CHEMICAL EQUILIBRIUM - Los Angeles Mission College€¦ · CHEMICAL EQUILIBRIUM • Reactions that...

Chemistry 102 Chapter 14

1

CHEMICAL EQUILIBRIUM

• Reactions that can go in both directions are called reversible reactions. • These reactions seem to stop before they go to completion. • When the rate of the forward and reverse reactions become equal, an equilibrium system is

established.

Stepwise view to an equilibrium system

Step 1 A + B → C + D Fast No Reaction ← C + D

Step 2 A + B → C + D

Forward reaction slows down. There are fewer A and B molecules available.

A + B ← C + D Reverse reaction starts slowly at first. There are few C and D molecules available.

Step 3 A + B → C + D

Forward reaction slows down further as the number of A and B molecules decreases.

A + B ← C + D Reverse reaction speeds up as the number of C and D molecules increases.

Step 4 A + B forward

reverse → ← C + D

RATE OF FORWARD REACTION = RATE OF REVERSE REACTION


2

CHEMICAL EQUILIBRIUM

Characteristics of a Chemical Equilibrium System: 1. A mixture of Reactants and Products is present

2. The composition of the reaction mixture no longer changes:

• Concentration of reactants is constant

• Concentration of products is constant

NOTE: Concentration of reactants ≠ Concentration of products

3. A Chemical Equilibrium is a Dynamic Equilibrium; both reactions (forward and reverse) are still going on

4. The Dynamic Equilibrium may be controlled (shifted to the right to favor products, or shifted to the left to favor reactants) by changing the conditions for the reaction.

Definition of Chemical Equilibrium: • A state reached by a reaction mixture when the rate of forward reaction and the rate of

reverse reactions become equal.


3

COMPOSITION OF AN EQUILIBRIUM MIXTURE

• In an equilibrium mixture, the amount of one substance is determined experimentally and the amounts of the other substances are calculated.

Example 1:

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

When 1.000 mol CO and 3.000 mol H2 are placed in a 10.00 L vessel at 927 °C and allowed to come to equilibrium, the mixture is found to contain 0.387 mol H2O. How many moles of each substance are present at equilibrium?

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

Initial 1.000 3.000 0 0

∆ – 0.387 – 3(0.387) + 0.387 + 0.387

Equilibrium (1.000–0.387) (3.000–1.161) +0.387 +0.387

Equilibrium 0.613 mol 1.839 mol 0.387 mol 0.387 mol

Example 2: Consider the following dynamic equilibrium:

CO (g) + H2O (g) → ← CO2 (g) + H2(g) Suppose you start with a gaseous mixture containing 1.00 mol of CO and 1.00 mol of H2O. When equilibrium is reached at 1000 °C, the mixture contains 0.43 mol H2. What is the molar composition of the equilibrium mixture?

CO (g) + H2O (g) → ← CO2 (g) + H2 (g)

Initial 1.00 mol 1.00 mol 0 0

∆ – x – x + x + x

Equilibrium (1.00–x) (1.00–x) +0.43 +0.43



4

Example 3: 0.750 mol of N2 and 2.250 mol of H2 are placed into a reaction vessel at 450 °C and 10.0 atm.

N2 (g) + 3 H2 (g) → ← 2 NH3 (g)

What is the composition of the equilibrium mixture if you obtain 0.060 mol of NH3 from it?

N2 (g) + 3 H2 (g) → ← 2 NH3 (g)

Initial

∆

Equilibrium

Equilibrium


5

THE EQUILIBRIUM CONSTANT

• Consider the following equilibrium system:

N2O4 (g) forward

reverse → ← 2 NO2 (g)

NOTE: Both forward and reverse reactions are elementary reactions.

At Equilibrium: Rate (forward reaction) = Rate(reverse reaction)

kf [N2O4] = kr [NO2] 2

2 f 2

r 2 4

k [NO ] = = constant k [N O ]

f c

r

k = K = equilibrium constant k

implies molar concentration

In General: aA + bB → ← cC + dD

c d

c a b

[C] [D] K = [A] [B]

[C], [D] = molar concentrations (molarities) of the products at equilibrium [A], [B] = molar concentrations (molarities) of the reactants at equilibrium

• This relationship was postulated by two chemists in 1864 and called the law of mass action.


