LA ANTICIPACION, PREINDICES DE MOVIMIENTO Y SUS FACTORES DE ESTUDIO

Many industrial processes depend upon the magnitude of the equilibrium constant. If a process is not likely to go to products conditions, then experiments will be done to determine how to make the reaction more favorable for the production of products. The magnitude of the equilibrium constant can allow us to determine whether a reaction is product-favored or reactant

The equilibrium constant tells us about the composition of an equilibrium mixture: its magnitude, in essence, lets us determine whether a reaction is more likely to occur in the forward direction or reverse direction. Thus, it tells us whether the equilibrium m

oduct at equilibrium.

1, tells us that the reaction is favored to the right, or product side

reaction will produce significant product, and the equilibrium mixture will contain a small amount of reactant. We can say that the or the reaction is “product-favored.”

< 1, tells us that the reaction is favored to the left, or reactant side

roduce significant product, and the equilibrium mixture will contain a larger amount of reactant. We can say that the or the reaction is “reactant-favored.”

Remember, the direction from which the equilibrium constant is determined must be specified. It is not useful to simply say that the . We must say that the equilibrium constant of the forward reaction is 1.49 x 10

Determine and discuss the utility of the following reactions in producing product by the reactions shown. That is, are favored or reactant-favored under the specified conditions?

2NO(g) Kc = 1 x 10^-30 at 298 K

2HI(g) Kp = 794 at 298 K

for the reactions shown above?

91 Many industrial processes depend upon the magnitude of the equilibrium constant. If a process is not likely to go to products at certain

make the reaction more favorable for the production of products. The favored or reactant-favored.

quilibrium mixture: its magnitude, in essence, lets us determine whether a reaction is more likely to occur in the forward direction or reverse direction. Thus, it tells us whether the equilibrium mixture will

1, tells us that the reaction is favored to the right, or product side. This means that the reactant. We can say that the

< 1, tells us that the reaction is favored to the left, or reactant side. This means that the amount of reactant. We can say that the

. It is not useful to simply say that the . We must say that the equilibrium constant of the forward reaction is 1.49 x 10⁸.

reactions in producing product by the reactions shown. That is, are

92 Part 6: MANIPULATING EQUATIONS AND Kc VALUES

We will consider at least three ways in which equations and their equilibrium constants can be manipulated:

• Changing a chemical equation – for example, by multiplying through by a value – increases the value of the equilibrium constant by a power equal to the change in the equation. This is easily seen by rewriting the equilibrium expression using the new stoichiometry. For example, multiplying the equation N2(g) + O2(g)⇋ 2 NO^(g) by two yields 2 N2(g) + 2 O2(g)⇋ 4 NO^(g), which causes the equilibrium expression to increase by the power of two, as in:

Kc (N2(g) + O2(g)⇋ 2 NO^(g)) = 1 x 10^-30 at 298 K, while Kc (2 N2(g) + 2 O2(g)⇋ 4 NO^(g)) = 1 x 10^-60 at 298 K

• Combining equations to arrive at an equation that is the sum of those combined requires that the individual equilibrium constants are multiplied by one another. This is easily seen by setting two expressions equal to a common species and rearranging the expressions. For example, we might be interested in the value of the equilibrium expression for the reaction 2 NOBr(g) + Cl2(g) ⇋ 2 NO^(g) + 2 BrCl(g). Although we do not have this value, we do know the values of the equilibrium constants for two reactions that sum to provide the reaction of interest:

2 NOBr(g)⇋ 2NO^(g) + Br2(g) Kc = 0.42 Br2(g) + Cl2(g) ⇋ 2BrCl^(g) Kc = 7.2

Multiplying these values of Kc gives (0.42 • 7.2) = 3.0 as the equilibrium constant for the reaction of interest.

• Reversing an equation would require using the inverse of the given Kc, as we saw in Part 4. This can easily be seen because to reverse a reaction requires the “flipping” of products and reactants.

Practice 5.4

2 A ⇋ 2 B + C (Kc,1 = 4.0) and C + D ⇋ 2 E (Kc,2 = 6.0)

Show that doubling a reaction’s stoichiometry results in squaring the value of its equilibrium constant.

Show that summing the two reactions above to yield 2 A + D ⇋ 2 B + 2 E gives Kc,3 = Kc,1 • Kc,2

93 Part 7: HETEROGENEOUS EQUILIBRIA

So far we have only considered the equilibrium established between species in the same phase – e.g., aqueous or gas. However, we should also consider the equilibria that may be established when the reactants and products may not be in a single phase. For example, equilibrium is established in the following reaction between all three species:

CaCO3(s)⇋ CaO^(s) + CO2(g)

For reasons we shall not consider⁴, it is possible to write an equilibrium expression for reactions occurring with heterogeneous species without including pure solids or liquids. Thus, the equilibrium expression for the above reaction is

Kp = PCO₂

The same could be said for a solvent like water.

