Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition

Chemical Equilibrium

CHAPTER 15

Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop

2

CHAPTER 15 Chemical Equilibrium

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Learning Objectives:

Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium

Constant (K) Reaction Quotient (Q) Kc vs Kp

ICE Tables Quadratic Formula vs Simplifying Assumptions LeChatelier’s Principle

3



Lecture Road Map:

① Dynamic Equilibrium

② Equilibrium Laws

③ Equilibrium Constant

④ Le Chatelier’s Principle

⑤ Calculating Equilibrium

4Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Dynamic Equilibrium



5

Dynamic Eq Equilibrium

• Chemical equilibrium exists when– Rates of forward and reverse reactions are equal– Reaction appears to stop – Concentration of reactants and products do not

change over time• Remain constant• Both forward and reverse reaction never cease

• Equilibrium signified by double arrows ( )


N2O4 2 NO2

• Initially have only N2O4

– Only forward reaction

– As N2O4 reacts NO2 forms

• As NO2 forms

– Reverse reaction begins to occur

– NO2 collide more frequently as concentration of NO2 increases

• Eventually, equilibrium is reached

– Concentration of N2O4 does not change

– Concentration of NO2 does not change

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 6





N2O4 2NO2

Closed system• Equilibrium can be

reached from either direction

• Independent of whether it starts with “reactants” or “products”

• Always have the same composition at equilibrium under same conditions



N2O4 2NO2

Reactants ProductsEquilibrium

Dynamic Eq Mass Action Expression


• Simple relationship among [reactants] and [products] for any chemical system at equilibrium

• Called the mass action expression– Derived from thermodynamics

• Forward reaction: A B • Reverse reaction: A B • At equilibrium: A B

Dynamic Eq Reaction Quotient


• Uses stoichiometric coefficients as exponent for each reactant

• For reaction: aA + bB cC + dD

Reaction quotient– Numerical value of mass action expression– Equals “Q ” at any time, and– Equals “K ” only when reaction is known to be at

equilibrium

12

Ex. 1 H2(g) + I2(g) 2HI(g) 440˚C

Exp’t Initial Amts

Equil’m Amts

Equil’m [M]

I 1.00 mol H2

0.222 mol H2

0.0222 M H2

10 L

1.00 mol I2

0.222 mol I2

0.0222 M I2

0.00 mol HI

1.56 mol HI 0.156 M HIII 0.00 mol H2

0.350 mol H2

0.0350 M H2

10 L 0.100 mol I2

0.450 mol I2

0.0450 M I2

3.50 mol HI

2.80 mol HI 0.280 M HI

13

Ex. 1 H2(g) + I2(g) 2HI(g) 440 ˚C

Exp’t Initial Amts

Equil’m Amts

Equil’m [M]

III 0.0150 mol H2

0.150 mol H2

0.0150 M H2

10 L 0.00 mol I2 0.135 mol I2 0.0135 M I21.27 mol HI 1.00 mol HI 0.100 M HI

IV 0.00 mol H2 0.442 mol H2

0.0442 M H2

10 L 0.00 mol I2 0.442 mol I2

0.0442 M I2

4.00 mol HI 3.11 mol HI 0.311 M HI

Equilibrium Concentrations (M )

Exp’t [H2] [I2] [HI]

I 0.0222 0.0222 0.156

II 0.0350 0.0450 0.280

III 0.0150 0.0135 0.100

IV 0.0442 0.0442 0.311

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Mass Action Expression

= same for all data sets at equilibrium

4.49)0222.0)(0222.0(

)156.0( 2

8.49)0450.0)(0350.0(

)280.0( 2

4.49)0135.0)(0150.0(

)100.0( 2

5.49)0442.0)(0442.0(

)311.0( 2

Average = 49.5


15

GroupProblem

Write mass action expressions for the following:

• 2NO2(g) N2O4(g)

• 2CO(g) + O2(g) 2CO2(g)

16

GroupProblem


Which of the following is the correct mass action expression for the reaction:

Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)?