6

EQUILIBRIUM CONSTANT

• The Equilibrium Constant, Kc, is a constant for a particular reaction at a given temperature.

NOTE: 1. If the overall reaction occurs by a multistep mechanism, one can show that Kc for the

overall reaction equals a product of ratios of the individual rate constants.

2. It is common practice to write Kc without units

Example 1: Write the equilibrium constant expression for the equation shown below:

2 NO2 (g) + 7 H2 (g) → ← 2 NH3 (g) + 4 H2O(g)

Kc =

Example 2: The Equilibrium Constant for a reaction is:

4 5 3 2

c 4 6 2

[NH ] [O ] K = [NO] [H O]

What is the Equilibrium Constant expression when the equation for this reaction is halved and then reversed?

Original Equation:

4 NO + 6 H2O → ← 4 NH3 + 5 O2

Halved and Reversed:

Kc =


7

CALCULATING KC FOR REACTIONS

• Molar concentrations (molarities) of products and reactants at equilibrium must be substituted in the expression of Kc

Example 1: CO (g) + 3 H2 (g) → ← CH4 (g) + H2O (g)

When 1.00 mol CO and 3.00 mol H2 are placed in a 10.0 L vessel at 927 °C and allowed to come to equilibrium, the mixture is found to contain 0.387 mol H2O. Calculate the equilibrium constant for this reaction.

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

Initial 1.00 mol 3.00 mol 0 0

∆ – 0.387 – 1.16 + 0.387 + 0.387


Equilibrium amounts: Equilibrium Concentrations(M) CO: 0.613 mol 0.613 mol/10.00 L = 0.0613 M H2 : 1.84 mol 1.84 mol/10.00 L = 0.184 M CH4: 0.387 mol 0.387 mol/10.00 L = 0.0387 M H2O: 0.387 mol 0.387 mol/10.00 L = 0.0387 M

2 4 c 3 3

2

[H O] [CH ] (0.387 M)(0.387 M) K = = = 3.93 [CO] [H ] (0.0613 M(0.184 M)

Example 2: An 8.00 L reaction vessel at 491 °C contained 0.650 mol H2, 0.275 mol I2 and 3.00 mol HI. Assuming that the substances are at equilibrium, find the value of Kc at 491 °C. The equilibrium is:

H2(g) + I2(g) → ← 2HI(g)

Moles at Equilibrium: 0.650 0.275 3.00

Molarities at Equilibrium: 0.650 moles 0.275 moles 3.00 moles (moles/L)

8.00 L 8.00 L 8.00 L

Molarities at Equilibrium: 0.08125 M 0.03438 M 0.375 M

2 2

c 2 2

[HI] (0.375 M) K = = = 50.3 [H ] [I ] (0.08125 M)(0.03438 M)


8

Example 3: H2S, a colorless gas dissociates on heating: 2 H2S → ← 2 H2 (g) + S2 (g) When 0.100 mol of H2S was put in a 10.0 L vessel and heated to 1132 °C, it gave an equilibrium mixture containing 0.0285 mol H2. What is the value of Kc at this temperature?

Note: It is convenient to express equilibrium amounts of reactants and products in molar concentrations (molarities in mol/L)

Molarity of H2S = Molarity of H2 =

2 H2S (g) → ← 2 H2 (g) + S2 (g)

Initial

∆

Equilibrium

Equilibrium

Kc =

Example 4: In the contact process, sulfuric acid is manufactured by first oxidizing SO2 to SO3, which is then reacted with water. The reaction of SO2 with O2 is:

2 SO2 (g) + O2 (g) → ← 2 SO3 (g)

A 2.00 L flask was filled with 0.0400 mol SO2 and 0.0200 mol O2. At equilibrium, at 900 K, the flask contained 0.0296 mol SO3. Obtain the value of Kc.

Molarity of SO2 = Molarity of O2 =

2 SO2 (g) + O2 (g) → ← 2 SO3 (g)

Initial

∆

Equilibrium

Equilibrium

Kc =


9

GASPHASE EQUILIBRIA (KP)

• GasPhase Equilibria refers to equilibrium systems where all reactants and products are gases.

