2][NO]2 = 1.0 X 10-5

[.20 -x]

3. Use equilibrium expression and substitute values.

4. There are 3 cases: (1) perfect square, (2) simplifying assumption (both easy to solve) (3) quadratic equation (harder)

=[.80 -x]

[2x]2Kc = [N2] [O2][NO]2 = 1.0 X 10-5

[.20 -x]

3. Use equilibrium expression and substitute values.

4. We can use a simplifying assumption here: RULE: If 100 x K < [A]0 then .80 - x = .80100 X 10-5 = 10-3 < [A]0 .......it’s ok to rid x.

=Kc = [N2] [O2][NO]2 = 1.0 X 10-5

[.80][2x]2

[.20] [.16][2x]2=

x = (1.0 X 10-5 (.16) 1/2 = 6.3 X 10-4 4

)([N2] = .80 - x = 0.80 - 6.3 X 10-4 = 0.80M[O2] = -.20 - x = 0.20 - 6.3 X 10-4 = 0.20M[NO] = 2x = 6.3 X 10-4 X 2 = 1.3 X 10-3M

Br2 (g) 2Br (g)At 1280˚C the equilibrium constant (Kc) for the reaction

is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Using the Quadratic Formula

Kc is 1.1 X 10-3

Let x be the change in concentration of Br2

[Br]2[Br2]

Kc =

The expression is not a perfect square: we therefore must use quadratic equation to solve it

Br2 (g) 2Br (g)1. Write Balanced Equation

2. Write Equilibrium Expression

3. Set Up ICE TableBr2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

[0.063] [0.012]

-x +2x

0.063 - x 0.012 + 2x

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

[Br]2[Br2]

Kc = =

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x4x2 + 0.0491x + 0.0000747 = 0

-b ± b2 – 4ac "2ax =

x = -0.00178x = -0.0105

At equilibrium, [Br] = 0.012 + 2x = -0.009 M and 0.00844 M

At equilibrium, [Br2] = 0.062 – x = 0.06378 M

ax2 + bx + c =0

Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

[0.063] [0.012]

-x +2x

0.063 - x 0.012 + 2x

Two solutions

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x4x2 + 0.0491x + 0.0000747 = 0

-b ± b2 – 4ac "2ax =

Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

x = -0.00178x = -0.0105

At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 MAt equilibrium, [Br2] = 0.062 – x = 0.0648 M

ax2 + bx + c =0

Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It decomposes by the reaction,

COCl2(g) CO(g) + Cl2(g) Kc = 8.3 x 10-4 (at 360 oC)

Calculate [CO], [Cl2], and [COCl2] at equilibrium when the initial amount of phosgene gas is 0.100 mol in a 10.0 L flask.

[COCl2] = 0.100 mol/10.0 L = 0.0100 M

Kc = 8.3 x 10-4 (at 360 oC)

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

Kc = 8.3 x 10-4 =[CO] [Cl2][COCl2]

Initial (M)

Change (M)

Equilibrium (M)

0.0 0.0

-x +x

0.010 - x x

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

0.010

+x

x

Kc = 8.3 x 10-4 =(x) (x)

(0.010 - x)

(continued) Kc = 8.3 x 10-4 =(x) (x)

(0.010 - x)

Solving for x yields x = 2.5 x 10-3 M and 0.0100 - x = 7.5 x 10-3 M.

Kc = 8.3 x 10-6 - 8.3 x 10-4 x = x2

Kc = 8.3 x 10-4 (0.010 - x) = x2

Kc = x2 + 8.3 x 10-4 x - 8.3 x 10-6 = 0

ax2 + bx + c =0

-b ± b2 – 4ac "2ax =

a = 1b = 8.3 x 10-4 c = - 8.3 x 10-6

[COCl2] [CO] + [Cl2]

(continued) Kc = 8.3 x 10-4 =(x) (x)

(0.010 - x)

Kc = 8.3 x 10-6 - 8.3 x 10-4 x = x2

Kc = 8.3 x 10-4 (0.010 - x) = x2

Kc = x2 + 8.3 x 10-4 x - 8.3 x 10-6 = 0

ax2 + bx + c =0-b ± b2 – 4ac "

2ax =

a = 1b = 8.3 x 10-4 c = - 8.3 x 10-6

4 a c = 4 x 1 x - 8.3 x 10-6

b2 = (8.3 x 10-4)2 = 6.89 X 10-7

c = - 8.3 x 10-6

c = - 8.3 x 10-6 Solving for x yields x = 2.5 x 10-3 M and 0.0100 - x = 7.5 x 10-3 M.