Equilibrium Laws



18

Equilibrium Equilibrium Laws

• For reaction

H2(g) + I2(g) 2HI(g) at 440 ˚C

at equilibrium write the following equilibrium law

• Equilibrium constant = Kc = constant at given T

• Use Kc since usually working with concentrations in mol/L

• For chemical equilibrium to exist in reaction mixture, reaction quotient Q must be equal to equilibrium constant, Kc


19

Equilibrium Predicting Equilibrium Laws

For general chemical reaction:• dD + eE fF + gG

– Where D, E, F, and G represent chemical formulas– d, e, f, and g are coefficients

• Mass action expression is

• Note: Exponents in mass action expression are stoichiometric coefficients in balanced equation.

• Equilibrium law is:


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Equilibrium Predicting Equilibrium Laws

• Only concentrations that satisfy this equation are equilibrium concentrations

• Numerator– Multiply concentration of products raised to their

stoichiometric coefficients• Denominator

– Multiply concentration reactants raised to their stoichiometric coefficients

is scientists’ convention


21

Equilibrium Example

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

Kc = 4.26 × 108 at 25 °C

What is equilibrium law?


22

Equilibrium Operations

Various operations can be performed on equilibrium expressions

1. When direction of equation is reversed, new equilibrium constant is reciprocal of original

A + B C + D

C +D A + B


23


1. When direction of equation is reversed, new equilibrium constant is reciprocal of original

3H2(g) + N2(g) 2 NH3(g) at 25˚C

2NH3(g) 3H2(g) + N2(g) at 25 ˚C

8

23

2

23 1026.4

][N][H

][NHcK


24


2. When coefficients in equation are multiplied by a factor, equilibrium constant is raised to a power equal to that factor.

A + B C + D

3A + 3B 3C + 3D


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2. When coefficients in equation are multiplied by factor, equilibrium constant is raised to power equal to that factor

3H2(g) + N2(g) 2NH3(g) at 25 ˚C

Multiply by 3

9H2(g) + 3N2(g) 6NH3(g)


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3. When chemical equilibria are added, their equilibrium constants are multiplied

A + B C + D

C + E F + G

A + B + E D + F + G

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3. When chemical equilibria are added, their equilibrium constants are multiplied

][CO][NO]][CO[NO

][NO

][NO][NO

3

222

2

3

2 NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

NO2(g) + CO(g) NO(g) + CO2(g)

Therefore

][CO][NO][NO][CO

2

2


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GroupProblem For: N2(g) + 3H2(g) 2NH3(g)

Kc = 500 at a particular temperature.

What would be Kc for following?

• 2NH3(g) N2(g) + 3H2(g)

• 1/2N2(g) + 3/2H2(g) NH3(g)

22.4

0.002


Equilibrium Constant



30

Equilibrium Constant Kc

• Most often Kc is expressed in terms of a ratio of concentrations of products and reactants as shown on previous slides

• Sometimes partial pressures, in atmospheres, may be used in place of concentrations


31

Equilibrium Kp

• Based on reactions in which all substances are gaseous

• Gas quantities are expressed in atmospheres in mass action expression

• Use partial pressures for each gas in place of concentrations

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


32

Equilibrium Relationship between Kp and Kc

• Start with ideal gas law

PV = nRT• Rearranging gives

• Substituting P/RT for molar concentration into Kc results in pressure-based formula

• ∆n = moles of gas in product – moles of gas in reactant


33

GroupProblem

Consider the reaction: 2NO2(g) N2O4(g)

If Kp = 0.480 for the reaction at 25 ˚C, what is value of Kc at same temperature?

n = nproducts – nreactants = 1 – 2 = –1

Kc = 11.7


34

GroupProblem

Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25 ˚C, what would be the Kp?