• Concentrations of gases can be expressed in terms of partial pressures, since the concentration of a gas is proportional to its partial pressure.

n mol P PV = nRT ( )= V L RT

n molar concentration of a gas = = P( V

1RT

)

• Kp is the equilibrium constant for a gaseous reaction expressed in terms of partial pressures. • Kp has a value different from Kc

Kp = Kc(RT) ∆n

sum of coefficients sum of coefficients ∆n = of – of

gaseous products gaseous reactants

Example 1:

CO (g) + 3 H2 (g) → ← CH4(g) + H2O(g)

4 2

2

CH H O p 3

CO H

P P K =

P P Kc = 3.92 (at 1200K)

∆n = (1+1) – (1+3) = – 2

Kp = Kc(RT) ∆n = 3.92 [(0.0821) (1200)] −2 = 4.04 x 10 −4

Example 2: The equilibrium constant Kc equals 10.5 for the following reaction at 227 °C.

CO(g) + 2 H2 (g) → ← CH3OH (g) What is the value of Kp at this temperature?

∆n =

Kp = Kc(RT) ∆n =

Constant at a given

temperature


10

OTHER EQUILIBRIUM CONSTANT

• If a given chemical equation can be obtained by taking the sum of other equations, the Equilibrium Constant for the overall equation equals the product of the equilibrium constants of the other equations.

Koverall = K1K2…

Example 1: The following equilibria occur at 1200K

CO(g) + 3 H2 (g) → ← CH4 (g) + H2O(g) K1 = 3.92

CH4(g) + 2 H2S (g) → ← CS2 (g) + 4 H2 (g) K2 = 3.3x10 4

CO(g) + 2H2S (g) → ← CS2 (g) + H2O (g) + H2 (g) K3 = ???

4 2 2 2 1 2 3 2

2 4 2

[CH ] [H O] [CS ] [H K = K = [CO] [H ] [CH ] [H S]

4 ]

4 1 2

[ CH K K = 2

2

] [H O] [CO] [ H

2 2 3

[CS ] [H x ]

4 ] 4 [ CH 3 2

2

= K ] [H S]

• The equilibrium constant for an overall equation is equal to the product of the equilibrium constants of the individual equations.

K3 = K1 x K2

• Because an equilibrium can be approached from either direction, the direction in which we write the chemical equation is arbitrary. For example, for the reaction shown below:

N2O4 (g) → ← 2 NO2 (g)

2 2

c 2 4

[NO ] K = = 0.212 (at 100 C) [N O ]

o

• When considering the equilibrium in the reverse direction: 2 NO2 (g) → ← N2O4 (g)

2 4 c 2

2

[N O ] 1 K = = = 4.72 (at 100 C) [NO ] 0.212

o

Kfwd = 1/Krev


11

CLASSIFICATION OF CHEMICAL EQUILIBRIA

• Chemical Equilibria can be classified according to the physical state of the reactants and products present:

I. Homogeneous Equilibrium An equilibrium that involves reactants and products in a single phase.

Example: Catalytic methanation:

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

II. Heterogeneous Equilibrium An equilibrium involving reactants and products in more than one phase

Example:

3 Fe (s) + 4 H2O (g) → ← Fe3O4 (s) + 4 H2 (g)

4 2

c 4 2

[H ] K = [H O]

NOTE: The concentrations of solids are omitted.

Reason: The concentration of a pure solid or pure liquid is a constant at a given temperature.

• Equilibrium Constant expression can be written with pure solids included:

4 2 4 4

2

3 c 3

[H ] [Fe O ] K = [Fe] [H O]

By rearrangement: 3

c

4

3 4

2 4

2

[Fe] K = [Fe

[H ] [H O O] ]

constant variable factors factors

3

c 3 4

4 ' 2 c 4

2

[Fe] K = [Fe O ]

[H ] K = [H O]

• The concentrations of pure solids and liquids are incorporated in the value of Kc ’ .

• The concentration of solvent is also omitted from the expression of Kc for a homogeneous reaction (if constant).

• The equilibrium is not affected by pure solids, pure liquids, or solvents.


