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

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 van’t Hoff Equation

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

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


Calculations Example Continued

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


Calculations Overview

• 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


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


36


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

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

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