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Chemistry 12 Unit II – Dynamic Equilibrium Notes II.1 – The Concept of Dynamic Equilibrium

• A reversible reaction is a chemical reaction where products can react to form the reactants and vice versa. A reversible reaction is represented by a double sided arrow ⇌.

• A reversible reaction results when there are two opposing reactions, formation of products and the reformation of reactants. Not all reactions are reversible reactions.

• A reversible reaction is said to be at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction.

• In order for equilibrium to exist, the reversible reaction must take place in a closed system so that the products that are formed from reactants will be available for the reverse reaction.

• Open system… Allows both energy and matter transfer.

• Closed system… Allows only energy transfer but not matter.

• Isolated system… Allows neither energy or matter transfer.

• Dynamic equilibrium: an equilibrium in which microscopic changes occur, but macroscopic changes do not.

• Macroscopic changes are visible or large scale changes. These include colour intensity, concentration, pH, temperature and pressure.

• Microscopic changes which occur at the atomic or molecular level and cannot be seen.

• So to clarify, dynamic equilibrium implies that all observable properties are constant, but at the microscopic level, there is a back and forth reaction between reactants and products. This back and forth reaction is in perfect balance, or equal.

• How to recognize whether a reaction has reached an equilibrium?

à a constant macroscopic property(s) of the system is observed. There should be no changes in macroscopic properties (pressure, colour etc).

• If an equilibrium system is disrupted by opening it and removing chemicals, the remaining chemicals will re- establish equilibrium if the system is closed again.

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Examples:

1. January 1998

2. April 2001

• Read the following and try to answers the questions that follow:

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• From doing the above exercise, we can draw the following conclusions:

o Temperature affects equilibrium equilibrium equilibri equilibrium um.

o A new equilibrium is attained at equ a new temperat equilibrium ur.

o Since both the hot and cold tubes become the same color at room olor

o When a system is at equilibrium, no MACROSCOPIC changes or.

Example: June 2009

August 2005

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II.2 – The Characteristics of Equilibrium

• From unit I, we know the rate is proportional to concentration.

• If [ ] is high, the rate is fast.

• If [ ] is low, the rate is slow.

• The values of kf and kr are specific constants at a given temperature.

• Now complete the graph Ex 6 and 7 from p. 40-41 Hebden. Note that x-axis needs to be extended to 20 minutes.

• In a closed system at equilibrium (constant temperature) we find:

1. Ratef = Rater (i.e., rate of decrease of reactant = rate of increase of product)

2. Ratef and Rater remain constant 3. Mostly [Reactant] ≠ [Product]

4. [Reactant] and [Product] remain constant 5. Macroscopic properties remain constant 6. If you disturb an equilibrium, it will try to reach equilibrium again 7. [Product] eq ÷ [Reactant]eq = keq reaches a certain value no matter how the

equilibrium is reached

• Remember that of two types of equilibria (static and dynamic), a chemical equilibrium is dynamic because macroscopic changes are constant but at the microscopic level, there is a continuous back and forth reaction between reactants and products.

Examples:

1. August 2001

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2. August 1999

3. January 2004

• It can be difficult to discuss reaction rates when establishing equilibrium. Generally, reaction rates depend on contraction.

• High concentration means a fast rate, low concentration means a slower rate. But what happens when the concentration decreases…what would happen to the reaction rate?

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• The question below is from a June 2002 provincial and one that causes many students grief.

• In the above question, we start with nothing but react, thus I have no product. What happens as the

reaction starts? What do you think is happening to the HBr concentration? What about the concentration of the products?

Try these ones…

***Do Hebden Questions #8-13 ***

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II.3 – Predicting Whether A Reaction is Spontaneous or Not Reaction kinetics vs Thermodynamic

• Reaction kinetics is interested in how high Ea is. The lower the Ea, the faster the reaction rate. • Thermodynamics is interested in whether a chemical reaction can “go” or not by analyzing which

species is energetically more stable. • Eg., Diamond vs. graphite (more stable)

• II.3 is introduction to thermodynamics where you are responsible for predicting whether a reaction is

spontaneous or not.

• Spontaneous change… a change that occurs by itself. • To do this, we must consider two aspects:

• Enthalpy (∆H)… Energy related • Low-energy side is favored over high-energy side because low energy leads to more

stability. • In general, we can say: Enthalpy favors a minimum value. In other words,

“things tend to become more stable as time passes”. • Entropy (∆S)… Randomness and state related

• Highly RANDOM states are favored over highly ORDERED states because there are more random states possible (Think of your bedroom!!!)