12

Examples: Identify each of the following equilibriums as homogeneous or heterogeneous and write Kc expressions for each:

1. FeO (s) + H2 (g) → ← Fe (s) + H2O (g)

2. CH4 (g) + 2 H2S (g) → ← CS2 (g) + 4 H2O (g)

3. Ti (s) + 2 Cl2 (g) → ← TiCl4 (l)

4. Given the equilibrium shown below, for which Kp = 0.0752 at 480 °C:

2 Cl2 (g) + 2 H2O (g) → ← 4 HCl (g) + O2 (g)

What is the value of Kp for the reaction shown below:

2 HCl (g) + ½ O2 (g) → ← Cl2 (g) + H2O (g)


13

MEANING OF THE EQUILIBRIUM CONSTANT

• The magnitude of the equilibrium constant indicates the extent to which the forward and reverse reactions take place.

Kc >>> 1

amount of products at equilibrium

amount of reactants at equilibrium

Ø Products are favored at equilibrium

N2 (g) + 3 H2 (g) → ← 2 NH3 (g) Equilibrium Conc’s: 0.010 M 0.010 M 2.02 M

2

2

8 2

3 c 3 2

2

[NH ] (2.02 M) K = = = [N ][H ] (0.010 M)(

4 0.010 )

.1x10 M

Kc <<< 1



Ø Reactants are favored at equilibrium

N2 (g) + O2 (g) → ← 2 NO (g) Equilibrium Conc’s: 1.0 M 1.0 M 6.8x10 −16 M

2 16 2

c 2 2

31 [NO] (6.8x10 M) K = = = [N ][O ] (1.0 M)(

4.6x M)

10 1.0

Kc ≈ 1 (0.5 – 10)


≅


Ø Neither reactants, nor products are predominant

CO (g) + 3 H2 (g) → ← CH4 (g) + H2O (g) Equilibrium Conc’s: 0.0613 M 0.184 M 0.0387 M 0.0387 M

4 2 c 3 3

2

[CH ][H O] (0.0387 M)(0.0387 M) K = = = [CO][H ] (0.0613 M)(0.184 M)

3.93


14

Example 1: The Equilibrium Constant for the reaction below equals 4.0 x 10 13 at 25 0 C.

2 NO (g) + O2 (g) → ← 2 NO2 (g) (a) Does the equilibrium mixture contain predominantly reactants or products?

(b) If [NO] = [O2] = 2.0 x 10 −6 M at equilibrium, what is the equilibrium concentration of [NO2]?

2 2 2 2

c 2 c 2 2 2

2 2 13 6 2 + 2 c

2

2

[NO ] K = [NO ] = K [NO] [O ] [NO] [O ]

[NO ] = K [NO] [O ] = (4.0x10 )(2.0x10 ) ( 1.8x 2.0 10 x10 )= M

2 NO (g) + O2 (g) → ← 2 NO2 (g)

Equilibrium Conc’s: 2.0 x10 −6 M 2.0 x10 −6 M 1.8 x10 −2 M

much smaller much larger

Example 2: Sulfur dioxide reacts with oxygen at 25 °C to reach equilibrium. Kc for this reaction at this temperature is 8.0x10 35 .

2 SO2 (g) + O2 (g) → ← 2 SO3 (g) a) From the magnitude of Kc, which reaction is favored (forward or reverse) when

equilibrium is established?

b) If an equilibrium mixture is 1.0 M SO3 and has equal concentrations of SO2 and O2, what is the concentration of SO2 in the mixture? Does this result agree with what you expect from the magnitude of Kc?


15

PREDICTING THE DIRECTION OF REACTION

• Consider the following reaction and the following composition of gas mixture present.

CO (g) + 3 H2 (g) → ← CH4 (g) + H2O (g) Kc = 3.93 (1200 K)

Composition of gas mixture: 0.0200 M 0.0200 M 0.00100 M 0.00100 M

(NOT AT EQUIL)

At 1200K and in the presence of a catalyst, what would be more likely ? 1. Would forward reaction be favored? (would products be favored?)

OR 2. Would reverse reaction be favored? (would reactants be favored?)

• To answer this, a new quantity is introduced: Qc = the Reaction Quotient

= has the same form as Kc but the concentrations used to calculate its value are not equilibrium concentrations.