A. 0.99

B. 2.0

C. 24

D. 2400

E. None of these

Δn = (4 – 3) = 1

Kp = Kc(RT)Δn

Kp= 0.99 × (0.082057 × 298.15)1

Kp = 24


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Equilibrium Homogeneous and Hetergeneous

Homogeneous reaction/equilibrium– All reactants and products in same phase– Can mix freely

Heterogeneous reaction/equilibrium– Reactants and products in different phases– Can’t mix freely– Solutions are expressed in M– Gases are expressed in M

– Governed by Kc


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Equilibrium Heterogeneous

2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

• Equilibrium Law =

• Can write in simpler form• For any pure liquid or solid, ratio of moles to

volume of substance (M ) is constant– e.g. 1 mol NaHCO3 occupies 38.9 cm3

2 mol NaHCO3 occupies 77.8 cm3

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2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

– Ratio (n/V ) or M of NaHCO3 is constant (25.7 mol/L) regardless of sample size

– Likewise can show that molar concentration of Na2CO3 solid is constant regardless of sample size

• So concentrations of pure solids and liquids can be incorporated into equilibrium constant, Kc

• Equilibrium law for heterogeneous system written without concentrations of pure solids or liquids


38


Write equilibrium laws for the following:

Ag+(aq) + Cl–(aq) AgCl(s)

H3PO4(aq) + H2O H3O+(aq) + H2PO4

–(aq)

39

Interpreting KC

• Large K (K >>1)–Means product rich

mixture–Reaction goes far

toward completione.g.2SO2(g) + O2(g)

2SO3(g)

Kc = 7.0 1025 at 25 °C

40

Interpreting KC

• Small K (K << 1)–Means reactant rich

mixture–Only very small

amounts of product formed

e.g. H2(g) + Br2(g)

2HBr(g) Kc = 1.4 10–21 at 25 °C

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Interpreting KC

• K 1–Means product and

reactant concentrations close to equal

–Reaction goes only about halfway

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• Size of K gives measure of how reaction proceeds

• K >> 1 [products] >> [reactants]

• K = 1 [products] = [reactants]• K << 1 [products] <<

[reactants]


Le Chatelier’s Principle


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Le Chatelier Definition


• Equilibrium positions – Combination of concentrations that allow Q = K– Infinite number of possible equilibrium positions

• Le Châtelier’s principle– System at equilibrium (Q = K) when upset by

disturbance (Q ≠ K) will shift to offset stress• System said to “shift to right” when forward reaction

is dominant (Q < K) • System said to “shift to left” when reverse direction

is dominant (Q > K)


• Q = K reaction at equilibrium

• Q < K reactants go to products– Too many reactants– Must convert some reactant to product to

move reaction toward equilibrium

• Q > K products go to reactants– Too many products– Must convert some product to reactant to

move reaction toward equilibrium

Le Chatelier Q & K Relationships


Le Chatelier Change in Concentration

Cu(H2O)42+(aq) + 4Cl–(aq) CuCl4

2–(aq) + 4H2O

blue yellow• Equilibrium mixture is blue-green

• Add excess Cl– (conc. HCl)– Equilibrium shifts to products

– Makes more yellow CuCl42–

– Solution becomes green



Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O

blue yellow

• Add Ag+ – Removes Cl–: Ag+(aq) + Cl–(aq) AgCl(s)– Equilibrium shifts to reactants

– Makes more blue Cu(H2O)42+

– Solution becomes increasingly more blue

• Add H2O?


Le Chatelier Change in Concentration: Example

For the reaction 2SO2(g) + O2(g)

2SO3(g)

Kc = 2.4 × 10–3 at 700 °C

Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture?

A.Towards the products

B.Towards the reactants

C.No change will occur



• When changing concentrations of reactants or products– Equilibrium shifts to remove reactants or products that

have been added– Equilibrium shifts to replace reactants or products that

have been removed


Le Chatelier Change in Pressure or Volume

• Consider gaseous system at constant T and n

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

• If volume is reduced– Expect pressure to increase– To reduce pressure, look at each side of reaction– Which has less moles of gas– Reactants = 3 mol + 1 mol = 4 mol gas– Products = 2 mol gas– Reaction favors products (shifts to right)



Consider gaseous system at constant T and n

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

• If pressure is increased, what is the effect on equilibrium?