• The amount of randomness in a system is ENTROPY. In general, we can say: Entropy favors a maximum value. In other words, “things tend to become jumbled up as time passes”.

• Some state of matter is more random than others: Gas >> Aqueous Solution > Liquid >> Solid

Rules:

1. Determine which side of the reaction is more stable (less energetic) to determine the direction based on ∆H

2. For ∆S, determine which side of the reaction is more random by identifying the side with the most number of particles with the most random state.

Examples to determine spontaneity of a chemical reaction: Case 1

C2H2(g) + 2Cl2(g) ________C2H2Cl4(l) + 386kJ Minimum enthalpy favors product side. Maximum entropy favors reactant side. Thus, the reaction is predicted to be at equilibrium. Case 2 2C(s) + O2(g) ________ 2CO(g) where ∆H = -221kJ Minimum enthalpy favors product side. Maximum entropy favors product side. Thus, the reaction is predicted to go 100% and is spontaneous.

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Case 3 4Au(s) + 3O2(g) ________ 2Au2O3(s) where ∆H = +162kJ Minimum enthalpy favors reactant side. Maximum entropy favors reactant side. Thus, the reaction is predicted to go 0% and is non-spontaneous.

Try these…

Predict whether each of the following reactions is spontaneous, non-spontaneous, or reaches equilibrium.

1. Zn(s) + 2 HCl(aq) ______ ZnCl2(aq) + H2(g) ΔH = -152 kJ

2. 3C(s) + 3H2(g) + h e a t ______ C3H6(g)

3. N2(g) + 3H2(g) ______ 2NH3(g) + 92.4 kJ

Predict whether each of the following reactions is exothermic or endothermic. 1. A(s) + B(l) ⇌ C(g)

2. A(g) + B(l) à C2(g)

***Do Hebden Questions #14-16 ***

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II.4 - Le Châtelier's Principle • Le Châtelier's Principle: If a closed system at equilibrium is subjected to a change,

processes will occur that tend to counteract that change. Or “Whatever we do, nature tried to undo”. nature tried to undo”. nature tried to undo”. nature tried

• When an equilibrium system is disrupted, the reaction rates are affected. However, eventually the equilibrium will be re-established through SHIFTING.

• Shift to the right = forward reaction rate gets faster in order to re-establish the equilibrium • Shift to the left = reverse reaction rate gets faster in order to re-establish the equilibrium

• Le Châtelier's Principle gives us the ability to quickly and easily predict the effect that any change of conditions will have on equilibrium.

• Anything that changes the equilibrium is referred to as a STRESS. Stresses are factors that disrupt and affect the position of an equilibrium (i.e. whether the equilibrium favours the reactants or products.)

Examples of stresses are changing temperature, concentrations, pressure etc…

***Note: Catalysts DO NOT affect the position of an equilibrium.*** Later you will learn that solids and pure liquid substance also DO NOT affect as well.

Examples of shifting: 3H2(g) + 2N2(g) ⇌2NH3(g) + 92 kJ

Stress Shifting H2 N2 NH3 [H2] is increased -----------------------

[NH3] is increased ----------------------- [NH3] is decreased -----------------------

Temperature is increased Temperature decreased Pressure is increased Volume is increased

Le Châtelier and Graphs

Consider the following equilibrium system:

2 NO(g) + Cl2(g)⇌2 NOCl(g) + 76kJ

We will use Le Chatelier’s Principle to predict the effect of changing the temperature, concentration and pressure conditions of this equilibrium.

A. The Effect of Temperature Changes

If temperature is INCREASED at t1, the reaction will shift left to reduce heat. As a result of shifting, the reverse reaction rate will be faster to consume NOCl and to produce more NO and Cl2. By t2, the equilibrium is re-established.

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Note: The change at t1 is gentle and the following change follows stoichiometry; however, the system will NOT retain the original concentration at t2.

B. The Effect of Concentration Changes

If [Cl2] is DECREASED at t1 by removing the species, the reaction will shift left to increase Cl2. As a result of shifting, the reverse reaction rate will be faster to consume NOCl and to produce more NO and Cl2. By t2, the equilibrium is re-established.

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Note: The change at t1 is sudden for Cl2 and the following change follows stoichiometry; however, the system will NOT retain the original concentration at t2.