• Pure solids and pure liquids do not appear in equilibrium expressions. However, you must not interpret this to mean that pure solids and liquids are not present at equilibrium or are not part of the chemical equation.

Practice 5.5

Determine the value of the equilibrium constant for the reaction: 2HF(aq) + C2O4

2-(aq) ⇋ 2F^-(aq) + H2C2O4(aq) given the following data:

HF(aq)⇋ H^+(aq) + F^–(aq) Kc = 6.8 x 10^-4 H2C2O4(aq)⇋ 2H⁺ + C2O42-(aq) Kc = 3.8 x 10^-6

What do the values of Kc in the original equations tell us about the direction of the equilibria for the two reactions shown?

Use the values of the equilibrium constants to justify that the two acids are weak.

Is the reaction of interest likely to occur to a great extent to products? Justify your answer.

4 This stems from the law of mass action; the values are essentially 1 for these phases.

94 Write the equilibrium expressions for the following heterogeneous reactions.

CO2(g) + H2(g)⇋ CO^(g) + H2O(l)

SnO2(s) + 2CO(g)⇋ Sn^(s) + 2CO2(g)

Sn(s) + 2H⁺(aq)⇋ Sn^2+(aq) + H2(g)

Consider for a moment the underlying concept behind the magnitude of Kc. Discuss how this might be related to the spontaneity of the reactions we saw in voltaic cells.

Part 8: CALCULATING EQUILIBRIUM CONSTANTS

There are two methods of calculating an equilibrium constant, and we will look at how to do this now. The purpose of calculating Kc

might be to determine the favorability of a process at a certain temperature.

• If you know the chemical equation and the values of the concentrations or partial pressures of all species at equilibrium, then you can directly calculate the equilibrium constant. Determine the balanced equation and write the expression for the equilibrium constant. Then, substitute values.

95 Practice 5.6

Calculate the equilibrium constants Kp and Kc for the production of ammonia from its elements at 745 K. The equilibrium pressures of the species are given here:

N2: 2.46 atm H2: 7.38 atm NH3: 0.166 atm

Calculate the equilibrium constant for the ionization of acetic acid at 298 K. The molar concentrations of the species present at equilibrium are given here:

[CH3COOH]: 0.0165 M [H⁺]: 0.000544 M [CH3COO^-]: 0.000544 M

When species’ equilibrium concentrations or partial pressures are unknown, the equilibrium expression and the stoichiometry of the reaction can provide a method for determining unknown values as long as some data can be obtained. Set up a table of values that you can complete for initial and equilibrium values. Once completed, the equilibrium values will provide the values to use in the equilibrium expression used to determine Kc.

1. Tabulate the known initial and equilibrium concentrations of the species in the equilibrium constant expression. Do not include pure solid, liquid or solvent species.

2. Calculate the change in concentrations of the species for which both initial and equilibrium concentrations are known.

3. Use the stoichiometry of the reaction – the coefficients – to determine the changes in all species from the initial conditions to the equilibrium.

4. Calculate the equilibrium concentrations. Use these values to calculate K^c.

Over the next few pages we shall complete several items that exemplify the material discussed above. In each case, various data will be available; this should give you a good mix of problems from which you can model solutions to a wide variety of equilibrium situations.

Please do not attempt to memorize a few cases that appear to recur – rather, be sure that you can analyze various scenarios in order to be successful in solving equilibrium problems!

Figure 30. Given initial data and determining product data, we can determine Kc.

96 Practice 5.7

Determine the value of Kc for the equilibrium established between ammonia and water to produce the ions

ammonium and hydroxide. The initial concentration in 5.0 L of water is 0.0124 M ammonia. The [OH^-] at equilibrium is 0.000464 M at 298 K.

What we have: We have the equilibrium concentration of a product, and we have the initial concentration of a reactant.

What we need: We need the equilibrium concentrations of all of the reactants and products in order to determine Kc. What we do: We set up a table that allows for a convenient method of determining the equilibrium

concentrations of other species.

Set up a table (widely called an “ICE” chart) and insert the known values for the species. An equation may need to be written and balanced first. Tables will include the initial concentrations, the change in concentrations, and the equilibrium concentrations.

NH3(aq) H2O(l) ⇋ NH4+(aq) OH–(aq)

Initial

Change X

Equilibrium

(Notice that we do not consider the liquid water in this heterogeneous equilibrium problem. Pure solids, liquids and solvents are not considered; here, liquid water is the solvent for ammonia.)

Determine the change values for which initial and equilibrium values are known. Here, we can calculate ∆[OH^-].

NH3(aq) H2O(l) ⇋ NH4+(aq) OH–(aq)

Initial 0.0124

X

0.0 0.0

Change

Equilibrium 4.64 x 10^-4

Use the stoichiometry to determine the change and equilibrium concentrations of the other species.