PLAN: Find the initial molar concentrations of all components and use these to calculate Qc. Compare Qc to Kc, determine in which direction the reaction will progress, and draw up expressions for the equilibrium concentrations.

Predicting reaction direction and calculating equilibrium concentrationsThe research and development unit of a chemical company is studying the reaction of CH4 and H2S, two components of natural gas. CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

In one experiment, 1.00 mol of CH4, 1.00 mol of CS2, 2.00 mol of H2S and 2.00 mol of H2 are mixed in a 250 mL vessel at 960 oC. At this temperature, Kc = 0.036.

(a) In which direction will the reaction proceed to reach equilibrium?

(b) If [CH4] = 5.56 M at equilibrium, what are the equilibrium concentrations of the other three substances?

SOLUTION: USE INITIAL CONDITIONS TO DETERMINE Q

Qc = [CS2][H2]4

[CH4][H2S]2= [4.0][8.0]4

[4.0][8.0]2= 64.0

Qc of 64 is >> than Kc = 0.036 The reaction will progress to the left.

[CH4]initial = 1.00 mol/0.25 L = 4.00 M

[H2S]initial = 2.00 mol/0.25 L = 8.00 M

[H2]initial = 2.00 mol/0.25 L = 8.00 M

[CS2]initial = 1.00 mol/0.25 L = 4.00 M

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

SOLUTION: USE AN ICE TABLE TO SOLVE

At equilibrium [CH4] = 5.56 M, so 5.56 = 4.00 + x; thus, x = 1.56 M

Therefore:[H2S] = 8.00 + 2x = 11.12 M

[CS2] = 4.00 - x = 2.44 M

[H2] = 8.00 - 4x = 1.76 M

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)concentrations

initial

change

equilibrium

4.00 8.00 4.00 8.00

+ x + 2x - 4x

4.00 + x 8.00 + 2x

- x

4.00 - x 8.00 - 4x

______________________________________________________

1. Concentration changes2. Pressure or Volume changes (gas)3. Temperature Changes4. Addition of a catalyst (not)

There are 4 Possible Disturbances to consider

Le Châtelier’s PrincipleIf a chemical system at equilibrium experiences a change in concentration, temperature, volume, or partial pressure, then the equilibrium shifts to counteract the imposed change according to the law of mass action.

Le Châtelier’s Principle•1. Changes in Concentration

ChangeShifts the Equilibrium

Increase concentration of product(s) leftDecrease concentration of product(s) right

Decrease concentration of reactant(s)Increase concentration of reactant(s) right

left

aA + bB cC + dD

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

Examine Qc to predict what will happen

Le Chatlier Principles: Effect of a Change in Concentration

To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O2;

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

What happens to(a) [H2O] if O2 is added? (b) [H2S] if O2 is added?

(c) [O2] if H2S is removed? (d) [H2S] if sulfur is added?

PLAN: We can write and compare Q with K when the system is disturbed to see in which direction the reaction will progress.

Q =[H2O]2

[H2S]2[O2]

(b) When O2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H2S] decreases.

(c) When H2S is removed, Q increases and the reaction progresses to the left to come back to K. So [O2] decreases.

(d) Sulfur is not part of the Q (K) expression because it is a solid. Therefore, as long as some sulfur is present the reaction is unaffected. [H2S] is unchanged.

(a) When O2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H2O] increases.

To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O2;

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

Le Chatlier Principles: Effect of a Change in Concentration 2. Pressure and Volume Effects In Gases

1N2O4 (g) 2NO2 (g)

There are two cases for changes in pressure and volume:

1. !n = 0 (mol of product gases - mol of reactant gases = 0)

2. "n # 0

Cl2(g) + I2(g) 2ICl(g) any change in P or V has no affect on the equilibrium!

If pressure is increased (volume decreased) the system will shift towards the side with fewer number of moles. If pressure is decreased (volume increased) then equilibrium shifts towards side with greater moles.

Le Châtelier’s Principle

Summary: Changes in Volume and Pressure For anything change to occur in equilibrium to there must be a difference in the number of moles on each side of the equation!

Stress or Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gasDecrease pressure Side with most moles of gas

Decrease volumeIncrease volume Side with most moles of gas

Side with fewest moles of gas

A (g) + B (g) C (g)

A (g) + B (g) C (g) + D(g)

Change will occur

No Change

N2O4 (g) 2NO2 (g)

Qp = Kp =(PNO2)2

PN2O4Qp =

(2× PNO2)2

(2× PN2O4)= 4Kp > Kp

Equilibrium will shift to the left (note it shifts to the side with fewer number of moles)

Figure 17.8 The effect of pressure (volume) on an equilibrium system.