3 3 4 2

c 2 2 3 2

[CH ][H O] (1.00x10 M)(1.00x10 M) Q = = = 6.25 [CO][H ] (2.00x10 M)(2.00x10 M)

Qc = 6.25 Kc = 3.92 should decrease to

• For the reaction mixture to reach equilibrium: Qc 3.92

Qc decreases if: Ø [CH4], [H2O] decreases Ø [CO], [H2] increases

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

Ø The reaction goes to the left Ø Reverse reaction is favored Ø Formation of reactants is favored

Qc = 0 Kc Qc = ∞ Forward Rxn Reverse Rxn

Reactants Equilibrium Products

Increasing Qc


16

PREDICTING THE DIRECTION OF REACTION

• In General: For a reaction mixture that is NOT AT EQUILIBRIUM :

aA + bB → ← cC + dD

c d

c a b

[C] [D] Q = [A] [B] Compare: Qc to Kc

• If: Qc > Kc Reaction will go the left ( )

• If: Qc < Kc Reaction will go to the right ( )

• If: Qc = Kc Reaction mixture is at equilibrium ( )

Examples: The following reaction has an equilibrium constant , Kc, equal to 3.59 at 900 °C, and the following composition of reaction mixture:

CH4 (g) + 2 H2S (g) → ← CS2 (g) + 4 H2 (g) 1.26 M 1.32 M 1.43 M 1.12 M

(a) Is the reaction mixture at equilibrium?

4 4 2 2

c 2 2 4 2

[H ] [CS ] (1.12 M) (1.43 M) Q = = = 1.02 [CH ][H S] (1.26 M)(1.32 M)

Qc = 1.02 Kc = 3.59 Qc ≠ Kc Not at equilibrium

(b) If not at equilibrium, in which direction will the reaction go to reach equilibrium?

Qc < Kc

Ø Reaction will go to the right Ø Forward reaction is favored Ø Formation of products is favored


17

CALCULATING EQUILIBRIUM CONCENTRATIONS

• We can use equilibrium constant to calculate the equilibrium concentration of all the substances in the mixture. Some of the variations that these problems occur are shown in the examples below:

Example 1: Nitric oxide, NO, is formed in automobile exhaust by the reaction of N2 and O2 (from air):

N2 (g) + O2 (g) → ← 2 NO (g)

Kc for this reaction equals 0.0025 at 2127 °C. If an equilibrium mixture at 2127 °C contains 0.023 mol N2 and 0.031 mol O2 per liter, what is the equilibrium concentration of NO?

2 2

c c 2 2 2 2

3 c 2 2

[NO] K = [NO] = K [N ][O ] [N ][O ]

[NO] = K [N ][O ] = (0.0025)(0.023)(0.031)= 1.3x10 M

Example 2: The equilibrium shown below has a Kp value of 1.45x10 –5 at 500 °C. In an equilibrium mixture of the three gases at this temperature, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this mixture?

N2 (g) + 3 H2 (g) → ← 2 NH3 (g)

Kp =


18

Example 3: For the reaction shown below, initially a mixture contains 0.850 mol each of N2 and O2 in a 8.00 L vessel. Find the composition of the mixture when equilibrium is reached at 3900 °C. Kc = 0.0123 at 3900 °C

N2 (g) + O2 (g) → ← 2 NO (g)

First: Obtain starting molar concentrations in mol/L

0.850 mol [N2] = [O2] = = 0.10625 M

8.00 L

N2 (g) + O2 (g) → ← 2 NO (g)

Initial 0.10625 0.10625 0

∆ –x –x +2x

Equilibrium 0.10625–x 0.10625–x 2x

2 2

c 2 2

[NO] (2x) K = 0.0123 = = [N ][O ] (0.10625 x)(0.10625 x)

Taking square roots of both sides: 2x 0.1109= (0.10625 x)

Rearranging: 0.10625 2x x = = 18.03 x 0.1109

Rearranging further: 0.10625 = 19.03 x x = 5.583 x 10 −3 M

Equilibrium concentrations are:

[O2] = [N2] = 0.10625 – x = 0.10625 – 0.005583 = 0.101 M

[NO] = 2x = 2 x 5.583 x 10 −3 M = 0.0112 M


19

Example 4: Ammonium hydrogen sulfide decomposes at room temperature as shown below:

NH4HS (s) → ← NH3 (g) + H2S (g) Kp= 0.108 at 25°C

A sample of ammonium hydrogen sulfide is placed in a flask at 25°C. After equilibrium has been reached, what is the total pressure of the flask? (Note: solids have no pressure)

Kp = 0.108 =

Example 5: At relatively high temperatures, the following reaction can be used to produce methyl alcohol:

CO (g) + 2 H2 (g) → ← CH3OH (l) Kc= 13.5

If the concentration of CO at equilibrium were found to be 0.010 M, what would be the equilibrium concentration of H2?