– nreactant = 1 + 1 = 2

– nproduct = 2

– Predict no change or shift in equilibrium



2NaHSO3(s) NaSO3(s) + H2O(g) + SO2(g)

• If you decrease volume of reaction, what is the effect on equilibrium?– Reactants: All solids, no moles gas– Products: 2 moles gas– Decrease in V, causes an increase in P– Reaction shifts to left (reactants), as this has fewer

moles of gas



• Reducing volume of gaseous reaction mixture causes reaction to decrease number of molecules of gas, if it can– Increasing pressure

• Moderate pressure changes have negligible effect on reactions involving only liquids and/or solids– Substances are already almost incompressible

• Changes in V, P and [X ] effect position of equilibrium (Q), but not K


Le Chatelier Change in Temperature

Ice water

Boiling water

Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O

blue yellow– Reaction endothermic– Adding heat shifts equilibrium toward products– Cooling shifts equilibrium toward reactants



Hf°=+6 kJ (at 0 °C)

– Energy + H2O(s) H2O(l )

– Energy is reactant– Add heat energy, shift reaction right

3H2(g) + N2(g) 2NH3(g) Hrxn= –47.19 kJ

– 3 H2(g) + N2(g) 2 NH3(g) + energy

– Energy is product– Add heat, shift reaction left

H2O(s) H2O(l)



• Increase in temperature shifts reaction in direction that produces endothermic (heat absorbing) change

• Decrease in temperature shifts reaction in direction that produces exothermic (heat releasing) change



• Changes in T change value of mass action expression at equilibrium, so K changed– K depends on T– Increase in temperature of exothermic reaction

makes K smaller• More heat (product) forces equilibrium to

reactants– Increase in temperature of endothermic reaction

makes K larger• More heat (reactant) forces equilibrium to

products


Le Chatelier Change with Catalyst

• Catalyst lowers Ea for both forward and reverse reaction

• Change in Ea affects rates k r and k f equally

• Catalysts have no effect on equilibrium


Le Chatelier Addition of Inert Gas at Constant Volume

Inert gas – One that does not react with components of reaction

e.g. argon, helium, neon, usually N2

• Adding inert gas to reaction at fixed V (n and T), increase P of all reactants and products

• Since it doesn’t react with anything– No change in concentrations of reactants or products– No net effect on reaction


Le Chatelier How To Use Le Chatelier’s Principle

1. Write mass action expression for reaction

2. Examine relationship between affected concentration and Q (direct or indirect)

3. Compare Q to K– If change makes Q > K, shifts left– If change makes Q < K, shifts right– If change has no effect on Q, no shift expected


GroupProblem

Consider:

H3PO4(aq) + 3OH–(aq) 3H2O(l) + PO43–(aq)

What will happen if PO43– is removed?

Q is proportional to [PO43–]

Decrease [PO43–], decrease in Q

Q < K equilibrium shifts to right


GroupProblem

The reaction

H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq)

is exothermic.

What will happen if system is cooled?

Since reaction is exothermic, heat is product Heat is directly proportional to Q Decrease in T, decrease in Q Q < K equilibrium shifts to right

heat


GroupProblem

The equilibrium between aqueous cobalt ion and the chlorine ion is shown:

[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O

pink blue

It is noted that heating a pink sample causes it to turn violet.

The reaction is:

A. endothermic

B. exothermic

C. cannot tell from the given information


GroupProblem

The following are equilibrium constants for the reaction of acids in water, Ka. Which reaction proceeds the furthest to products?