+

lower P(higher V)

more moles of gas

higher P(lower V)

fewer moles of gas

Le Chatlierʼs Principle: Affect of volume (pressure) on the equilibrium position

How would you change the volume of each of the following reactions to increase the yield of products?

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

(b) S(s) + 3F2(g) SF6(g)(c) Cl2(g) + I2(g) 2ICl(g)

PLAN:

When gases are present, a change in volume will affect the concentration of the gas. If the volume decreases (pressure increases), the reaction will shift to fewer moles of gas and vice versa.

Le Chatlierʼs Principle: Affect of volume (pressure) on the equilibrium position

How would you change the volume of each of the following reactions to increase the yield of products?

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

(b) S(s) + 3F2(g) SF6(g)

(c) Cl2(g) + I2(g) 2ICl(g)

(a) CO2 is the only gas present. To increase its yield, decrease the pressure which = increasing the volume.

(b) There are more moles of gaseous reactants than products, so decrease the volume (increase the pressure) to shift the reaction to the right.

(c) There are an equal number of moles of gases on both sides of the reaction. Therefore, a change in volume will have no effect.

Summary Changes in TemperatureChange Exothermic Rx

Increase temperature K decreasesDecrease temperature K increases

Endothermic Rx

K increasesK decreases

3. Changes in TemperatureThere are two cases. Think of heat as reactant or as product!

2. When ∆H > 0 = Endothermic => think heat is reactant

1. When ∆H < 0 => Exothermic => think heat is a product

heat + A (g) + B (g) C (g) + D (g)

A (g) + B (g) C (g) + D (g) + heat

Kc = [C]c[D]d

[A]a[B]b

What happens to K can be analyzed using mass-action!

#H = + 180.5 kJ

Is the amount of NO(g) formed from the given amounts of N2 and O2 greater at higher of lower temperatures?

N2 (g) + O2 (g) 2NO (g)

TRANSLATION: "H = 180.5 kJ =>ENDO = HEAT IS REACTANT!

Heat + N2 (g) + O2 (g) 2NO (g)

Increasing the temperature by adding more heat to this reaction will increase the amount of NO formed and decrease the reactant side. Kc will increase in value because NO2 increases!

PRODUCT-FAVORED SHIFT

Kc = [NO]2

[N2][O2]

How does an increase in temperature affect the concentration of the underlined substance and Kc for the following reactions?

(a) CaO(s) + H2O(l) Ca(OH)2(aq) !Ho = -82 kJ

(b) CaCO3(s) CaO(s) + CO2(g) !Ho = 178 kJ

(c) SO2(g) S(s) + O2(g) !Ho = 297 kJ

PLAN:

Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on Kc.

How does an increase in temperature affect the concentration of the underlined substance and Kc for the following reactions?

(a) CaO(s) + H2O(l) Ca(OH)2(aq) !Ho = - 82 kJ

(b) CaCO3(s) CaO(s) + CO2(g) !Ho = 178 kJ

(c) SO2(g) S(s) + O2(g) !Ho = 297 kJ

(a) CaO(s) + H2O(l) Ca(OH)2(aq) + heat

Increase in temp will shift reaction left, decrease [Ca(OH)2], and decrease Kc.

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

The reaction will shift right, resulting in an increase in [CO2] and increase in Kc.

(c) SO2(g) + heat S(s) + O2(g)The reaction will shift right, resulting in an decrease in [SO2] and increase in Kc.

For the endothermic reaction, X(g) + Y2 <==> XY(g) + Y(g) the following pictures depict molecular scenes reactions mixtures at various stages in the reaction. Green = X

1) If Kc = 2 at the temperature of the reaction which scene represents the mixture at equilibrium?

2) What will have to happen to the other two scenes to reach equilibrium (shift towards products or towards reactants?)

3) How will a rise in temperature alter the equilibrium [Y2]?

11 2 3

A catalyst lowers Ea for both forward and

reverse reactions and only achieves

equilibrium at a faster rate.

4. Adding a Catalyst to a system at equilibrium• does not shift the position of an equilibrium system• does not alter the value of Kc or Kp

UncatalyzedPathway

CatalyzedPathway

Catalyst lowers Ea for both forward and reverse reactions.

uncatalyzed catalyzed

Catalyst lowers Ea for both forward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

4. Adding a Catalyst• does not change value of Kc • does not shift the position of an equilibrium system• system will reach equilibrium faster (kinetics)

Le Châtelier’s Principle

Summary of Le Châtelier’s Principle

Shift Equilibrium

Concentration yes

Pressure yes

Volume yes

Temperature yes

Catalyst no

Change Kc or Kp

no

no

no

yes

no

Disturbance