20

CALCULATIONS INVOLVING QUADRATIC EQUATIONS

• Some of the problems involving equilibrium involve use of the quadratic equation in order to determine the unknown variable. The example below shown one such problem.

Example 6: PCl5 decomposes when heated: PCl5 (g) → ← PCl3 (g) + Cl2 (g) If the initial concentration of PCl5 is 1.00 M, what is the equilibrium composition of the gaseous mixture at 160 °C? Kc at 160 °C is 0.0211

PCl5 (g) → ← PCl3 (g) + Cl2 (g)

Initial 1.00 M 0 0

∆ –x +x +x

Equilibrium 1.00–x x x

3 2 c

5

[PCl ][Cl ] (x)(x) K = = [PCl ] (1.00

= 0.0211 x)

(0.0211)(1.00 – x) = x 2 0.0211 – 0.0211 x = x 2

x 2 + 0.0211 x – 0.0211 = 0 Quadratic Equation

2 1 2 ax + bx + c = 0 x ,x =

± 2 b b 4ac 2a

x 2 + 0.0211 x – 0.0211 = 0

x = 2 0.0211 ± (0.0211) 4( 0.0211)

= 2

0.0211 ±0.2913 2

• In theory: There are two mathematical solutions

• In practice: Only one solution makes physical sense


21

1 x = 0.0211 + 0.29 0. 13 = 2

1351 2 x = 0.0211 0.2913 = 2

0.1562

correct reject (concentration cannot

be negative) Equilibrium Concentrations are:

[PCl5] = 1.00 M – x = 1.00 M 0.1351 M = 0.86 M

[PCl3] = [Cl2] = x = 0.135 M

Example 7: Calculate the composition of the gaseous mixture obtained when 1.25 mol of carbon dioxide is exposed to hot carbon at 800 °C in a 1.25L vessel. The equilibrium constant K c at 800 °C is 14.0 for the reaction:

CO2 (g) + C (s) → ← 2 Cl (g)

Initial

∆

Equilibrium

[CO2] =

Kc = 14.0 =


22

LE CHATELIER’S PRINCIPLE

• The effect of changes on the equilibrium can be predicted using the Le Chatelier’s principle.

• Le Chatelier’s principle states that: “If a stress is applied to a system at equilibrium, the system will respond in such a way as to relieve the stress and restore a new equilibrium under a new set of conditions”.

• Note sequence of events: 1. Stress applied 2. Equilibrium system response (equilibrium shift) 3. New equilibrium

• Stress is a change in any of the following: A) Concentration of Reactants or Products B) Pressure C) Temperature

A) Effect of Concentration Change on Equilibrium (Adding or Removing Reactants or Products)

Consider: A + B → ← C + D

Stress: increased to B*

Response: Ø Forward reaction speeds up Ø Equilibrium shifts to the right Ø Products are favored

New Equil.: A’ + B’ → ← C’ + D’

decreased increased increased increased A’ < A B<B’<B* C’>C D’>D


23

EFFECT OF CONCENTRATION CHANGE ON EQUILIBRIUM

In General:

• Adding Reagent Ø Equilibrium always shifts in the direction that tends to reduce the concentration

of the added reacting species.

Ø When concentration of reactant is increased equilibrium shifts forward ( → )

Ø When concentration of product is increased equilibrium shifts reverse (← )

• Removing Reagent Ø Equilibrium always shifts in the direction that tends to increase the concentration

of the removed reacting species.

