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Chapter 14Chemical

Equilibrium

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Describing Chemical Equilibrium1. Chemical Equilibrium—A Dynamic Equilibrium2. The Equilibrium Constant3. Heterogeneous Equilibria; Solvents in Homogeneous

Equilibria

Using the Equilibrium Constant4. Qualitatively Interpreting the Equilibrium Constant5. Predicting the Direction of Reaction6. Calculating Equilibrium Concentrations

Contents and Concepts

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Changing Reaction Conditions: Le Châtelier’s Principle

7. Removing Products or Adding Reactants

8. Changing the Pressure and Temperature

9. Effect of a Catalyst

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Learning Objectives

Describing Chemical Equilibrium1. Chemical Equilibrium—A Dynamic

Equilibriuma. Define dynamic equilibrium and chemical

equilibrium.b. Apply stoichiometry to an equilibrium

mixture.

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2. The Equilibrium Constanta. Define equilibrium-constant expression

and equilibrium constant.b. State the law of mass action.c. Write equilibrium-constant expressions.d. Describe the kinetics argument for the

approach to chemical equilibrium.e. Obtain an equilibrium constant from

reaction composition.

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2. The Equilibrium Constant (cont)

f. Describe the equilibrium constant Kp; indicate how Kp and Kc are related.State the law of mass action.

g. Obtain Kc for a reaction that can be written as a sum of other reactions of known Kc values.

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3. Heterogeneous Equilibria; Solvents in Homogeneous Equilibria

a. Define homogeneous equilibrium and heterogeneous equilibrium.

b. Write Kc for a reaction with pure solids or liquids.

Using the Equilibrium Constant

4. Qualitatively Interpreting the Equilibrium Constant

a. Give a qualitative interpretation of the equilibrium constant based on its value.

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5. Predicting the Direction of Reactiona. Define reaction quotient, Q.b. Describe the direction of reaction after

comparing Q with Kc.c. Use the reaction quotient.

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6. Calculating Equilibrium Concentrationsa. Obtain one equilibrium concentration given

the others.b. Solve an equilibrium problem (involving a

linear equation in x).c. Solve an equilibrium problem (involving a

quadratic equation in x).

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Changing the Reaction Conditions; Le Châtelier’s Principle

7. Removing Products or Adding Reactantsa. State Le Châtelier’s principle.b. State what happens to an equilibrium when

a reactant or product is added or removed.c. Apply Le Châtelier’s principle when a

concentration is altered.

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8. Changing the Pressure and Temperature

a. Describe the effect of a pressure change on chemical equilibrium.

b. Apply Le Châtelier’s principle when the pressure is altered.

c. Describe the effect of a temperature change on chemical equilibrium.

d. Apply Le Châtelier’s principle when the temperature is altered.

e. Describe how the optimum conditions for a reaction are chosen.

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9. Effect of a Catalyst

a. Define catalyst.

b. Compare the effect of a catalyst on rate of reaction with its effect on equilibrium.

c. Describe how a catalyst can affect the product formed.

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Chemical reactions often seem to stop before they are complete.

Actually, such reactions are reversible. That is, the original reactants form products, but then the products react with themselves to give back the original reactants.

When these two reactions—forward and reverse—occur at the same rate, a chemical equilibrium exists.

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The graph shows how the amounts of reactants and products change as the reaction approaches equilibrium.

CO(g) + 3H2(g) CH4(g) + H2O(g)

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This graph shows how the rates of the forward reaction and the reverse reaction change as the reaction approaches equilibrium.

CO(g) + 3H2(g) CH4(g) + H2O(g)

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Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.

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We can apply stoichiometry to compute the content of the reaction mixture at equilibrium.

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When heated PCl5, phosphorus pentachloride, forms PCl3 and Cl2 as follows:

PCl5(g) PCl3(g) + Cl2(g)

When 1.00 mol PCl5 in a 1.00-L container is allowed to come to equilibrium at a given temperature, the mixture is found to contain 0.135 mol PCl3. What is the molar composition of the mixture?

