Unit A: EquilibriumGeneral Outcomes:• Explain that there is a balance of opposing reactions in chemical equilibrium systems.

• Determine quantitative relationships in simple equilibrium systems.

Unit Introduction Read p. 670

• Chemistry involves the study of change with chemicals.

• Lots happening at a molecular level.

• chemical rxn equilibrium is always a dynamic balance between opposing changes

• rxns occur at different but equivalent rates within a closed system

• observe net effect directly

• i.e. measurable property does not fluctuate

• Unit A explores the nature of dynamic equilibrium in chemical systems by focussing on

acid-base solutions, allowing you to describe, explain, and predict new chemical systems

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Prior Knowledge Activityp. 672 #1-7

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EQUILIBRIUM SYSTEMS

Chapter 15

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• Static Equilibrium – no movement b/c opposing forces act simultaneously

• Example: book sitting on a level desk• gravity pulls down and desk pushes up

• Chemical Equilibrium always in a dynamic system since balance exists between

opposing agents of change (opposite processes occurring at the same rate) even

though entities are too small to see

• Example: juggler – some objects move upwards and some downwards• No net change – rate of upward and downward movement equal at any given time

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Today’s Objectives

• Define equilibrium and state the criteria that apply to a chemical system in equilibrium

• i.e. closed system, constancy of properties, as well as equal rates of forward and reverse reactions.

Section 15.1 (pp. 676-689)

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Closed Systems At Equilibrium

• matter cannot enter or leave the system

• Example: test tube or beaker (no gas produced in rxn)

• easier to study chemical properties when separate rxn from its surroundings.

• The use of a controlled system is an integral part of scientific study

• Example: soft drink in closed bottle (closed system in equilibrium) See Figure 1 – p. 676

• nothing appears to change until opened where reduced pressure releases CO2(g)

• Can re-carbonate a flat drink by adding pressurized CO2(g) (reverse rxn)

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Closed Systems At Equilibrium

• Fundamental Collision-Reaction Theory

• initially used four assumptions for Stoich calculations

• fast, spont, quant, stoich

• Not always true

• Example: corrosion not instantaneous

• Chapter 15 examines assumption of quantitative reaction

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INVESTIGATION 15.1p. 700Prelab

Discuss procedure

• determine quantity to mix

• same concentrations used

• select different volumes to react

• include choice in prediction

• Record Observations using ink

• submit original observations sheet with report (initialed by instructor)

• informal lab report

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HOMEWORK

• INV 15.1 Prelab – p. 700

• prepare procedure and observation table

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INVESTIGATION 15.1p. 700Post-lab

Discuss anomaly that occurred

• unexpected result that contradicts previous rules or experience

• Direct evidence of both reactants present after rxn appears to have stopped

• i.e. rxn not quantitative

Read “Did You Know?” – p. 677 (Anomalies – Signals for Change)

Discuss Practice Q – p. 677 #2

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HOMEWORK

• INV 15.1 Informal Report – p. 700

• DUE: W, Sept 23 by 4pm

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Today’s Objectives

• Define equilibrium and state the criteria that apply to a chemical system in equilibrium

• i.e. closed system, constancy of properties, as well as equal rates of forward and reverse reactions

• Identify, write, and interpret chemical equations for systems at equilibrium.

• Define Kc to predict the extent of the reaction and write equilibrium-law expressions for given chemical equations, using lowest whole-number coefficients.

Section 15.1 (pp. 676-689)

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Chemical Equilibrium

• Not all reactions are quantitative (reactants products)

• Evidence: For many reactions reactants are present even after the reaction appears to have stopped

• Recall the necessary conditions for a chemical reaction:

• Particles must collide with the correct orientation and have sufficient energy

• If product particles can collide effectively, a reaction is said to be reversible

• Reaction Rate depends on temperature, surface area, and concentration

>99%

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Chemical Equilibrium

• Consider the following reversible reaction:

• The final state of this chemical system can be explained as a competition between:

• Assuming that this system is closed (so the reactants and products cannot escape)

and will eventually reach a Dynamic Equilibrium • opposing changes occur simultaneously at the same rate

The collisions of reactants

to form products (forward)

The collisions of products to

re-form reactants (reverse)

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Chemical Equilibrium

• INV 15.1 anomaly explained by this theory with the idea that reverse rxn occurs, where products react to re-form original reactants

• Equilibrium established from the balance between forward (written left to right)

and reverse rxns (written right to left) competing for collisions

• use double headed, equilibrium arrow ⇌

• the final state of this chemical system can be explained as a competition between collisions of reactants to form products and collisions of products to re-form reactants

• This competition requires a closed system

• here bound by the volume of the liquid phase in the beaker

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Chemical Equilibrium• Consider the following hypothetical system:

AB + CD AD + BC forward reaction, therefore

AD + BC AB + CD reverse reaction

• Initially, only AB and CD are present. • The forward reaction is occurring exclusively at its highest rate.