Ø When concentration of reactant is decreased equilibrium shifts reverse (← )

Ø When concentration of product is decreased equilibrium shifts forward ( → )

Example 1:

H2 (g) + I2 (g) → ← 2 HI (g) (700K) Starting conc’s 1.00 M 1.00 M 0

Change: –0.79 M –0.79 M + 1.58 M

Equil Conc’s: 0.21 M 0.21 M 1.58 M Stress: + 0.20 M

Response: Equilibrium shifts forward to use up added reactant

H2 (g) + I2 (g) → ← 2 HI (g) (700K) New Equil. Conc’s: 0.15 M 0.35 M 1. 70 M

0.21M<0.35 M<0.41 M


24

Example 2:

Cl2 (g) + 2 H2O (l) → ← HOCl (aq) + H3O + (aq) + Cl − (aq)

Reagent Species Change in concentration Equilibrium Shift

Cl2 increase

Cl2 decrease

H2O increase

H2O decrease

HOCl increase

HOCl decrease

Cl – increase

Cl – decrease

Example 3: CO (g) + 3 H2 (g) → ← CH4 (g) + H2O (g)

Molar Equil. Composition: 0.613 mol 1.839 mol 0.387 mol 0.387 mol

What changes in the amount of reagents would produce more CH4?

Ø increase amount of CO or H2 Ø remove CH4 or H2O

Ø Most practical and least expensive solution is to remove H2O (cool reaction mixture to condense water vapor)

CO (g) + 3 H2 (g) → ← CH4 (g) + H2O (g) Molar Equil. Composition: 0.613 mol 1.839 mol 0.387 mol 0.387 mol

Stress: − 0.387 mol

Response: Equilibrium shifts

New Equilibrium Composition: 0.491 mol 1.473 mol 0.509 mol 0.122 mol decreased decreased increased decreased

• Equilibrium shift can also be predicted from an evaluation of Qc (Reaction Quotient)


25

EFFECT OF PRESSURE CHANGE ON EQUILIBRIUM

• A change in pressure has an effect on equilibrium only when the following two conditions exist:

1. At least one of the reacting species (reactant or product) is a gas.

2. Total number of moles of gaseous reactants ≠ Total number of moles of

gaseous products

• The Pressure of a gas may be: Ø increased by decreasing the volume, at constant temperature

(achieved by decreasing the size of the reaction vessel)

Ø decreased by increasing the volume, at constant temperature (achieved by increasing the size of the reaction vessel)

• To predict the equilibrium shift caused by a change in pressure, consider the following:

The effect of increasing the pressure (decreasing the volume) of the equilibrium system

= Increasing the concentration of

gaseous reacting species (reactants and products)

Example 1: (One reacting species is a gas)

CaCO3 (s) → ← CaO (s) + CO2 (g)

Stress Pressure Change

Direction of Equilibrium shift

Amounts of Reacting Species CaCO3 CaO CO2

Decrease in volume increased Reverse (← )

(removes gas) increased decreased decreased

Increase in volume decreased Forward ( → )

(adds gas) decreased increased increased


26

Example 2: (All reacting species are gases)

• Number of moles of reactants ≠ Number of moles of products

N2 (g) + 3 H2 (g) → ← 2 NH3 (g)

4 moles 2 moles

Stress: Pressure is increased (Volume is decreased)

Response: Equilibrium shifts to the side that has the fewer number of moles ( → ) Reason: Fewer moles of gas will exert less pressure

(provided temperature is constant)

New Equil: N2 (g) + 3 H2 (g) → ← 2 NH3 (g) decreased decreased increased



Amounts of Reacting Species N2 H2 NH3

Decrease in volume increased Forward ( → )

(fewer mol of gas) decreased decreased increased

Increase in volume decreased Reverse (← )

(more mol of gas) increased increased decreased

Example 3: (Some (not all) reacting species are gases)

C (s) + 1 CO2 (g) → ← 2 CO (g)



Amounts of Reacting Species C CO2 CO

Decrease in volume

Increase in volume


27

Example 4: (Evaluating Qc to predict Equilibrium Shift)

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

4 moles 2 moles

Stress: Equilibrium mixture is compressed to half its volume, at constant temperature

Pressure Change: Pressure is doubled

Result: Partial pressure of all gaseous reactants (concentrations) are doubled

4 2 c 3

2

[CH ][H O] K = [CO][H ]

4 2 c 3

2

( [CH ])( [H O] Q = ( [CO])( [H

2 2 2 ]) 2

c 4 2 c c

2 3

K [CH ][H O] Q = Q 4 16

= [CO][H ] 4

Qc < Kc

Ø Reaction proceeds in the forward direction Ø Equilibrium shifts → Ø Products are favored