A. Ka = 2.2 × 10–3

B. Ka = 1.8 × 10–5

C. Ka = 4.0 × 10–10

D. Ka = 6.3 × 10–3


Calculating Equilibrium



Calculations Overview

• For gaseous reactions, use either KP or KC

• For solution reactions, must use KC

• Either way, two basic categories of calculations

1. Calculate K from known equilibrium concentrations or partial pressures

2. Calculate one or more equilibrium concentrations or partial pressures using known KP or KC


Calculations Kc with Known Equilibrium Concentrations

• When all concentrations at equilibrium are known– Use mass action expression to relate

concentrations to KC

• Two common types of calculationsA. Given equilibrium concentrations, calculate KB. Given initial concentrations and one final

concentration• Calculate equilibrium concentration of

all other species• Then calculate K


Calculations

Ex. 3 N2O4(g) 2NO2(g)

• If you place 0.0350 mol N2O4 in 1 L flask at equilibrium, what is KC?

• [N2O4]eq = 0.0292 M

• [NO2]eq = 0.0116 M

KC = 4.61 10–3

Kc with Known Equilibrium Concentrations


For the reaction: 2A(aq) + B(aq) 3C(aq) the equilibrium concentrations are: A = 2.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of Kc at this temperature?

A. 14

B. 0.15

C. 1.5

D. 6.75

GroupProblem


Calculations

Ex. 4 2SO2(g) + O2(g) 2SO3(g)

At 1000 K, 1.000 mol SO2 and 1.000 mol O2 are placed in a 1.000 L flask. At equilibrium 0.925 mol SO3 has formed. Calculate K C for this reaction.

• First calculate concentrations of each– Initial

– Equilibrium

Kc with Known Equilibrium Concentrations


Calculations Example Continued

• Set up concentration table– Based on the following:

• Changes in concentration must be in same ratio as coefficients of balanced equation

• Set up table under balanced chemical equation– Initial concentrations

• Controlled by person running experiment– Changes in concentrations

• Controlled by stoichiometry of reaction– Equilibrium concentrations

EquilibriumConcentration = Initial

Concentration– Change in

Concentration



2SO2(g)

+ O2(g) 2SO3(g)

Initial Conc. (M) 1.000 1.000

0.000

Changes in Conc. (M)

Equilibrium Conc. (M)

[SO2] consumed = amount of SO3 formed

= [SO3] at equilibrium = 0.925 M

[O2] consumed = ½ amount SO3 formed = 0.925/2 = 0.462 M

[SO2] at equilibrium = 1.000 – 0.975 = 0.075

[O2] at equilibrium = 1.00 – 0.462 = 0.538 M

–0.925 –0.462 +0.925

0.075 0.538 0.925



• Finally calculate KC at 1000 K

][O][SO

][SO

22

2

23

c K

]538.0[]075.0[

]925.0[2

2cK

Kc = 2.8 × 102 = 280


Calculations ICE Table Summary

ICE tables used for most equilibrium calculations:1. Equilibrium concentrations are only values used in mass

action expression Values in last row of table

2. Initial value in table must be in units of mol/L (M) [X]initial = those present when reaction prepared

No reaction occurs until everything is mixed

3. Changes in concentrations always occur in same ratio as coefficients in balanced equation

4. In “change” row be sure all [reactants] change in same directions and all [products] change in opposite direction. If [reactant]initial = 0, its change must be an increase (+) because

[reactant]final cannot be negative

If [reactants] decreases, all entries for reactants in change row should have minus sign and all entries for products should be positive


Calculations Calculate [X ]equilibrium from Kc and [X ]initial

• When all concentrations but one are known

– Use mass action expression to relate Kc and known concentrations to obtain missing concentrations

Ex. 5 CH4(g) + H2O(g) CO(g) + 3H2(g)

• At 1500 °C, Kc = 5.67. An equilibrium mixture of gases had the following concentrations: [CH4] = 0.400 M and [H2] = 0.800 M and [CO] = 0.300 M.

What is [H2O] at equilibrium ?


Calculations

Ex. 5 CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = 5.67

[CH4] = 0.400 M; [H2] = 0.800 M; [CO] =0.300 M

• What is [H2O] at equilibrium?

• First, set up equilibrium

• Next, plug in equilibrium concentrations and Kc

[H2O] = 0.0678 M

27.2154.0

5.67)([0.400]800][0.300][0.

