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1 Chemical Equilibrium Chapter 15 AP CHEMISTRY http://206.180.99.10/~robertsj/AP%20C hapter%20Notes/ap%20ch13.ppt
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1 Chemical Equilibrium Chapter 15 AP CHEMISTRY robertsj/AP%20Chapter%20Note s/ap%20ch13.ppt.

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Page 1: 1 Chemical Equilibrium Chapter 15 AP CHEMISTRY robertsj/AP%20Chapter%20Note s/ap%20ch13.ppt.

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Chemical Equilibrium

Chapter 15

AP CHEMISTRYhttp://206.180.99.10/~robertsj/AP%20Chapter%20Notes/ap%20ch13.ppt

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Chemical Equilibrium

The state where the concentrations of all reactants and products remain constant with time.

On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

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Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2:

N2O4(g) 2NO2(g).

At some time, the color stops changing and we have a mixture of N2O4 and NO2.

Chemical equilibrium is the point at which the concentrations of all species are constant.

The Concept of Equilibrium

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QuickTime™ and aSorenson Video decompressorare needed to see this picture.

NO2 - N2O4 Demonstration

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The Concept of Equilibrium

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The Concept of EquilibriumAs the substance warms it begins to decompose:

N2O4(g) 2NO2(g)

A mixture of N2O4 (initially present) and NO2 (initially formed) appears brown.

When enough NO2 is formed, it can react to form N2O4:

2NO2(g) N2O4(g).

At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4:

The double arrow implies the process is dynamic.

N2O4(g) 2NO2(g)

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Notes on Equilibrium Expressions (EE)

•K does not include any pure solids or liquids

•The expression shows products divided by reactants

•Like the rate constant, k, the units of K depend on the experiment being performed

•For the reverse reaction K’ = 1/K (reactants and products switch)

Kl m

j k C DA B

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Notes on Equilibrium Expressions (EE)

•For a reaction multiplied by an integer, “n”, Knew = (Korig)n

•For a given reaction, K is dependent only on temperature

Kl m

j k C DA B

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Equilibrium Expression

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

K NO H O

NH O2

2

24 6

34 7

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Heterogeneous Equilibria

. . . are equilibria that involve more than one phase.

CaCO3(s) CaO(s) + CO2(g)

K = [CO2]

The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

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Practice Problem:Consider the reaction represented by the equation:

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

In trial #1, you start with 6.00 M Fe3+(aq) and 10.0 M SCN-(aq), and at equilibrium the concentration of FeSCN2+(aq) is 4.00 M.

What is the value of the equilibrium constant for this reaction?

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

Initial Change Equilibrium

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Practice Problem

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)Equilibrium 2.00 6.00 4.00

0.33 K ]0.2][0.6[

]00.4[ K

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Practice Problem #2:

Using the previous reaction:

Fe3+(aq) + SCN-(aq) FeSCN2+(aq) and the K value we determined: K = 0.33 determine if the following concentrations are at equilibrium:

Initial:10.0 M Fe3+(aq), 8.00 M SCN-(aq), and 2.00 M FeSCN2-

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Reaction Quotient

H2(g) + F2(g) 2HF(g)

Q HF

H F2 2

02

0 0

•After the equilibrium constant (K) is known, we can use it to determine if a reaction is at equilibrium

•The reaction quotient, Q, has the same form as the equilibrium constant expression EXCEPT initial concentrations are used instead of equilibrium concentrations

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Predicting the Direction of a Reaction Using Reaction Quotient

•If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K).

•If Q < K then the forward reaction must occur to reach equilibrium.

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Solving Equilibrium Problems

1. Balance the equation.

2. Write the equilibrium expression.

3. List the initial concentrations.

4. Calculate Q and determine the shift to equilibrium.

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Solving Equilibrium Problems(continued)

5. Define the change needed to reach equilibrium.

6. Substitute equilibrium concentrations into equilibrium expression and solve.

7. Check calculated concentrations by calculating K.

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K vs. Kp

For

any reaction:

Kp = K(RT)n

n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.

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Le Châtelier’s Principle

. . . if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

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C

Consider the production of ammonia

As the pressure increases, the amount of ammonia present at equilibrium increases.

As the temperature decreases, the amount of ammonia at equilibrium increases.

Can this be predicted?

Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

N2(g) + 3H2(g) 2NH3(g)

Le Châtelier’s Principle

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C

Le Châtelier’s Principle

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Effects of Changes on the System

1. Concentration: The system will shift away from the added component.

2. Temperature: K will change depending upon the temperature (treat the energy change as a reactant).

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Increase of Pressure to an Equilibrium.

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Effects of Changes on the System (continued)

3. Pressure:

a. Addition of inert gas does not affect the equilibrium position.

b. Decreasing the volume shifts the equilibrium toward the side with fewer moles.

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Change in Reactant or Product Concentrations

Consider the Haber process

If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier).

That is, the system must consume the H2 and produce products until a new equilibrium is established.

Therefore, [H2] and [N2] will decrease and [NH3] increases.

Le Châtelier’s Principle

N2(g) + 3H2(g) 2NH3(g)

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Change in Reactant or Product ConcentrationsLe Châtelier’s Principle

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The Haber ProcessLe Châtelier’s Principle

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•N2 and H2 are pumped into a chamber.

•The pre-heated gases are passed through a heating coil to the catalyst bed.•The catalyst bed is kept at 460 - 550 C under high pressure.

•The product gas stream (containing N2, H2 and NH3) is passed over a cooler to a refrigeration unit.

•In the refrigeration unit, ammonia liquefies but not N2 or H2.

•The unreacted nitrogen and hydrogen are recycled with the new N2 and H2 feed gas.

•The equilibrium amount of ammonia is optimized

The Haber Process for producing NH3

Le Châtelier’s Principle