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We will organize this problem by using the chemical reaction to set up a table of initial, change, and equilibrium amounts.

Initially we had 1.00 mol PCl5 and no PCl3 or Cl2.

The change in each is stoichiometric: If x moles of PCl5 react, then x moles of PCl3 and x moles of Cl2 are produced.

For reactants, this amount is subtracted from the original amount; for products, it is added to the original amount.

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We were told that the equilibrium amount of PCl3 is 0.135 mol. That means x = 0.135 mol.

We can now find the amounts of the other substances.

PCl5(g) PCl3(g) + Cl2(g)

Initial 1.00 mol 0 0

Change –x +x +x

Equilibrium 1.00 – x x x

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Moles PCl5 = 1.00 – 0.135 = 0.87 mol

(2 decimal places)

Moles PCl3 = 0.135 mol

(given with 3 significant figures)

Moles Cl2 = 0.135 mol

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The Equilibrium Constant, KcThe equilibrium constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration term to a power equal to its coefficient in the balanced chemical equation.

The equilibrium constant, Kc, is the value obtained for the Kc expression when equilibrium concentrations are substituted.

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For the reaction

aA + bB cC + dD

The equilibrium constant expression is

Kc = ba

dc

BA

DC

?

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Methanol (also called wood alcohol) is made commercially by hydrogenation of carbon monoxide at elevated temperature and pressure in the presence of a catalyst:

2H2(g) + CO(g) CH3OH(g)

What is the Kc expression for this reaction?

COH

OHCH2

2

3cK

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When we are given some information about equilibrium amounts, we are able to calculate the value of Kc.

We need to take care to remember that the Kc expression uses molar concentrations.

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Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen:

2CO2(g) 2CO(g) + O2(g)

At 3000 K, 2.00 mol CO2 is placed into a 1.00-L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO2 remains.

What is the value of Kc at this temperature?

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We can find the value of x.

2.00 – 2x = 0.90

1.10 = 2x

x = 0.55 mol

0.55 mol1.10 mol0.90 mol

x2x2.00 – 2xEquilibrium

+x+2x–2xChange

002.00 molInitial

O2(g)2CO(g) +2CO2(g)

2x = 2(0.55) = 1.10 mol

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2CO2(g) 2CO(g) + O2(g)

22

22

CO

O COcK

M 1.10L 1.00

mol 1.10CO

M 0.90L 1.00

mol 0.90CO2

M 0.55L 1.00

mol 0.55O2

2

2

0.90

0.55 1.10cK

0.82cK

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In a heterogeneous equilibrium, in the Kc expression, the concentrations of solids and pure liquids are constant (due to these substances’ constant density).

As a result, we incorporate those constants into the value of Kc, thereby making a new constant, Kc. In other words, equilibrium is not affected by solids and pure liquids as long as some of each is present.

More simply, we write the Kc expression by replacing the concentration of a solid or pure liquid with 1.

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Write the Kc expression for the following reaction:

H2O(g) + C(s) CO(g) + H2(g)

OH

H CO

2

2cK

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Given:

aA + bB cC + dD; K1

When the reaction is reversed:

cC + dD aA + bB; K2

The equilibrium constant expression is inverted:

K2 = 1

1

DC

BA

Kdc

ba

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Given:

aA + bB cC + dD; K1

When the reaction is doubled:

2aA + 2bB 2cC + 2dD; K2

The equilibrium constant expression, K2 , is the square of the equilibrium constant expression, K1:

K2 =

ba

dc

22

22

BA

DC

2

BA

DC

ba

dc

21K

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For the reaction

aA(g) + bB(g) cC(g) + dD(g)

The equilibrium constant expressions are

Kc = and Kp = ba

dc

PP

PP

BA

DC ba

dc

BA

DC

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How are these related?