O][H3

2

Calculate [X ]equilibrium from Kc and [X ]initial


Calculations Calculating [X ]Equilibrium from Kc

When Initial Concentrations Are Given

• Write equilibrium law/mass action expression• Set up Concentration table

– Allow reaction to proceed as expected, using “x” to represent change in concentration

• Substitute equilibrium terms from table into mass action expression and solve


Calculations Calculate [X]equilibrium from [X]initial and KC

Ex. 6 H2(g) + I2(g) 2HI(g) at 425 ˚C

KC = 55.64

If one mole each of H2 and I2 are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?

Step 1. Write Equilibrium Law


Calculations

Step 2: Construct an ICE table

• Initial [H2] = [I2] = 1.00 mol/0.500 L =2.00 M

• Amt of H2 consumed = Amt of I2 consumed = x

• Amount of HI formed = 2x

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

Initial 2.00 2.00 0.000

Change

Equilibrium

– x +2x– x

+2x2.00 – x 2.00 – x

Calculate [X]equilibrium from [X]initial and KC


Calculations

Step 3. Solve for x• Both sides are squared so we can take square root of both

sides to simplify

)00.2(2

459.7x

x

xx 2)00.2(459.7

xx 2459.7918.14

58.1459.9918.14 x

x459.9918.14



Calculations

Step 4. Equilibrium Concentrations

• [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M

• [HI]equil = 2x = 2(1.58) = 3.16


Initial 2.00 2.00 0.00

Change

Equilibrium

– 1.58 +3.16– 1.58

+3.160.42 0.42




Ex. 7 H2(g) + I2(g) 2HI(g) at 425 ˚C

KC = 55.64

• If one mole each of H2, I2 and HI are placed in a 0.500 L flask at 425 ˚C, what are the equilibrium concentrations of H2, I2 and HI?

• Now have product as well as reactants initially



Calculations


Initial 2.00 2.00 2.00

Change

Equil’m

– x +2x– x

2.00 + 2x2.00 – x 2.00 – x

2

22

)00.2(

)200.2()00.2)(00.2(

)200.2(64.55

x

xxx

x

2

2

)00.2(

)200.2(64.55

x

xK

Step 2. Concentration Table




)00.2(200.2

459.7xx

xx 200.2)00.2(459.7

xx 200.2459.7918.14

37.1459.9918.12

x

x459.9918.12

Step 3. Solve for x

[H2]equil = [I2]equil = 2.00 – x

= 2.00 – 1.37 = 0.63 M

[HI]equil = 2.00 + 2x = 2.00 + 2(1.37) = 2.00 + 2.74 = 4.74 M


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

Kc = 0.0123 at 3900 ˚C

If 0.25 moles of N2 and O2 are placed in a 250 mL container, what are the equilibrium concentrations of all species?

A. 0.0526 M, 0.947 M, 0.105 M

B. 0.947 M, 0.947 M, 0.105 M

C. 0.947 M, 0.105 M, 0.0526 M

D. 0.105 M, 0.105 M, 0.947 M

GroupProblem


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

• Initial 1.00 1.00 0.00• Change – x – x + 2x• Equil 1.00 – x 1.00 – x + 2x

GroupProblem



Example: Quadratic Equation

Ex. 8

CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) + H2O(l)acetic acid ethanol ethyl acetate

KC = 0.11

An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?




Step 1. Write equilibrium law

• Need to find equilibrium values that satisfy this

Step 2: Set up concentration table using “x” for unknown– Initial concentrations– Change in concentrations– Equilibrium concentrations

11.0H]COOH][CHH[C

]HCCO[CH

2352

5223 cK




Step 2 Concentration Table

• Amt of CH3CO2H consumed = Amt of C2H5OH consumed = – x

• Amt of CH3CO2C2H5 formed = + x

• [CH3CO2H]eq and [C2H5OH ] = 0.810 – x

• [CH3CO2C2H5] = x

(M) CH3CO2H(aq) +

C2H5OH(aq)

