Chemical Equilibrium

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# Chemical Equilibrium - PowerPoint PPT Presentation

Chemical Equilibrium. Chapter 14. Henri L. le Chatlier 1850-1936. Adapted thermodynamics to equilibria; formulated the principle known by his name. The Concept of Equilibrium. Consider colorless N 2 O 4 . At room temperature, it decomposes to brown NO 2 : N 2 O 4 ( g )  2NO 2 ( g ).

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### Chemical Equilibrium

Chapter 14

Henri L. le Chatlier

1850-1936.

equilibria; formulated the

principle known by his name.

The Concept of Equilibrium

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

N2O4(g)  2NO2(g).

(colorless)

(brown)

As NO2 is produced, it reacts back to form N2O4:

N2O4(g) 2NO2(g)

At some point, the forward and reverse rates become the same. This is equilibrium.

Furthermore, it is a dynamic equilibrium.

The double arrow implies a dynamic equilibrium

The Concept of Equilibrium

• Consider

Forward reaction: A  B Rate = kf[A]

Reverse reaction: B  A Rate = kr[B]

• At equilibrium kf[A] = kr[B].
• For an equilibrium we write

A

B

• As the reaction progresses
• [A] decreases to a constant,
• [B] increases from zero to a constant.
• When [A] and [B] are constant, equilibrium is achieved.

The Concept of Equilibrium

Consider reaction:

A(g) B(g)

PA

At beginning: PA = PAo

PB = 0

PB

As reaction proceeds,

Equilibrium occurs when there is no further change in concentration as time progresses.

The Concept of Equilibrium

Alternatively:

As reaction proceeds,

forward rate, kf[A] decreases

reverse rate, kr[B] increases

until the two rates become the same. This is equilibrium.

Or, when kf[A] = kr[B], equilibrium is achieved.

The Equilibrium Constant

When dealing with gases, partial pressures can conveniently be used.

Consider

• If we start with a mixture of N2 and H2 (in any proportions), the reaction will reach equilibrium with a constant set of pressures of N2, H2 and NH3.
• If we start with just NH3 and no N2 or H2 , the reaction will proceed in reverseand N2 and H2 will be produced until equilibrium is achieved, with a constant set of pressures of N2, H2 and NH3.

The Equilibrium Constant

• No matter the starting composition of reactants and products, the same ratio of concentrations (for solutions) or pressures (for gases) is achieved at equilibrium.
• For a general reaction

aA + bB pP + qQ

the equilibrium constant expression is

where Keq is the equilibrium constant (in terms of

concentration or pressure)

The Magnitude of Equilibrium Constants

• The equilibrium constant, Keq, is the ratio of products to reactants.
• Therefore, the larger the Keq the more products are present at equilibrium.
• Conversely, the smaller the Keq the more reactants are present at equilibrium.
• If Keq >> 1, then products dominate at equilibrium and equilibrium lies to the right.
• If Keq << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.
• Keq = kf/kb

[B]

[A]

For A  B, Keq =

An equilibrium is established by placing 2.00 moles of frozen N2O4(g) in a 5.00 L and heating the flask to 407 K. It was determined that at equilibrium the amount of the NO2(g) is 1.31 mole. What is the value of the equilibrium constant?

[NO2]2

Kc =

[N2O4]

Use

molarities

in table

Set up a table.

Let x = [NO2 ] produced.

N2O4(g) 2 NO2(g)

[Initial] (mol/L) 0.400 M 0 M

[change]

[Equilibrium] 0.262

given

given 2.00/5.00 = 0.400 M

-½ x = -0.131

+ x = 0.262

0.269

K = (0.262)2/0.269 = 0.255

Given 1.31/5.00 = 0.262 M

N

O

(

g

)

2

N

O

(

g

)

2

4

2

We just calculated that at 407 Kelvin , Kc = 0.255 for

For the reverse reaction, calculate the new K by

taking the reciprocal: Kc = 1/0.255 = 3.92

Proof:

Kc = [N2O4]/[NO2]2

(0.269)/(0.262)2 =3.92

For the “half” reaction, calculate the square root:

Kc = (0.255)1/2 = 0.505

½ N

O

g

)

N

O

(

g

)

(

2

4

2

Proof:

Kc = [NO2]/[N2O4]1/2 =

(0.262)/(0.269)1/2 =

= 0.505

Be careful! Kc values must be

evaluated for the reaction as written!

