EQUILIBRIUM TOPIC 7 - Weebly · TOPIC 7 EQUILIBRIUM 7.1 EQUILIBRIUM. UNDERSTANDING/KEY IDEA 7.1.A A...

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TOPIC 7 EQUILIBRIUM 7.1 EQUILIBRIUM

Transcript of EQUILIBRIUM TOPIC 7 - Weebly · TOPIC 7 EQUILIBRIUM 7.1 EQUILIBRIUM. UNDERSTANDING/KEY IDEA 7.1.A A...

Page 1: EQUILIBRIUM TOPIC 7 - Weebly · TOPIC 7 EQUILIBRIUM 7.1 EQUILIBRIUM. UNDERSTANDING/KEY IDEA 7.1.A A state of equilibrium is reached in a closed system when the rates of the forward

TOPIC 7 EQUILIBRIUM7.1EQUILIBRIUM

Page 2: EQUILIBRIUM TOPIC 7 - Weebly · TOPIC 7 EQUILIBRIUM 7.1 EQUILIBRIUM. UNDERSTANDING/KEY IDEA 7.1.A A state of equilibrium is reached in a closed system when the rates of the forward

UNDERSTANDING/KEY IDEA 7.1.A

A state of equilibrium is reached in a closed system when the rates of the forward and reverse reactions are equal.

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• The system is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged.

• A reversible reaction is a chemical reaction in which the products can react to re-form the reactants.

Arrows going both directions ( ⇄ ) indicates equilibrium in a chemical equation.

2HgO(s) ⇄ 2Hg(l) + O2(g)

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EQUILIBRIUM IN PHYSICAL SYSTEMS

• A physical system is one in which a chemical reaction is not occurring.

• Equilibrium will only occur in a “closed” system.• A “closed” system is one in which no matter can

be exchanged with the surroundings.• A typical example is the

evaporation-condensation cycle of a volatile liquid in a closed container.– When the rate of evaporation equals the rate of

condensation, the system is at equilibrium.

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EQUILIBRIUM IN CHEMICAL SYSTEMS

• A chemical system is one in which a chemical reaction is occurring.

• For the system to reach equilibrium, the forward reaction must equal that of the reverse reaction.

• See the graphic on the next slide.

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2NO2(g) ⇄ 2NO(g) + O2(g)

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CHARACTERISTICS OF THE EQUILIBRIUM STATE

• 1. Equilibrium is dynamic – the reaction has not stopped but both reactions or processes are still occurring on the microscopic level even though it looks like the reaction or process has stopped.

• 2. Equilibrium is achieved in a closed system so that no matter can escape to the surroundings.

• 3. The concentrations of reactants and products remain constant at equilibrium – they are being produced or consumed at an equal rate.

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• 4. There is no macroscopic change in properties – meaning the observable properties do not change at equilibrium because they depend on the concentration of the mixture components.

• 5. Equilibrium can be reached from either direction – meaning that at the same conditions equilibrium will be reached whether the reaction started with all reactants, all products or a mixture of both.

• 6. It is important to note that even though the concentrations of reactants and products are constant at equilibrium, this does not imply they are equal.

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Equilibrium Position• The equilibrium position is the proportion of the

reactants and products in the equilibrium mixture.• A reaction with an equilibrium mixture with mostly

products is said to “lie to the right”.• A reaction with an equilibrium mixture with mostly

reactants is said to “lie to the left”.• We can quantitatively compare equilibrium

mixtures with different conditions by using the equilibrium constant Kc.

• There can be numerous equilibrium positions, but only one equilibrium constant, Kc, at a specified temperature.

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UNDERSTANDING/KEY IDEA 7.1.B

The equilibrium law describes how the equilibrium constant (Kc) can be determined for a particular chemical reaction.

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Equilibrium Constant KcFor the reaction:

Where Kc is the equilibrium constant, and is unitless. The only thing that can change Kc for a reaction is a change in temperature.

jA + kB ⇄ lC + mD

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APPLICATION/SKILLS

Be able to deduce the equilibrium constant (Kc) from an equation for a homogeneous reaction.

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Writing an Equilibrium Expression

2NO2(g) ⇄ 2NO(g) + O2(g)Remember that concentrations are written with square brackets [ ]. Kc is calculated using molar

concentrations. Solids and liquids are not included in the K expression.

Kc = ???

Write the equilibrium expression for the reaction:

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UNDERSTANDING/KEY IDEA 7.1.C

The magnitude of the equilibrium constant indicates the extent of a reaction at equilibrium and is temperature dependent.

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Product Favored EquilibriumLarge values for K signify the reaction is “product favored”

When equilibrium is achieved, most of the reactants have been converted to products.

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Reactant Favored EquilibriumSmall values for K signify the reaction is “reactant favored”

When equilibrium is achieved, very little reactant has been converted to product.

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UNDERSTANDING/KEY IDEA 7.1.D

The reaction quotient (Q) measures the relative amount of products and reactants present during a reaction at a particular point in time. Q is the equilibrium expression with non-equilibrium concentrations. The position of the equilibrium changes with changes in concentration, pressure and temperature.