CH3CO2C2H5(aq) + H2O(l)

I 0.810 0.810 0.000

C

E

–x +x– x

+x0.810 – x 0.810 – x




Step 3. Solve for x• Rearranging gives

• Then put in form of quadratic equation

ax2 + bx + c = 0

• Solve for the quadratic equation using

xxx )62.16561.0(11.0 2

011.01782.007217.0 2 xxx

007217.01782.111.0 2 xx

aacbb

x2

42



Example: Quadratic Equation Step 3. Solve for x

• This gives two roots: x = 10.6 and x = 0.064• Only x = 0.064 is possible

– x = 10.6 is >> 0.810 initial concentrations – 0.810 – 10.6 = negative concentration,

which is impossible

)11.0(2)07217.0)(11.0(4)1782.1()1782.1( 2

x

22.0164.11782.1

22.0)032.0()388.1(1782.1

x




Step 4. Equilibrium Concentrations

[CH3CO2C2H5]equil = x = 0.064 M

[CH3CO2H]equil = [C2H5OH]equil = 0.810 M – x = 0.810 M – 0.064 M = 0.746 M

CH3CO2H(aq) +

C2H5OH(aq)

CH3CO2C2H5(aq) + H2O

I 0.810 0.810 0.000

C

E

–0.064 +0.064– 0.064

+0.0640.746 0.746



Example: Cubic

When KC is very smallEx. 9 2H2O(g) 2H2(g) + O2(g)

• At 1000 °C, KC = 7.3 10–18

• If the initial H2O concentration is 0.100 M, what will the H2 concentration be at equilibrium?


182

2

22

2 103.7O][H

][O][H cK



Example: CubicStep 2. Concentration Table

• Cubic equation – tough to solve• Make approximation

– KC very small, so x will be very small– Assume we can neglect x – Must prove valid later

Conc (M ) 2H2O(g)

2H2(g)

+O2(g)

Initial 0.100 0.00 0.00

Change

Equil’m

– 2x +x+2x

+x+2x 0.100 – 2x

2

3

2

218

)2100.0(

4

)2100.0(

)2(103.7

x

x

x

xx



Example: Cubic

Step 3. Solve for x• Assume (0.100 – 2x) 0.100

• Now our equilibrium expression simplifies to

Conc (M) 2H2O (g) 2H2 (g)

+ O2 (g)

Initial 0.100 0.00 0.00

Change

Equil’m

– 2x +x+2x+x+2x 0.100

010.04

)100.0(

)2(103.7

3

2

218 xxx

)103.7(010.04 183 x = 7.3 × 10–20


96


Example: Cubic

Step 3. Solve for x

• Now take cube root

• x is very small • 0.100 – 2(2.6 10–7) = 0.09999948 • Which rounds to 0.100 (3 decimal places)

• [H2] = 2x = 2(2.6 10–7) = 5.2 10–7 M

2020

3 108.14103.7

x

73 20 106.2108.1 x


Calculations Simplifications: When Can You Ignore x In Binomial (Ci – x)?

• If equilibrium law gives very complicated mathematical problems and if K is small– Then the change (x term) will also be small and we can

assume it can be ignored when added or subtracted from the initial concentration, Ci.

• How do we check that the assumption is correct?– If the calculated x is so small it does not change the

initial concentration

(e.g. 0.10 Minitial – 0.003 Mx-calc = 0.10)

– Or if the answer achieved by using the assumption differs from the true value by less than five percent. This often occurs when Ci > 100 x Kc


For the reaction 2A(g) B(g)

given that Kp = 3.5 × 10–16 at 25 ˚C, and we place 0.2 atm A into the container, what will be the pressure of B at equilibrium?

2A B

I 0.2 0 atm

C –2x +x

E 0.2 – 2x x ≈0.2 x = 1.4 × 10–17

[B]= 1.4 × 10–17 atm

Proof: 0.2 - 1.4 × 10–17 = 0.2

GroupProblem

Documents

Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition