ch 13 chemical equilibrium. the concept of equilibrium chemical equilibrium occurs when a reaction...

CH 13

Chemical Equilibrium

The Concept of EquilibriumThe Concept of Equilibrium

• Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

• As a system approaches equilibrium, both the forward and reverse reactions are occurring.

• At equilibrium, the forward and reverse reactions are proceeding at the same rate.

Chemical equilibrium

• Dynamic processDynamic process • Rate of forward Rxn = Rate of reverse RxnRate of forward Rxn = Rate of reverse Rxn

HH22OO(l) (l) HH22OO((g)g)

(reactant) (product)(reactant) (product)

Dynamic

EquilibriumConcentration of reactants Concentration of reactants and products remain constantand products remain constant over time.over time.

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

At Equilibium

SO2(g)+O2(g)

Initially

SO3(g)

Initially

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

At Equilibium

SO2(g)+O2(g)

Initially

SO3(g)

Initially

Depicting EquilibriumDepicting Equilibrium

• Consider the reversible reaction between N2O4 and NO2:

• In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow.

N2O

4 2NO

2

A System at EquilibriumA System at Equilibrium

• Once equilibrium is achieved, the amount of each reactant and product remains constant.

The Equilibrium ConstantThe Equilibrium Constant

• Mathematically, the rate of a chemical reaction is described though its rate law.

• For the forward reaction:

• The forward rate law is given as:

Where: kf is the rate constant

[N2O4] is the concentration of N2O4

N2O

4 2NO

2

Rate = kf [N

2O

4]

The Equilibrium ConstantThe Equilibrium Constant

• For the reverse reaction:

• And the rate law is:

Rate = kr [NO

2]2

2NO2 N

2O

4

The Equilibrium ConstantThe Equilibrium Constant

• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Rewriting, it becomes:

kf

kr

[NO

2]2

[N2O

4]

The Equilibrium ConstantThe Equilibrium Constant

• The ratio of the rate constants is a constant, at that temperature, thus:

• Where Keq is the “Equilibrium Constant” for the reversible reaction.

K

eq

kf

kr

[NO

2]2

[N2O

4]

The Equilibrium ConstantThe Equilibrium Constant

• To generalize this expression, consider the general chemical reaction:

• The equilibrium expression for this reaction would be:

• This is the generalized equilibrium constant based on concentration of the components in the system.

aA bB cC dD

K

c

[C]c [D]d

[A]a[B]b

[C]c [D]d

[A]a [B]bKeq =

Equilibrium constant (K)

• Equilibrium expressionEquilibrium expression

(for any reaction at constant temperature)

aA + bB cC + dD

moles per litermoles per liter

coefficients

productsreactants

[C]c [D]d

[A]a [B]bKeq =

aA + bB cC + dDproductsreactants

Equilibrium constant (K)

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

Keq = [ NH3 ] 2

[ N2 ] [ H2 ] 3

Equilibrium constant (K)

Equilibrium Expression

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

K NO H O

NH O2

2

24 6

34 7

The Equilibrium ConstantThe Equilibrium Constant

• Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written as a pressure-dependent constant:

• Since, from the Ideal Gas Law:

K

p

pCc p

Dd

pAa p

Bb

PV nRT

p

i n

iV RT c

iRT

Relationship between Relationship between KKcc and and KKpp

• Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes:

• Where: δn = (moles of product) – (moles of reactant)

K

pK

cRT n

Relationship between Kc and Kp

• Relationship between concentration and pressure obtained from the ideal gas law. • Recall PV = nRT or

• Substitute for P in equilibrium expression. Consider the reaction:aA + bB cC + dD

• Use this relationship to relate KP and Kc

E.g. Determine Kp for the synthesis of ammonia at 25°C;N2(g) + 3H2(g) 2NH3(g) Kc = 4.1x108

E.g. 2 Determine Kp for the synthesis of HI at 458°C;½ H2(g) + ½ I2(g) HI(g) Kc = 0.142.

RT]A[

RTV

nP A

A

nc

)ba()dc(cP

baba

dcdc

ba

dc

bB

aA

dD

cC

P

RTKRTKK

RT]B[]A[

RT]D[]C[

RT]B[RT]A[

RT]D[RT]C[

PP

PPK

• It does not matter whether we start with N2 and H2 or whether we start with NH3. We will have the same proportions of all three substances at equilibrium.