We know

From the ideal gas law, we know that

So,

A

A[A]V

n

RT

P

V

n A

A

A

RTV

nP

A

AA

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When you express an equilibrium constant for a gaseous reaction in terms of partial pressures, you call it the equilibrium constant, Kp.

In general, the value of Kp is different from that of Kc.

We will explore this relationship on the next slides. Recall the ideal gas law and the relationship between pressure and molarity of a gas:

RTV

n

V

nRTP

nRTPV

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ba

dc

p PP

PPK

BA

DC bbaa

ddcc

RTRT

RTRT

BA

DC

n

ba

dc

p RTK BA

DC

b-a-dc

ba

dc

p RTK BA

DC

Kp = Kc (RT)n

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For catalytic methanation,

CO(g) + 3H2(g) CH4(g) + H2O(g)

the equilibrium expression in terms of partial pressures becomes

and

22

RT

KRTKK c

cp

3HCO

OHCH

2

24

PP

PPKp

?

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The value of Kc at 227°C is 0.0952 for the following reaction:

CH3OH(g) CO(g) + 2H2(g)

What is Kp at this temperature?

Kp = 0.0952(RT)n

whereT = 227 + 273 = 500. KR = 0.08206 L atm/(mol K)n = 2

Kp = 1.60 × 102

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We can use the value of the equilibrium constant in several ways.

First, we can qualitatively describe the content of the reaction mixture by looking at the magnitude of Kc.

Second, we can determine the direction in which a reaction will proceed by comparing Kc to the value of the reaction quotient, Q, which has the same expression as Kc but uses nonequilibrium values.

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Finally, we can determine equilibrium concentrations given the initial concentrations and the value of Kc.

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When Kc is very large (>102), the equilibrium mixture is mostly products.

When Kc is very small (<10-2), the equilibrium mixture is mostly reactants.

When Kc approaches 1, the equilibrium mixture contains appreciable amounts of both reactants and products.

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Kc = 0.82 for a reaction. Describe the

composition of the equilibrium mixture.

Because Kc < 100 and > 0.01, at equilibrium there will be substantial amounts of both reactants and products.

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Reaction Quotient, Q

The reaction quotient has the same form as the equilibrium constant, but uses initial concentrations for its value.

When Kc > Q, the reaction proceeds to the right.

When Kc < Q, the reaction proceeds to the left.

When Kc = Q, the reaction is at equilibrium.

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Qc must move toward Kc.

Here the numerator must increase; more products must be produced.

Here the denominator must

increase; more reactants must be

produced.

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Calculating Equilibrium Concentrations

1. When all but one equilibrium concentration and the value of Kc are known.

2. When the value of Kc and the initial concentrations are known.

a. When the Kc expression is a perfect square: solving a linear equation.

b. When the Kc expression is not a perfect square: solving a quadratic equation.

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Nickel(II) oxide can be reduced to the metal by treatment with carbon monoxide.

CO(g) + NiO(s) CO⇌ 2(g) + Ni(s)

If the partial pressure of CO is 100. mmHg and the total pressure of CO and CO2 does not exceed 1.0 atm, will this reaction occur at 1500 K at equilibrium? (Kp = 700. at 1500 K.)

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forward. proceed willreaction The

ppK Q

6.6100

660

CO

CO2 P

PQp

mmHg 100.CO P

mmHg 660. 2CO P

?

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Nitrogen and oxygen form nitric oxide.

N2(g) + O2(g) 2NO(g)

If an equilibrium mixture at 25°C contains 0.040 M N2 and 0.010 M O2, what is the concentration of NO in this mixture? The equilibrium constant at 25°C is 1.0 × 10-30.

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0.010 0.040

NO10 1.0

230

0.0100.04010 1.0 NO -302

342 10 4.0 NO

M1710 2.0 NO

N2(g) + O2(g) 2NO(g)

22

2

O N

NOcK

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When the initial concentration and the value of Kc are known, we return to the stoichiometric chart of initial, change, and equilibrium (ICE) amounts or concentrations to find the equilibrium concentrations.