CONCLUSIONS: At constant temperature:

1. If the pressure is increased (Volume is decreased), the reaction shifts in the direction of fewer moles of gas

2. If the pressure is decreased (Volume is increased), the reaction shifts in the direction of more moles of gas.


28

EFFECT OF TEMPERATURE CHANGE ON EQUILIBRIUM

1. Effect on Rates of Reactions An increase in temperature:

Ø increases the rate of both forward and reverse reactions Ø causes the system to reach equilibrium faster

2. Effect on Equilibrium Ø Depends on the type of reaction (exothermic or endothermic)

N2 (g) + 3 H2 → ← 2 NH3 (g) ∆H = –91.8 kJ

Meaning:

N2 (g) + 3 H2 exothermic

endothermic → ← 2 NH3 (g) + 91.8 kJ

• An increase in temperature favors the endothermic reaction (← ) Reason: Ø the endothermic reaction absorbs heat and thus lowers the temperature Ø decreases Kc of the equilibrium system

• A decrease in temperature favors the exothermic reaction ( → ) Reason: Ø the exothermic reaction gives off heat and thus increases the temperature Ø increases Kc of the equilibrium system

N2 (g) + 3 H2 (g) → ← 2 NH3 (g) ∆H = –91.8 kJ

Stress Equilibrium Shift

Amounts of Reacting Species Kc N2 H2 NH3 Increase in temp. (Heat added) ← increased increased decreased decreased

Decrease in temp. (Heat removed) → decreased decreased increased increased

CONCLUSIONS: 1. Equilibrium Shift with Temperature Change

In order to increase the amount of products obtained: (a) the temperature must be increased if the reaction is endothermic ( ∆H 0 > 0) (b) the temperature must be decreased if the reaction is exothermic ( ∆H 0 < 0)

2. Effect of Temperature on Kc If the amount of products is to be increased, Kc should also increase:

(a) For an endothermic reaction, Kc is larger at higher temperatures (b) For an exothermic reaction, Kc is larger at lower temperatures.


29

EFFECT OF CATALYS ON EQUILIBRIUM

• A catalyst is a substance that increases the rate of a reaction but is not consumed by it.

2 SO2(g) + O2 (g) → ← 2 SO3 (g) Kc = 1.7 x 10 26

• Kc is very large, therefore the reaction should go almost to completion. • However, both forward and reverse reactions are very slow (if uncatalyzed) and it takes a very long

time to reach equilibrium with very little SO 3 obtained in a reasonable amount of time. • The use of a Pt catalyst speeds up both forward and reverse reactions and as a result equilibrium is

reached fast. • The equilibrium mixture contains mostly SO 3 due to the large K c .

A CATALYST: 1. Has no effect on the equilibrium composition 2. Speeds up the attainment of equilibrium 3. Is very useful for reactions:

Ø that are slow, if uncatalyzed Ø that have a large Kc,

4. Is not useful for reactions with a small Kc 5. May have an effect on the amount of product obtained if it affects the rate of one reaction out of

several possible reactions:

• GOAL: production of NO (used to obtain HNO3) by the oxidation of NH3

(a) 4 NH3 (g) + 5 O2 (g) → ← 4 NO + 6 H2O (g) Large Kc (slow)

(b) 4NO → ← 2 N2 (g) + 2 O2 (g) Large Kc (slow)

Overall: 4 NH3 (g) + 3O2 (g) → ← 2N2 (g) + 6 H2O (g)

NOTE: No NO is produced

SOLUTION: • Use of a Pt catalyst speeds up reaction (a) but not reaction (b). • Therefore use of Pt catalyst at moderate temperatures produces NO selectively, since

decomposition of NO is too slow at these temperatures to be significant.

RESULT: Equilibrium mixture will contain mostly NO

CHEMICAL EQUILIBRIUM - Los Angeles Mission College€¦ · CHEMICAL EQUILIBRIUM • Reactions that...

Documents

Transcript of CHEMICAL EQUILIBRIUM - Los Angeles Mission College€¦ · CHEMICAL EQUILIBRIUM • Reactions that...