Chemical EquilibriumPart 1

Ch. 15 in Textbook

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I. Static vs. Dynamic

Generally speaking, an equilibrium is a state of balance.

A static equilibrium is one in which there is no motion.

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A dynamic equilibrium is one in which there is motion despite there being no net change.

A chemical equilibrium is an example of a dynamic equilibrium.

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II. Chemical Equilibria

Suppose we have the gaseous reactants I2 and H2. They undergo a synthesis reaction to form HI:

H2 + I2 → 2HI As HI accumulates, some

molecules have enough energy to decompose to H2 and I2:

2HI → H2 + I2

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As this process continues, eventually the rate of the forward reaction equals the rate of the reverse reaction.

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Link

The final equilibrium mixture will contain both reactants and products.

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Although the amounts are NOT necessarily equal, the amounts must eventually remain constant since the forward and reverse reactions are occurring simultaneously and at the same rate.

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Although there is no noticeable net change in reactants or products, the reaction is still proceeding in the forward and reverse directions, making chemical equilibrium a dynamic equilibrium.

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HW: 15.2Tutorial

III. Kinetics Flashback!!!!

Given: A B Assume that the

forward and reverse reactions are both elementary steps.

Rate (forward) = kf [A] Rate (reverse) = kr [B] At equilibrium: kf [A]

= kr [B]

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Rearranging: [B]/[A] = kf/kr = constant

What is the meaning of this?!

At a given temp. the products and reactants will ALWAYS be in the same ratio at equilibrium, no matter the starting point…

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HW: 15.3

IV. The Law of Mass Action

This proportion can be represented by the equilibrium expression.

The Law of Mass Action states that the equilibrium expression depends on the equilibrium concentrations of reactants and products.

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For the reaction:aA + bB pP + qQ

the equilibrium expressionis written as: K = [P]p [Q]q

[A]a [B]b

where K is the equilibrium

constant.Oudaily.com

Ex) What is the equilibrium expression for the synthesis of hydrogen iodide?

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HW: 15.6, 15.20

In KINETICS, the rate law did NOT depend on the stoichiometry of the net equation, only on the rate-determining step.

In EQUILIBRIUM, the equilibrium expression DOES depend on the stoichiometry of the net equation.

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Yes, typo

V. The Equilibrium Constant, K,…

A) Dependence …does not depend on the

reaction mechanism. …does not depend on the

initial concentrations of reactants and products.

…depends on the equilibrium concentrations of reactants and products.

..also depends on temperature! (more on that later…)

We generally omit the final units.

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B) Magnitude

Since products are divided by reactants in the expression: A larger K value (>>1)

means that products dominate the final reaction mixture and we say “the equilibrium lies to the right.”

A smaller K value (<<1) means that reactants dominate the final reaction mixture and we say “the equilibrium lies to the left.”

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HW: 15.10

C) Kc

This is the constant when concentrations are expressed in molarity.

It is the most common version of K.

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D) Kp

This is the constant when concentrations (of gases) are expressed in terms of partial pressures (atm).

K = (PP) p (PQ)q

(PA)a (PB)bLankapeacewatch.com

E) Relationship between Kc and Kp

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Note: My notes are WAY better…this PowerPoint slide doesn’t even have a stupid picture in it…or a picture of another chemistry PowerPoint slide…of another PowerPoint Slide…of another PowerPoint slide…ad infinitum.

E) Relationship between Kc and Kp (fo’ real dis time)

PV = nRT P = (n/V) RT For substance A:

PA = [A] RT Do this for all partial pressures in an

equilibrium mixture and we get: Kp = Kc (RT)Δn

where Δn is the moles of gaseous products -

the moles of gaseous reactants. Link

Ex) Given: 2SO3(g) 2SO2(g) + O2(g)

Write the Kp expression.

Calculate Kp if Kc = 4.08 x 10-3 at 1000 K.

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HW: 15.8 (a) & (b), 15.12

F) Direction

Equilibrium can be reached from any direction (all reactants or all products or any mixture of both).

Kc (forward) = 1/Kc (reverse)

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HW: 15.16

Breakdancing Baby Break

VI. Heterogeneous Equilibria

These involve one or more reactants or products present in a different phase.

Many equilibria do not involve dissolved species or gases.

The concentration of a pure solid or liquid is constant and therefore does not appear in the equilibrium expression.

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Ex) 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)

Write the Kp expression.

Pig Iron from

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Thus, only the partial pressures of water and hydrogen affect the equilibrium expression.

Don’t forget though, the solids must be present in order for the equilibrium to be established.

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HW: 15.18

VII. Calculating K

Use an ICE chart: Initial, Change, and Equilibrium.