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Hydrogen iodide decomposes to hydrogen gas and iodine gas.

2HI(g) H2(g) + I2(g)

At 800 K, the equilibrium constant, Kc, for this reaction is 0.016.

If 0.50 mol HI is placed in a 5.0-L flask, what will be the composition of the equilibrium mixture in molarities?

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2HI(g) H2(g) + I2(g)

Initial 0.10 M 0 0

Change –2x +x +x

Equilibrium 0.10 – 2x x x

M 0.10L 5.0

mol 0.50HI 0

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The Kc expression is, Kc =

Substituting:

Because the right side of the equation is a perfect square, we can take the square root of both sides.

2

2

2 20.1020.10 0.016

x

x

x

xx

2

22

HI

IH

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Solving: 0.126(0.10 – 2x) = x 0.0126 – 0.252x = x

0.0126 = 1.252xx = 0.010 M

Substituting: Mx 0.010IH 22

Mx 0.080.0200.1020.10HI

x

x

20.10 0.126

22

20.10 0.016

x

x

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When the Kc expression is not a perfect square, the equation must be rearranged to fit the quadratic format:

ax2 + bx + c = 0

The solution is a

acbbx

2

4

2

?

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N2O4 decomposes to NO2. The equilibrium reaction in the gas phase is

N2O4(g) 2NO2(g)

At 100°C, Kc = 0.36.

If a 1.00-L flask initially contains 0.100 M N2O4, what will be the equilibrium concentration of NO2?

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N2O4(g) 2NO2(g)

Initial 0.100 M 0

Change –x +2x

Equilibrium 0.100 – x 2x

Again, we begin with the table:

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The Kc expression is, Kc =

Substitute:

Rearrange:

Substitute:

00.0360.364x2 x

42

22

ON

NO

0.0360.364 cba

x

x

x

x

0.100

4

0.100

2 0.36

22

42

0.036440.360.36

2 x

0 2 cbxax

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Solve:

We eliminate the negative value because it is impossible to have a negative concentration.

Substitute to find the equilibrium concentration of NO2:

8

0.70560.36

-x

0.1050.045 x

0.150.06 xx

Mx 0.120.0622NO2

?

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Given:

H2(g) + F2(g) 2HF(g); Kc = 1.15 × 102

3.000 mol of each species is put in a 1.500-L vessel. What is the equilibrium concentration of each species?

First calculate the initial concentrations:

M 2.000 L 1.500

mol 3.000HFFH 00202

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

Initial 2.000 M 2.000 M 2.000 M

Change –x –x +2x

Equilibrium 2.000 – x 2.000 – x 2.000 + 2x

22

2

FH

HFcK

2

22

-2.000

22.000 10 1.15

x

x

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2

22

-2.000

22.000 10 1.15

x

x

x

x

-2.000

22.000 10.72

xx 22.000 -2.00010.72

xx 2 2.000 10.72 21.44

Mx 1.5312.72

19.44

x12.72 19.44

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Now compute the equilibrium concentrations:

Double-check by substituting these equilibrium concentrations into the Kc expression and solving. The answer should be the value of Kc.

This is within round-off error.

Mx 0.47 1.532.0002.000FH 22

Mx 5.06 3.06 2.000 2 2.000HF

22

2

22

2

10 1.16(0.47)

(5.06)

F H

HFcK

?

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The value of Kc at 227°C is 0.0952 for the following reaction:

CH3OH(g) CO(g) + 2H2(g)

What is Kp at this temperature?

Kp = 0.0952(RT)n

T = 227 + 273 = 500. KR = 0.08206 L atm/(mol K)n = 2

Kp = 1.60 × 102

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Le Châtelier’s Principle

When a system in chemical equilibrium is disturbed by a change in

• temperature, • pressure, or • concentration,

the system shifts in equilibrium composition in a way that tends to counteract this change of variable.