What Does the Value of What Does the Value of KK Mean? Mean?

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

• If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

The Concentrations of Solids and The Concentrations of Solids and Liquids Are Essentially ConstantLiquids Are Essentially Constant

• Both can be obtained by dividing the density of the substance by its molar mass — and both of these are constants at constant temperature.

• e.g., the concentration of water molecules can be calculated as:

H2O

MwtH2O

1000 g/L

18 g/mol55.6 mol/L

The Concentrations of Solids and The Concentrations of Solids and Liquids Are Essentially ConstantLiquids Are Essentially Constant

• Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression – they are taken to have invariant concentrations or “activities” equal to unity (1):

• And:

PbCl

2(s) Pb2

(aq) 2Cl

(aq)

K

c

[Pb2 ][Cl ]2

[PbCl2]

[Pb2 ][Cl ]2

Equilibrium CalculationsEquilibrium Calculations

• A closed system initially containing 0.001M H2 and 0.002 M I2 is allowed to reach equilibrium at 448oC.

• Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M.

• Calculate Kc at 448oC for the reaction taking place, which is:

H

2(g) I

2(g)2HI

(g)

What Do We Know?What Do We Know?

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At equilibrium

1.87 x 10-3

[HI] Increases by 1.87 x 10[HI] Increases by 1.87 x 10-3-3 MM

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At equilibrium

1.87 x 10-3

StoichiometryStoichiometry tells us [H tells us [H22] and [I] and [I22]]

decrease by half as much …decrease by half as much …

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

1.87 x 10-3

We can now calculate the equilibrium We can now calculate the equilibrium concentrations of all three compounds…concentrations of all three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

……and, therefore, the equilibrium and, therefore, the equilibrium constant is calculated as:constant is calculated as:

Kc

[HI]2

[H2][I

2]

1.87x10 3 26.5x10 5 1.065x10 3 51

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, so only gases and/or aqueous species are included in the expression.

Reaction Quotient

If the law of mass action is applied with initial concentrations, the expression becomes a value called the reaction quotient (Q). This helps to determine the direction of the move toward equilibrium.

If Q > K then the reaction will favor reactants

If Q < K then the reaction will favor products

If Q=K the reaction is at equilibrium already

Reaction Quotient (continued)

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

QHF

H F2 2

02

0 0

Compare Q with K to determine direction :

Q > K to left Q < K to right Q = K at Equilibrium

The Reaction Quotient (The Reaction Quotient (QQ))

• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.

• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

Q

[C]ic [D]

id

[A]ia[B]

ib

• If Q = K, then the system is at equilibrium.

• If Q > K, there is too much product and the equilibrium will shift to the left.

• If Q < K, there is too much reactant and the equilibrium will shift to the right.

aA bB cC dD

Le Châtelier’s PrincipleLe Châtelier’s Principle

“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

Le Chatelier’s principle

• Stress causesStress causes shift shift in equilibriumin equilibrium• Adding or removing reagent

• N2(g) + 3 H2(g) 2 NH3(g)

• Stress causesStress causes shift shift in equilibriumin equilibrium• Adding or removing reagent

• N2(g) + 3 H2(g) 2 NH3(g)

Add moreN2?

NN22

Reaction shifts to the right[NH3] inc, [H2] dec

Le Chatelier’s principleLe Chatelier’s principle

Adding or removing reagent

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

Adding or removing reagent

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

Add moreNH3?

NHNH33

Reaction shifts to the left[NN22] and [H2] inc

Le Chatelier’s principleLe Chatelier’s principle

Adding Pressure affects an equilibrium with gases

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

Add PP?

Increasing pressure causes the equilibrium to shift to the side with the least moles of gas.

4 mol

of reactants

2 molof products

Solving Equilibrium Problems From Initial Conditions

1. Balance the equation.

2. Write the equilibrium expression.

3. List the Initial concentrations.

4. Calculate Q and determine the shift (Change) to

equilibrium.

Solving Equilibrium Problems

5. Define Equilibrium concentrations.

6. Substitute equilibrium concentrations into equilibrium expression and solve.

7. Check calculated concentrations by calculating K.

Take note of ICE calculations

Example 13.9

Example 13.10

• Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K the value of K is 5.10. Find equilibrium concentrations if 1.00 mol of each component is mixed in a 1.00 L flask.

Example 13.12

Example 13.13—For Small Values of K