• As AB and CD react, their concentration decreases. • This causes the reaction rate to decrease as well.

• As AD and BC form, the reverse reaction begins to occur slowly.

• As AD and BC’s concentration increases, the reverse reaction speeds up.

• Eventually, both the forward and reverse reaction occur at the same rate ⟹ Dynamic Equilibrium

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Conditions of Dynamic Equilibrium

1. Can be achieved in all reversible reactions when the rates of the forward and reverse reaction become equal

• Represented by ⇌ rather than by

2. All observable properties appear constant

• colour, pH, etc.

3. Can only be achieved in a closed system

• no exchange of matter and must have a constant temperature

4. Equilibrium can be approached from either direction.

• This means that the equilibrium concentrations will be the same regardless if you started with all reactants, all products, or a mixture of the two

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• Equilibrium reached in closed chemical system

(no matter in or out) with constant macroscopic properties

(no observable change occurring, i.e. always looks the same)

despite continued microscopic changes

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Types Of Equilibrium

• all types are explained by a theory of dynamic equilibrium

• Phase Equilibrium single chemical substance exists in more than one phase, or state of matter, in a closed system

• Example: water in a sealed container evaporates until reaches the maximum vapour pressure, assuming that is remains at a constant temperature (See Figure 2 – p. 677)

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Types Of Equilibrium

• Solubility Equilibrium single chemical solute interacting with solvent, where excess solute is in contact with the saturated solution (See Figure 3 – p. 677)

• i.e. a saturated solution where rate of dissolving = rate of recrystallization

• Chemical Reaction Equilibrium involves several substances

• reactants and products in a closed system

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Modelling Dynamic Equilibrium

• Mini Investigation – p. 678

Cylinder #1

Volume (mL)

Cylinder #2

Volume (mL)

25.0 0.0

20.0 5.0

17.0 8.0

14.0 11.0

11.0 14.0

8.0 17.0

5.0 20.0

2.0 23.0

2.0 23.0

2.0 23.0

2.0 23.0

Assume that each time the large straw transfers

5 mL and the smaller straw transfers 2 mL.

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Chemical Reaction Equilibrium

• Combines ideas from different theories (atomic, kinetic molecular, collision-reaction) as well as the concepts of reversibility and dynamic equilibrium

• Describes and explains, but limited in predicting quantitative properties of equilibrium systems

Read “Did You Know?” – p. 680 (Lavoisier and Closed Systems)

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Hydrogen-Iodine Reaction Equilibrium System Figure 4 – p. 679

• Studied extensively b/c simple structure molecules and gas phase rxn

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NOTE: use term reagent instead of

reactant b/c of rxn reversibility

Hydrogen-Iodine Reaction Equilibrium System Figure 4 – p. 679

• rxn rapid at first, then dark purple color of iodine vapor gradually fades and remains constant

• Table 1 compiles data from three experiments that manipulate concentrations of the initial products and reactants, as well as react at 448C so that the system quickly reaches an observable equilibrium each time.

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Hydrogen-Iodine Reaction Equilibrium System Figure 4 – p. 679

• quantitatively graph the reaction progress by plotting the concentration/quantity of the reagents versus time (Figure 5)

• use data to describe equilibrium as a percent rxn and by the equilibrium constant• Percent Reaction describes equilibrium for only one specific system example• Equilibrium Constant describes all systems of the same reaction, at a given temperature

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The rate of reaction of the

reactants decreases as the

number of reactant molecules

decrease. The rate that the

product turns back to reactants

increases as the number of

product molecules increases.

These two rates become equal

at some point, after which the

quantity of each will not change.