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When a substance that is part of the equilibrium is added to the mixture, the equilibrium shifts to use it (in a direction that makes the substance a reactant).

When a substance that is part of the equilibrium is removed from the mixture, the equilibrium shifts to produce it (in a direction that makes the substance a product).

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Changes in the partial pressure of substances that are part of the equilibrium are handled in the same way as adding or removing a substance.

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The following reaction is at equilibrium:

COCl2(g) CO(g) + Cl2(g)

a. Predict the direction of reaction when chlorine gas is added to the reaction mixture.

b. Predict the direction of reaction when carbon monoxide gas is removed from the mixture.

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a. When we add Cl2, the reaction will shift in the reverse direction to use it. Note: reverse = left = .

b. When we remove CO, the reaction will shift in the forward direction to produce it. Note: forward = right = .

COCl2(g) CO(g) + Cl2(g)

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A change in the total pressure occurs because of a change in the volume of the reaction container.

When the size of the container decreases, the overall pressure increases. The reaction will shift to reduce the pressure—that is, it will shift toward the side of the reaction with fewer gas molecules.

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When the size of the container increases, the overall pressure decreases. The reaction will shift to increase the pressure—that is, it will shift toward the side with more gas molecules.

In the event that both sides of the equilibrium reaction have the same number of moles of gas, pressure has no effect on the equilibrium.

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In which direction will each reaction shift when the volume of the reaction container is increased?

a. CO(g) + 2H2(g) CH3OH(g)

b. 2SO2(g) + O2(g) 2SO3(g)

c. COCl2(g) CO(g) + Cl2(g)

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a.CO(g) + 2H2(g) CH3OH(g)This reaction shifts reverse = left =

b.2SO2(g) + O2(g) 2SO3(g)This reaction shifts reverse = left =

c. COCl2(g) CO(g) + Cl2(g)This reaction shifts forward = right =

When the container volume is increased, the total pressure is decreased. Each system will shift to produce more gas by shifting toward the side with more moles of gas.

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Changing the temperature changes the value of the equilibrium constant.

Changing the temperature can also cause a shift in the equilibrium.

The direction of each of these changes depends on the sign of Ho.

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For an endothermic reaction, Ho > 0 (positive), we consider that heat is a reactant.

For an exothermic reaction, Ho > 0 (negative), we consider that heat is a product.

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For an endothermic reaction, increasing the temperature increases the value of Kc.

For an exothermic reaction, increasing the temperature decreases the value of Kc.

Decreasing the temperature has the opposite effect.

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In addition to the value of Kc, we can consider the direction in which the equilibrium will shift.

When heat is added (temperature increased), the reaction will shift to use heat.

When heat is removed (temperature decreased), the reaction will shift to produce heat.

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Given:

2H2O(g) 2H2(g) + O2(g); Ho = 484 kJ

Would you expect this reaction to be favorable at high or low temperatures?

We rewrite the reaction to include heat:Heat + 2H2O(g) 2H2(g) + O2(g)

When heat is added, the reaction shifts forward = right = .

The reaction is favorable at high temperatures.

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The Fischer–Tropsch process for the synthesis of gasoline consists of passing a mixture of carbon monoxide and hydrogen over an iron–cobalt catalyst.

A typical reaction that occurs in the process is

8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)

Suppose the reaction is at equilibrium at 200°C, then is suddenly cooled to condense the octane, and then the remaining gases are reheated to 200°C. In which direction will the equilibrium shift?

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This is essentially removing octane, a product. This change causes the reaction to produce octane by shifting forward = right = .

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A typical reaction that occurs in the Fischer–Tropsch process is

8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)

In which direction will the equilibrium shift when the pressure is increased?

When the overall pressure is increased, the reaction will shift to reduce the pressure—that is, the reaction shifts to fewer gas molecules. In this case, the reaction will shift forward = right =