NH3(aq) H2O(l) ⇋ NH4+(aq) OH-(aq)

Initial 0.0124

X

0.0 0.0

Change + 4.64 x 10^-4

Equilibrium 4.64 x 10^-4

Use the equilibrium expression and the concentrations at equilibrium to determine Kc.

97

Determine the value of Kp for the equilibrium established between sulfur trioxide and its decomposition products, sulfur dioxide and oxygen gas. At 1000 K, a container was filled with sulfur trioxide gas to a pressure of 0.500 atm. At equilibrium, the partial pressure of sulfur trioxide is known to be 0.200 atm.

Set up a table and insert the known values for the species. An equation may need to be written and balanced first.

Determine the change values for which the initial and equilibrium values are known.

Use the stoichiometry to determine the change and equilibrium concentrations of the other species.

Use the equilibrium expression and the concentrations at equilibrium to determine Kp.

The pressure of gas PCl5 in an evacuated container was measured as 4.50 atm. After decomposition and the establishment of equilibrium, the total pressure in the container was measured as 7.00 atm. Use this information to determine the values of Kp and Kc for the decomposition of PCl5. The products are PCl3(g) and Cl2(g).

98 Part 9: REACTION QUOTIENTS

For any particular “random” mixture, the reaction may be moving toward equilibrium from reactants, moving toward equilibrium from products, or it may be at equilibrium. We may often be interested in discovering where a mixture lies on the “equilibrium continuum.”

• Determining Q, the reaction quotient

Using known values for a reaction mixture, we can determine the reaction quotient for a particular mixture. The reaction quotient is the mathematical result achieved by inserting known concentration or partial pressure values for a reaction mixture into an equilibrium expression. The reaction quotient is the ratio of products to reactants for a particular mixture of reactants and products.

We can compare this ratio to the known ratio provided as Kc, as we shall see on the next page. You should already see that when the concentrations’ quotient equals the value of Kc, then the reaction is at equilibrium. (Kc is really just a special case of Q.)

Practice 5.8

What is the reaction quotient, Q, when a mixture of 2.00 mol H2(g), 1.00 mol N2(g), and 2.00 mol NH3(g) is placed in a 2.00 L container? (Hint: We cannot use mol values in a Kc or Kp expression.)

• Using Q.

Once we have determined the reaction quotient, we can compare the value of the quotient to the equilibrium constant. The comparison of Q to Kc

allows us to determine whether the mixture described will move toward the production of reactants or products, or we will see that the mixture is at equilibrium.

Figure 31. The figure represents graphically the continuum from less-than-equilibrium concentrations to greater-than-equilibrium concentrations of products. It is important to note that a reaction will not remain on either side, but will instead work toward

equilibrium.

99

If Q < Kc, then the mixture has a larger than equilibrium reactant concentration, and the mixture will produce additional product

If Q = Kc, then the mixture is at equilibrium

If Q > Kc, then the mixture has a larger than equilibrium product concentration, and the mixture will produce additional reactant

It might be convenient to view the nature of Q in the form below to conceptualize its meaning:

Practice 5.9

After writing the balanced equation, determine the reaction quotient for the following mixture placed in a 0.50 L container. Then, determine whether the mixture is at equilibrium, will result in the production of reactants or will result in the production of products. The value of the

equilibrium constant is 51 at the temperature of the mixture.

0.02 mol HI(g), 0.01 mol H2(g) and 0.03 mol I2(g)

were mixed in a container. The reaction is the production of HI(g) from its elements.

When Q for a mixture is less than Kc, then the mixture is

“heavy” on reactants and will move toward products.

When Q for a mixture is equal to Kc, then the mixture is at equilibrium and will move in both directions at equal rates.

When Q for a mixture is greater than Kc, then the mixture is “heavy” on products and will move toward reactants.

A + B

⇋

C + D

100 Part 10: CALCULATING EQUILIBRIUM CONCENTRATIONS

In Part 8 we calculated the equilibrium constant for reactions in which we knew the equilibrium data for some species and the initial concentration for one or more species. We will now look at how to calculate the equilibrium concentrations of the species in a reaction when we know the equilibrium constant but do not know any equilibrium concentrations. Here, we are asking the question, “What will be the concentrations at equilibrium?”

We can calculate the concentrations of the species of a chemical reaction if we know the balanced chemical equation and the equilibrium constant. This provides the stoichiometry, the equilibrium expression and the ratio of products to reactants at equilibrium (i.e., Kc).

Behind the calculation of equilibrium concentrations is the observation that there is a predictable change in the concentrations based on the stoichiometry of an equation. Thus, we can utilize this to determine equilibrium concentrations. We shall see applications of this soon.