Percent Yield

• Percent YIELD product quantity at equilibrium compared to theoretical maximum

• Synonymous with percent reaction

• useful for communicating the equilibrium position

• Maximum possible yield is calculated using stoich

• theoretical value

• assume quantitative forward rxn with no reverse rxn

• Use %Yield to reference quantities of chemicals present in equilibrium system

• write %Yield above equilibrium arrows (⇌) in chemical equation

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Percent Yield

Example: If 2.50 mol of hydrogen gas reacts with 3.0 mol of iodine gas in a 1.00L vessel, what is the percent yield if 3.90 mol of hydrogen iodide is present at equilibrium

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%𝑌𝑖𝑒𝑙𝑑 =𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙

𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑚𝑎𝑥× 100%

2.5 𝑚𝑜𝑙 H2 ×2 𝑚𝑜𝑙 HI

1 𝑚𝑜𝑙 H2= 5𝑚𝑜𝑙 HI

⇒ % 𝑌𝑖𝑒𝑙𝑑 =3.9 𝑚𝑜𝑙

5 𝑚𝑜𝑙× 100% = 78%

• Scientist consider all rxns occurring in both directions

fwd

Reactants ⇌ Products

rev

• Think of all chemical equations as two sets of reagents that create two opposing reactions with different %Yields at equilibrium

• Reactants ≈ Products ⇒ both reagents

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Categories of Equilibria

• non-spont rxn strongly favors reactants (%Yield << 1%)

• Quantitative rxn favors products (%Yield >> 99%)

• observed to be complete

• generally written as a single arrow b/c reverse reaction negligible

• technically not wrong with have equilibrium arrow

• Redefine understanding of quantitative:

• rxn >99% complete, meaning that the reverse rxn happens so little that it can be ignored for all normal purposes since reverse rxn is non-spont

• consider for our purposes that if a rxn is quantitative, then (arbitrarily) assume that less than 0.1% of the original reactants remain un-reacted at equilibrium

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– p. 680

• if no L.R. then can NOT assume complete (quantitative) rxn

• stoich calculations differ slightly

• arrange values in an ICE table (initial, change, equilibrium)

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Categories of Equilibria

Sample Problem 15.1p. 681

Demonstrate

• all substances gas, therefore can apply mole ratio for stoich calculations with concentrations instead of chemical amounts (moles) b/c same volume for every gaseous substance in the closed container

• i.e. closed system ∴ volume common factor

• same applies to rxn with all (aq) entities dissolved in the same volume of solvent

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Example: Analyze Trial 2 data from Table 1 – p. 679

• balance rxn

• set up ICE table beneath

• solve for the change

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HOMEWORK

• Practice Qs - p. 682 #3-7

• Lab Exercise 15.A – p. 683 (Synthesis of Equilibrium Law)

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Equilibrium Constant, Kc

• Formally known as Keq

• Mathematical relationship between amount concentrations (mol/L) in a rxn

• constant value derived for a chemical system over a range of amount concentrations

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Equilibrium Lawp. 684

• generalized expression derived from analysis of numerous equilibrium systems• upper case = entities

• lower case = coefficients of balanced rxn eq’n

𝑎𝐴 + 𝑏𝐵 ⇌ 𝑐𝐶 + 𝑑𝐷

𝐾𝑐 =[𝐶]𝑐[𝐷]𝑑

[𝐴]𝑎[𝐵]𝑏

• use consistent units so they can be ignored in the Kc expression

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Equilibrium Law

• Consider the convention of Products over Reactants

Let x = coeff.

𝐾𝑐 =[𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠]𝑥

[𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]𝑥

• only applies if amount concentrations are observed to remain constant (at equilibrium), in a closed system, and at a given temperature

• very valuable constant in chemistry

• all Kc values (or calculations using it) must specify the rxn temperature at equilibrium

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Communication Example 1p. 684

Write the equilibrium law expression for the reaction of nitrogen monoxide gas with oxygen gas to form nitrogen dioxide gas.

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Equilibrium Law

• Kc describes the extent of the forward rxn

• reverse rxn written as reciprocal 1

𝐾𝑐

• Example: Kc for decomposition is reciprocal of Kc for formation rxn

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Communication Example 2p. 684

The value of Kc for the formation of HI(g) from H2(g) and I2(g) is 40, at a given temperature.

What is the value of Kc for the decomposition of HI(g) at the same temperature?