First, determine all initial concentrations.

Second, determine any available equilibrium concentrations.

Third, use the reaction stoichiometry to determine the change in concentration from the initial to equilibrium.

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Fourth, determine all equilibrium concentrations.

Fifth, plug and chug into equilibrium expression and find K. Nitrocotton.com

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

A vessel at 1000 K contains 6.09 x 10-3

M SO3. At equilibrium, the SO3 concentration is 2.44 x 10-3 M. What is the value of Kc? Portlandsentinel.com

HW: 15.24, 15.26

VIII. The Reaction Quotient (Q)

Given the initial concentrations of reactants and products, we can determine the direction the reaction will proceed in.

Substitute concentrations into the reaction quotient expression, which is the same as K except NOT at equilibrium.

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K uses equilibrium concentrations!!!!!!

Q uses initial concentrations!!!!!!

K-Dubs, 6th grade

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If Q>K, then there is an excess of products which will react to form reactants to establish equilibrium (shift left).

If Q<K, then there is an excess of reactants which will react to form products to establish equilibrium (shift right).

If Q=K, equilibrium is established.

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HW: 15.28

IX. Calculating Equilibrium Concentrations

Ex 1) At 500 K the value of Kp is 0.497 for the following reaction:PCl5(g) PCl3(g) + Cl2(g)If the partial pressure of PCl5 is 0.860 atm and the partial pressure of PCl3 is 0.350 atm at equilibrium, what is the partial pressure of Cl2 in the mixture?

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Ex 2) Using the equilibrium from Ex. 1, what are the equilibrium partial pressures of all species if PCl5 has an initial pressure of 1.66 atm? Desperatelyseeking

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HW: 15.34, 15.36, 15.38, 15.40

TO BE CONTINUED…

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Equilibrium Part 2

Ch. 15 in Textbook

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X) Le Chatelier’s PrincipleA) Stress

Once an equilibrium is established, the forward and reverse reactions will proceed at equal rates and the amounts of reactants and products will not change.

Equilibrium will continue indefinitely unless conditions are somehow changed to disrupt it.

A disturbance to the equilibrium is called a stress.

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B) The Principle

When you deal with stress in your own life (e.g. taking this class) you do something to rid yourself of the stress and re-establish your personal equilibrium.

Similarly, French industrial chemist Henri-Louis Le Chatelier stated that when a stress is applied to a chemical equilibrium, the equilibrium will shift its position to eliminate the disturbance.

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C) Changes in Concentration

Adding a reactant or a product to an equilibrium mixture causes the equilibrium to speed up in the direction that will remove the excess.

This shift is only temporary and a new equilibrium is established with new equilibrium concentrations (although no change in K).

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Similarly, removing a reactant or product from an equilibrium mixture causes the equilibrium to speed up in the direction that will make up for the deficit.

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How can we maximize our ammonia output in the Haber Process?

N2(g) + 3H2(g) ↔ 2NH3(g)

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Link

D) Changes In Volume/Pressure

The pressure may be increased by reducing the container volume or the addition of a non-reacting gas.

The equilibrium will speed up in the direction that will reduce the pressure through the production of fewer moles of gas.

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Similarly, the pressure may be decreased by increasing the container volume or the removal of a non-reacting gas.

The equilibrium will speed up in the direction that will increase the pressure through the production of greater moles of gas.

This shift is only temporary and a new equilibrium is established with new equilibrium concentrations (although no change in K).

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How can we maximize our ammonia output in the Haber Process?

N2(g) + 3H2(g) ↔ 2NH3(g)

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Link

E) Changes in Temperature

Increasing the temperature results in an excess of heat, causing the equilibrium to speed up in the direction that removes the heat (endothermic).

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Similarly, lowering the temperature results in a deficit of heat, causing the equilibrium to speed up in the direction that produces heat (exothermic).

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Unlike all other stresses, however, the temperature changes the value of K.

In other words, a completely different equilibrium mixture is created at the different temperature.

If a reaction shifts to the right, then K increases.

If a reaction shifts to the left, then K decreases.

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How can we maximize our ammonia output in the Haber Process?

N2(g) + 3H2(g) ↔ 2NH3(g) + heat

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F) Adding Catalysts A catalyst lowers the

activation energy for both the forward and the reverse reactions.

A catalyst therefore speeds up the forward and reverse reactions equally.

A catalyst speeds up the rate at which equilibrium is achieved, but the equilibrium mixture and K remain exactly the same.

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Haber found that iron mixed with metal oxides allowed the Haber Process to run at sufficiently low temperatures.

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HW: 15.44, 15.46

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