C. The Effect of Pressure Changes

Recall that pressure of a system can be changed through two methods. One is to add/remove gaseous substance which has the same effect as the concentration change. Another way to change pressure is to increase/decrease the volume of the reaction vessel.

In general:

If we decrease the volume, pressure is increased which simultaneously increase the concentration of all gasses in the system.

If we increase the volume, pressure is decrease which simultaneously decrease the concentration of all gasses in the system.

If pressure is DECREASED at t1 by increasing the volume, the concentration of Cl2, No and NOCl will all spike down. Then the reaction will shift left to increase pressure. As a result of shifting, the reverse reaction rate will be faster to consume NOCl and to produce more NO and Cl2. By t2, the equilibrium is re-established.

Note: The change at t1 is sudden for all gasses and the following change follows stoichiometry; however, the system will NOT retain the original concentration at t2.

D. The Effect of Adding Catalysts The addition of a catalyst has NO effect on the position of equilibrium; however, it does allow equilibrium to be reached faster.

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Recall from section I.11 in unit I, adding a catalyst speeds up both the forward and reverse rates by an equal amount. So the reaction at equilibrium will remain at equilibrium.

E. The Effect of Solid and Pure Liquid Solid and pure liquid substance have a constant “concentration”. This means even if you add/remove solid substance into a closed system at equilibrium, its “concentration” will remain the same and has NO effect on the position of equilibrium. Same is true for pure liquid. E.g., A(s) + B(l) ⇌ C(g) Addition of A will NOT shift the equilibrium to right. Removal of B will NOT shift the equilibrium to left. E.g., A(s) + B(l) ⇌ C(l) Addition of A will NOT shift the equilibrium to right. Removal of B will shift the equilibrium to left as B is no longer a pure liquid.

***Do Hebden Questions #17-24 ***

Examples:

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II.5 - Industrial Applications of Equilibrium Principles Equilibrium concepts are widely used in the manufacturing of chemicals. By changing pressure, concentration, or temperature we can shift an equilibrium reaction and cause it to produce more of a desired product. The following are only two of many examples:

1. The Haber process for Making Ammonia

In 1910, the German chemist, Fritz Haber (1868-1934), developed a process that combines nitrogen and hydrogen to make ammonia. Ammonia is used as a raw material for fertilizer, explosives and other important chemicals.

The process that makes ammonia occurs by the following equilibrium:

2N2(g) + 3 H2(g) ⇌ 2 NH3(g) + 92.4 kJ

To reach equilibrium quickly, high T and high P are required. However, once it reaches equilibrium, low T and high P would shift it to right to maximize the yield of ammonia.

2. Making Cement From Limestone

Limestone, CaCO3, can be used to produce “quicklime”, CaO, according to the reaction:

CaCO3(s) + 175 kJ ⇌ CaO(s) + CO2(g)

To reach equilibrium quickly, high T and high P are required. However, once it reaches equilibrium, low P and high T would shift it to right to maximize the yield of ammonia. The quicklime can then be added to a mixture of sand, clay, iron oxide and gypsum (CaSO4) to make Portland cement powder.

Examples: June 1999

June 2001

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II.6 - The Equilibrium Expression and the Equilibrium Constant • Earlier in the unit from ex. 6, we found that the ratio of [product] to [reactant] at equilibrium is

constant. • This ratio is constant at a given temperature and is called the “equilibrium constant” . It is

often represented by the symbol Keq.

• For the reaction: aA + bB ⇌ cC + dD We write: Keq = The formula is called the equilibrium expression. The value of Keq itself is called the equilibrium constant at a given temperature. There is no unit for Keq.

“Concentrations” which are constant are not included in the Keq expression. This includes: • Solid

o Solids cannot be appreciably compressed therefore their density and molar concentrations are constant.

• Pure liquids o Liquids cannot be appreciably compressed therefore their density and molar concentrations are

constant. However, if there is another liquid present, which can dilute the first liquid, then the liquid is not pure and can have its concentration changed by dilution. Therefore, a liquid is said to be pure if and only if it is the only liquid which exists on both sides of the entire equilibrium equation.