𝐾𝑐, 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 =1

𝐾𝑐,𝑓𝑜𝑟𝑤𝑎𝑟𝑑=

1

40= 0.025

OR

H2(g) + I2(g) ⇌ 2 HI(g)

⟹

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Formation (forward rxn) Decomposition (reverse rxn)

2 HI(g) ⇌ H2(g) + I2(g)

𝐾𝑐 =[HI]2

[H2][I2]= 40 𝐾𝑐 =

[H2][I2]

[HI]2=

1

40= 0.025

Equilibrium Law

• higher numerical value of Kc implies a greater tendency of the system to favor the forward rxn• ⬆Kc , ⬆products

• i.e. products favored at equilibrium

• Kc > 1 ⟹ equilibrium favors products

• Kc < 1 ⟹ equilibrium favors reactants

• Kc gives info on equilibrium position of rxn, NOT on rxn rate• consider in Thermochemistry Unit (Energetics)

• moderate concentration change of reactants or products results in small concentration change

is all other entities, therefore Kc remains same• Kc is affected by a large change in concentrations

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Equilibrium Law

• Adjust Kc to reflect pure substances,

• i.e. (s) or condensed (l) have fixed amount concentration

• concentration of condensed (s) and (l) states are NOT included in Kc calculations

• only include variable concentrations of dissolved (aq) solutions and gaseous mixtures

• consider net ionic equation

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Communication Example 3p. 685

Write the equilibrium law expression for the decomposition of solid ammonium

chloride to gaseous ammonia and gaseous hydrogen chloride.

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Omit solid from the

Equilibrium Law

expression

Communication Example 4p. 685

Write the equilibrium law expression for the reaction of zinc in

copper(II) chloride solution.

Zn(s) + CuCl2(aq) ⇌ ZnCl2(aq) + Cu(s)

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Omit solids and

spectator ions from

the Equilibrium Law

expression

Equilibrium Law

• Equilibrium depends on [reacting substances]• Represent in expression as actually exists

• Represent ions as individual entities

• Always write expression from balanced net ionic rxn equation

with simplest whole number coefficients, unless otherwise stated

• i.e. Kc Excludes: • (s) & (l) pure substances

• spectator ions

• units

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HOMEWORK

• Lab Exercise 15.B – p. 686 (Determining an Equilibrium Constant)

• DUE: R, Sept 24

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Today’s Objectives

• Calculate equilibrium constants and concentrations when

• concentrations at equilibrium are known

• initial concentrations and one equilibrium concentration are known

• equilibrium constant and one equilibrium concentration are known

Section 15.1 (pp. 676-689)

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Calculations in Equilibrium Systems

Example: In the following system: N2(g) + 3H2(g) ⇌ 2NH3(g) Kc = 1.2 at 375oC

0.249 mol N2(g), 3.21x10-2 mol H2(g), and 6.42x10-4 mol NH3(g) are combined in a 1.00 L vessel.

Is the system at equilibrium? If not, predict the direction in which the reaction must proceed.

𝐾𝑐 =[NH3]

2

[N2][H2]3 =

(6.42 x 10−4)2

0.249 (3.21 X 10−2)3= 0.0500 ≠ 1.2

Value < Kc ∴ more reactants than products

reaction must proceed to the right

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Calculations in Equilibrium Systems

Example: Find the equilibrium concentration of the ions that are formed when solid

silver chloride is dissolved in water. (Kc = 5.4x10-4)

AgCl(s) ⇌Ag+(aq) + Cl–(aq)

𝐾𝑐 = [Ag+][Cl−] = 5.4x10−4

at equilibrium: [Ag+] = [Cl–] = 𝑥

⇒ 𝑥2 = 5.4x10−4

𝑥 = 0.023 𝑚𝑜𝑙𝐿

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Predict Final Equilibrium Concentrations

• examine simple homogenous systems and algebraically predict reagent concentrations at equilibrium using a known value for Kc

and initial concentration values

• usually requires using an ICE table

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ICE Tables

Example: Consider the following equilibrium at 100oC: N2O4(g) ⇌ 2 NO2(g)

2.0 mol of N2O4(g) was introduced into an empty 2.0 L bulb. After equilibrium was established, only 1.6 mol of N2O4(g) remained. What is the value of Kc?