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The Reaction QuotientFor some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action.

jA + kB ⇄ lC + mD

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Significance of the Reaction Quotient

❖ If Q = K, the system is at equilibrium

❖ If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved

❖ If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

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APPLICATION/SKILLS

Be able to determine the relationship between different equilibrium constants for the same reaction at the same temperature.

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• The equilibrium constant Kc has a fixed value at the same temperature.

• You can derive Kc from different sets of data for the same reaction at constant temperature.

• All Kc values should be equal no matter how you arrived at the value since only temperature affects Kc.

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GUIDANCEBe familiar with the relationship between Kc values for reactions that are multiples or inverses of each other.

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Conclusions about Equilibrium Expressions❖ If an equation is written in reverse, the K

expression is the reciprocal of the original K.

2NO2(g) ⇄ 2NO(g) + O2(g)

2NO(g) + O2(g) ⇄ 2NO2(g)

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Conclusions about Equilibrium Expressions❖ When the balanced equation for a reaction

is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power.

2NO2(g) ⇄ 2NO(g) + O2(g)

NO2(g) ⇄ NO(g) + ½O2(g)

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• If you add two reactions together, you will multiply the two K expressions to obtain the new K’.

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UNDERSTANDING/KEY IDEA 7.1.E

A catalyst has no effect on the position of equilibrium or the equilibrium constant.

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ADDITION OF A CATALYST

• When a catalyst is added, both the rate of the forward and the rate of the reverse reactions are increased by the same amount so there is no effect on the equilibrium position.

• It only speeds up how fast Kc is achieved.

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APPLICATION/SKILLSBe able to apply Le Chatelier’s principle to predict the qualitative effects of changes of temperature, pressure, and concentration on the position of equilibrium and on the value of the equilibrium constant.

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LeChatelier’s PrincipleWhen a system at equilibrium is placed under stress,

the system will undergo a change in such a way as to relieve that stress.

Translated: The system undergoes a temporary shift in order to restore equilibrium.

1. When you add a substance or heat, the system shifts to the opposite side.

2. When you take out a substance or heat, the system shifts to the side of the take out.

3. When you increase pressure, the system shifts to the side with the least number of gaseous molecules.

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LeChatelier Example #1A closed container of ice and water is at equilibrium. Then, the temperature is raised.

Ice + Energy ⇄ Water

The system temporarily shifts to the _______ to restore equilibrium.right

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LeChatelier Example #2

A closed container of N2O4 and NO2 is at equilibrium. NO2 is added to the container.

N2O4 (g) + Energy ⇄ 2 NO2 (g)

The system temporarily shifts to the _______ to restore equilibrium.left

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LeChatelier Example #3

A closed container of water and its vapor is at equilibrium. Vapor is removed from the system.

water + Energy ⇄ vapor

The system temporarily shifts to the _______ to restore equilibrium.right

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LeChatelier Example #4

A closed container of N2O4 and NO2 is at equilibrium. The pressure is increased.

N2O4 (g) + Energy ⇄ 2 NO2 (g)

The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation.

left

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Industrial Applications

• ***Remember the only thing that can change the value of Kc is a change in temperature.

• In reactions involving the manufacture of a chemical, the goal is to obtain the highest yield possible.

• We use our knowledge of Le Chatelier’s principle and kinetics to maximize the yield.

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HABER PROCESS: Production of NH3

• N2(g) + 3H2(g) ⇄ 2NH3(g) ∆H = -93 kJ/mol• First we need to get information from the

balanced equation.– All products and reactants are in the gaseous

state with 4 moles on the left and 2 on the right.

– The reaction is exothermic so we want to re-write the equation as:

– N2(g) + 3H2(g) ⇄ 2NH3(g) + heat

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N2(g) + 3H2(g) ⇄ 2NH3(g) + heat• To maximize the amount of NH3 obtained we can do

several things:– 1. Remove NH3 as it is produced so the reaction will

continue to the right producing more NH3 to achieve equilibrium.

– 2. Increasing the pressure pushes the equilibrium to the side of the least gaseous molecules which is to the right; therefore, producing more NH3.

– 3. We want a lower temperature since the reaction is exothermic, but not too low which would slow things down too far.

– 4. Adding a catalyst will speed up the reaction and compensate for the lower temperature.

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CONTACT PROCESS: Production of sulfuric acid, H2SO4

• Sulfuric acid has the highest production of any chemical in the world.

• There are three steps in the production of H2SO4 with the second step being the rate determining step.

• 1. Combustion of sulfur to form sulfur dioxide.• 2. Oxidation of sulfur dioxide to form sulfur

trioxide.• 3. Combination of sulfur trioxide produced in

step 2 with water.

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2SO2(g) + O2(g) ⇄ 2SO3(g) ∆H=-196 kJ/mol

• Again, the reaction is exothermic and has fewer gaseous molecules on the right so we will basically follow the same industrial applications as with the Haber process.

• 1. We will not take out SO3 because it is what is needed for step 3 of the process.

• 2. Increase pressure to shift equilibrium to the right.

• 3. Lower temperature because the reaction is exothermic.

• 4. Add a catalyst to increase the rate of the reaction.