***Do Hebden Questions #31-35 *** Examples: Write the Keq expression for the following reactions:

1. H2(g) + F2(g) ⇌2HF(g)

2. H2O(g) + CO(g) ⇌H2(g) + CO2(s)

3. Br2(l) + H2(g)⇌2HBr(g)

4. CH3COCH3(l) + Cl2(g)⇌CH3COCH2Cl(l) + HCl(aq)

Examples:

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II.7 - Le Chatelier’s Principle and the Equilibrium Constant

• When the temperature is decreased and held at a certain value over time, the equilibrium will shift to the product side:

2 NO(g) + Cl2(g) 2 NOCl(g) + 76 kJ

A decrease in temperature will then cause an increase in [PRODUCT] and a decrease in [REACTANT], and since:

Keq = [Products] then Keq is increased. [Reactants]

**Only a temperature change can affect the value of Keq. *** Changes in pressure and concentration or the addition of a catalyst have no effect on Keq.

• Since

a LARGE value of Keq (greater than 1) implies that a LARGER amount of products (small amount of reactant) is present at equilibrium

a SMALL value of Keq (less than 1) implies that a SMALLER amount of products (large amount of reactant) is present at equilibrium

A value of Keq = 1 implies that [product] = [reactant] at equilibrium

Examples:

***Do Hebden Questions #36-45 ***

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II.8 - Equilibrium Calculations

For the calculation problems: 1. Write Keq expression (Watch out for (s) and pure (l)) 2. Create an ICE table and fill in the table with molar concentrations as much as possible (It may have “x”) 3. Substitute “E” values from the table into the Keq expression 4. Solve for the unknown

In the following examples, we will see the various types of calculations that can be performed based on the equilibrium expression and some experimental data.

Example A:

A 2.0 L bulb contains 6.00 mol of NO2(g), 3.0 mol of NO(g) and 0.20 mol of O2(g) at equilibrium. What is Keq for:

2 NO(g) + O2(g) ⇌ 2 NO2(g)

Example B:

Into a 2.00 L bulb was introduced 4.00 mol of NO2(g). After a while equilibrium was attained according to the equation:

2 NO(g) + O2(g) ⇌ 2 NO2(g)

At equilibrium, 0.500 mol of NO(g) was found. Find the value of Keq?

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Example C:

A certain amount of NO2(g) was introduced into a 5.00 L bulb. When equilibrium was attained according to the equation,

2 NO(g) + O2(g) ⇌ 2 NO2(g)

the concentration of NO(g) was 0.800 M. If Keq has a value of 24.0, how many moles of NO2 were originally put into the bulb? Example D:

2 NO(g) + O2(g) ⇌ 2 NO2(g) Keq = 49

If we introduce 2.0 mol of NO(g), 0.20 mol of O2(g), and 0.40 mol of NO2(g) into a 2.0 L bulb, which way will the reaction shift in order to reach equilibrium?

Note: This question is not asking us to find a numerical value; it is asking for a decision. To make this decision we must make a simple comparison between two numbers. This comparison will tell us one of three things:

The reaction may be:

1. At equilibrium 2. Shift to right to reach equilibrium 3. Shift to left to reach equilibrium

One of the numbers that we will compare is Keq, the other is Q.

Q is the REACTION QUOTIENT (sometimes called a ‘Trial Ion Product’) and is a “trial value” for Keq.

Comparison [Product]÷[Reactant] is To reach equilibrium, must shift Number line

Q = Keq just right Already at equilibrium

Q > Keq too big Left

Q < Keq too small Right

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Example E:

Keq = 3.5 for: SO2(g) + NO2(g) ) ⇌ SO3(g) + NO(g)

If 4.0 mol of SO2(g) and 4.0 mol of NO2(g) are placed in a 5.0 L bulb and allowed to come to equilibrium, what concentration of all species will exist at equilibrium? Example F:

A 1.0 L reaction vessel contained 1.0 mol of SO2, 4.0 mol of SO3 and 4.0 mol of NO at equilibrium according to:

SO2(g) + NO2(g) ) ⇌ SO3(g) + NO(g)

If 3.0 mol of SO2 is added to the reaction mixture, what will be the new concentration of NO when equilibrium is re-attained? Note: The unknown is the final [NO] when equilibrium is re-established. Therefore, all other concentrations must be given or be readily calculated AND you must have the value of Keq. But you are not given the value for Keq. You are told the amount of each gas previously present at equilibrium. This information will allow you to calculate Keq.

***Do Hebden Questions #47-49, 57, 52, 58, 61, 50, 51, 54, 60, 62, 55, 59, 63-66 ***