Solve for x: 1 – 𝑥 = 0.8 ⇒ 𝑥 = 0.2 , 2𝑥 = 0.4

𝐾𝑐 =0.42

0.8= 0.2

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mol/L N2O4(g) 2 NO2(g)

I: 1.0 0

C: – x + 2x

E: 1.0 – x = 0.80 2x

2.0 mol = 1.0 mol/L (I)

2.0L

1.6 mol = 0.8 mol/L (E)

2.0L

ICE Tables

STEPS:

• Always write out the equilibrium reaction and equilibrium law expression if not given.

• Draw an ICE Table (Initial, Change in, and Equilibrium concentrations)

• I + C = E

• Substitute known values

• Solve for x

• Solve for equilibrium concentrations

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ICE Tables

Example: A 10 L bulb is filled with 4.0 mol of SO2(g), 2.2 mol of O2(g), and 5.6 mol of SO3(g).

At equilibrium, the bulb was found to contain 2.6 mol of SO2(g). Calculate Kc for this reaction:

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2 SO2(g) + O2(g) ⇌ 2 SO3(g)

4.0 mol = 0.40 mol/L (I)

10.0L

2.2 mol = 0.22 mol/L (I)

10.0L

5.6 mol = 0.56 mol/L (I)

10.0L

2.6 mol = 0.26 mol/L (E)

10.0L

ICE Tables

Example: A 10 L bulb is filled with 4.0 mol of SO2(g), 2.2 mol of O2(g), and 5.6 mol of SO3(g).

At equilibrium, the bulb was found to contain 2.6 mol of SO2(g). Calculate Kc for this reaction:

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2 SO2(g) + O2(g) ⇌ 2 SO3(g)

I 0.4 0.22 0.56

C – 2x – x + 2x

E 0.26 0.22 – x 0.56 + 2x

𝐾𝑐 =0.72

(0.15)(0.26)2= 48

0.4 – 2x = 0.26

x = 0.07

0.01 – 0.05𝑥 = 2𝑥

ICE Tables

Example: Determine the equilibrium concentrations for all species present given that

the initial concentration of each reactant is 0.200 mol/L in the following reaction:

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N2(g) + O2(g) ⇌ 2 NO(g) Kc = 0.00250

I 0.2 0.2 0

C – x – x + 2x

E 0.2 – x 0.2 – x 2x

𝐾𝑐 = 0.0025 =(2𝑥)2

(0.2 − 𝑥)2square root both sides ⇒ 0.005 =

2𝑥

0.2 − 𝑥⇒

0.01 = 2.05𝑥

𝑥 = 0.00488

0.195 mol/L 0.195 mol/L 0.00976 mol/L

Sample Problem 15.2p. 686

• ICE table required to solve

• see handout

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ICE Tables Approximation Rule

When [reactants] > 1000Kc ⟹ disregard the change in concentration

Example: Calculate the concentration of gases produced when 0.100 mol/L COCl2(g)

decomposes into carbon monoxide and chlorine gas. (Kc for this reaction is 2.2x10-10)

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

I 0.1 0 0

C – x + x + x

E 0.1 – x x x

0.1 > 1000 x 2.2x10-10

∴ assume 0.1 – x ≈ 0.1i.e. negligible change

𝐾𝑐 = 2.2 × 10−10 =𝑥2

0.1

⟹ 𝑥 = 4.7 × 10−6 𝑚𝑜𝑙𝐿

0.100 mol/L 4.7x10-6 mol/L 4.7x10-6 mol/L

Predicting Equilibrium

• NOTE: If the initial reactant concentrations are not equal, solving for unknown change in concentration requires the application of the quadratic formula

• not seen in Nelson textbook

• can find quadric formula in data booklet

𝑥1,2 =−𝑏 ± 𝑏2 − 4𝑎𝑐

2𝑎

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Predicting Equilibrium

• predicting equilibrium by applying the quadratic formula

Example: In a closed, 500mL reaction vessel at a temperature of 198C

phosphorus trichloride gas and chlorine gas combined to form

phosphorus pentachloride gas. Initially 0.015mol of each reactant is

placed in the vessel. The Kc for this reaction is 2.00 at 198C. What

is the amount concentration of the product present at equilibrium?

See Handout

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HOMEWORK

• Section 15.1 Review Qs – p. 688 #1-11

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