Construct a table similar in format to that used for calculating Kc. This table will include the initial concentrations (if required), the change in the concentrations as the reaction reaches equilibrium and the final equilibrium values.

Use the stoichiometry of the equation to determine the changes in concentration of the species. Because we do not know the final values at equilibrium (this is what we are trying to determine) we must use a variable for the change.

Use the initial and change concentrations to construct expressions for the equilibrium values:

[initial] – [change] = [equilibrium]

Substitute the expressions obtained in the previous step into the equilibrium expression for the reaction. This may require the solving of a quadratic equation if your calculator cannot handle variables.

Practice 5.10

A 1.000-L flask is filled with 1.000 mol hydrogen and 1.000 mol iodine gas at 721 K. The value of the equilibrium constant is 50.5 at this temperature. What are the concentrations of the species at equilibrium? The reaction is shown below.

H2(g) I2(g) ⇋ 2HI(g)

Initial

Change

Equilibrium

101 For the equilibrium N2(g) + O2(aq)⇋ 2 NO^2(g) the equilibrium constant, Kp, is 1.0 x 10^-30 at 298 K. A gas cylinder is charged with NO (g) to 1.66 atm at 298 K. What are the equilibrium values of all species?

What is the value of Kc for the reaction?

Is the reaction shown above a redox reaction? Justify your response. If it is, identify the oxidation numbers on all atoms on both sides.

Part 11: LE CHÂTELIER'S PRINCIPLE

Henri-Louis Le Châtelier (pronounced “le-SHOT-lee-ay”) discovered that the disturbance of a system at equilibrium will cause the system to move again toward equilibrium. He suggested the following principle, known as Le Châtelier’s Principle:

If a system at equilibrium is disturbed by a change in temperature, pressure or reactant or product concentrations, then the system will respond by shifting its equilibrium position to counteract the effect of the disturbance.

It is important to note that a Le Châtelier response is an initial response that ceases when the disturbance has passed and the reaction again reaches equilibrium. While the system will reestablish equilibrium where Q = Kc, you should note that this does not mean the values of reactants’ and products’ concentrations remain the same – they will change, but their ratio will not change.

• Changes in Temperature

When temperature is increased, the equilibrium shifts in the direction that absorbs heat. Treat “heat/energy” as a reactant (endothermic) or product (exothermic) to consider its effect. For exothermic reactions the equilibrium shifts to form additional reactant; for endothermic reactions the equilibrium shifts to produce additional product.

102 Practice 5.11

Show the use or production of heat in an endothermic and an exothermic reaction. Provide a brief discussion as to how Le Châtelier’s Principle suggests that the equilibrium will shift to the right for endothermic reactions and to the left for exothermic reactions when a reaction mixture is heated or cooled.

• Changes in Volume and/or Pressure

When a gas phase reaction’s volume is decreased, the pressure of the equilibrium mixture must increase – thus, Le Châtelier’s Principle indicates that the equilibrium will shift to decrease the pressure to its equilibrium value. That is, the equilibrium will shift to decrease the gas molecules in the volume: the equilibrium will shift to the side of the reaction that possesses fewer molecules of gas.

When a gas phase reaction’s volume is increased, the pressure of the equilibrium mixture must decrease – thus, Le Châtelier’s Principle indicates that the equilibrium will shift to increase the pressure to its equilibrium value. That is, the equilibrium will shift to increase the gas molecules in the volume: the equilibrium will shift to the side of the reaction that possesses more molecules of gas.

Practice 5.12

Consider the equilibrium established by the reaction between N2O4(g) ⇋ 2 NO²(g). Discuss the effect of the following changes according to Le Châtelier's Principle. (Temperature is constant where pressure changes, and pressure is constant where temperature changes.)

Increase the pressure of the container at equilibrium

Decrease the volume of the container at equilibrium

Increase the temperature at equilibrium; the reaction is endothermic.

Increase the volume of the container at equilibrium

103

• Changes in Species Concentration

The addition of a species at equilibrium will cause a system to counteract the addition of the species: more of the species will be consumed.

The removal of a species at equilibrium will cause a system to counteract the removal of the species: more of the species will be produced.

Practice 5.13

Consider the equilibrium established by the exothermic reaction 3H2 + N2⇋ 2NH^3. Discuss the effect of the following changes according to Le Châtelier's Principle:

Remove ammonia upon production.

Add additional nitrogen gas.

Increase the volume of the container under constant temperature and pressure

Decrease the pressure of the equilibrium mixture at constant temperature

Increase the temperature of the equilibrium mixture; the reaction is exothermic.

Add neon gas to the mixture at constant volume.

Figure 32. The addition of nitrogen gas causes a response in the reaction between nitrogen and hydrogen. The system responds by reestablishing an equilibrium so that Q = Kc.

ADVANCED PLACEMENT CHEMISTRY

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