Heterogeneous Equilibria

• When all reactants and products are in one phase, the equilibrium is homogeneous.
• If one or more reactants or products are in a different phase, the equilibrium is heterogeneous.
• Consider:

solids

• experimentally, the amount of CO2 does not depend on the amounts of CaO and CaCO3. Solids and pure liquids are not considered in the equilibrium constant expression. Thus,

Kc = [CO2]

Calculating Equilibrium Constants

For 2 NO(g) + Cl2(g)  2NOCl(g),

calculate Keq given that:

PNO=0.095 atm; PCl2=0.171 atm and PNOCl= 0.28 atm

at equilibrium.

= (0.28)2/(0.095)2(0.171)

= (0.0784)/(0.009025)(0.171)

= (0.0784)/(0.00154)

= 50.8

Calculating Equilibrium Constants

1.374 g H2 and 70.31 g Br2 heated in 2-L vessel at 700 K are allowed to react. At equilibrium, the vessel contains 0.566 g H2.

Calculate Kc .

The reaction is:

H2 (g) + Br2 (g) 2 HBr (g)

What is known?

start: 1.374 g 70.31 g 0

at equil 0.566 g ? ?

Start by changing everything from grams to moles:

H2: 1.374 g = 0.687 mol; at equil H2: 0.566 g = 0.283 mol

Br2: 70.31 g = 0.440 mol

Now divide by 2 to give molarities (the vessel is 2-L)

Initial [H2]= 0.344 mol; Final [H2] = 0.142

Initial [Br2] = 0.220

1st step

2nd step

3rd step

Set up table (values are molarities):

H2 (g) + Br2 (g) 2 HBr (g)

Initial 0.344 0.220 0

Change

Final 0.142

Now fill in changes:

H2 (g) + Br2 (g) 2 HBr (g)

Initial 0.344 0.220 0

Change -0.202 -0.202 +0.404

Final 0.142

Now calculate final values:

H2 (g) + Br2 (g) 2 HBr (g)

Initial 0.344 0.220 0

Change -0.202 -0.202 +0.404

Final 0.142 0.018 0.404

4th step

Kc = [HBr]2/[H2][Br2]

= (0.404)2/(0.142)(0.018)

= 63.8

Applications of Equilibrium Constants

Predicting the Direction of Reaction

• We define Q, the reaction quotient, for a general reaction

as

• where [A], [B], [P], and [Q] are molarities at any time.
• Q = Keqonly at equilibrium.

Applications of Equilibrium Constants

Predicting the Direction of Reaction

• If Q > Keq 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 Keq).
• If Q < Keq then the forward reaction must occur to reach equilibrium.

Le Châtelier’s Principle

• Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will shift to remove the disturbance.
• Three types of disturbances:
• Amount of material. If you add a reactant or remove a product, the equilibrium shifts to the right. If you remove a reactant or add a product, the equilibrium shifts to the left.
• Pressures. By increasing the pressure you shift the equilibrium in the direction of the lesser number of gas moles; by decreasing the pressure you shift the equilibrium in the direction of the greater number of gas moles.
• Temperature. By increasing the temperature, you shift an exothermic reaction to the left, and an endothermic reaction to the right. By decreasing the temperature, you shift an exothermic reaction to the right, and an endothermic reaction to the left.

Le Châtelier’s Principle

• Consider the production of ammonia (the Haber process)

+ heat

The following have been observed for this exothermic reaction:

• If hydrogen is added, the equilibrium shifts to the right.
• If ammonia is removed, the equilibrium shifts to the right.
• As the pressure increases, the amount of ammonia present at equilibrium increases (reaction goes to right).
• As the temperature decreases, the amount of ammonia at equilibrium increases (reaction goes to right).
• If hydrogen is removed, the equilibrium shifts to the left.

Le Châtelier’s Principle

The Effect of Catalysts

• A catalyst lowers the activation energy barrier for the reaction.
• Therefore, a catalyst will decrease the time taken to reach equilibrium.
• A catalyst does not effect the composition of the equilibrium mixture.
• Catalysis is a kinetics matter, not one of equilibrium.
• Catalysis speeds up a reaction, but the reaction arrives at the same equilibrium point (it just gets there faster).

A catalyst changes the rate of a chemical reaction, by lowering the activation energy Ea., but the ΔH is not changed.

ΔH