16-3-2011

Chemical Equilibrium

The Concept of equilibrium

Equilibrium: is a state in which there are no observable changes in concentration with time.

Chemical Equilibrium is reached when: 1. The rates of forward and reverse reactions are equal.

2. The concentrations of the reactants and products no longer change with time.

The Concept of Equilibrium

1. The reaction system is closed.

2. Opposing reactions occur at equal rates, i.e. the rates of forward and reverse reactions are equal.

3. Dynamic process, i.e. never stops.

4. Concentrations of reactants and products are constant.

5. Equilibrium can be reached from either direction, reactants or products.

Physical versus chemical equilibrium

A physical equilibrium is a process where no chemical reaction takes place, like phase changes: A chemical equilibrium is an equilibrium process where reactants are converted to products:

Dynamic Equilibrium

When compounds react to form products, their concentrations decrease while the concentration of products increase with time until a point is reached where no extra change in concentrations takes place. This point is called the equilibrium point where the rate of the forward reaction is the same as that of the backward reaction.

This is also called a

dynamic equilibrium

since reactants continue to form products but at the same rate as products reform reactants.

equilibrium

N 2 O 4 (g) 2NO

equilibrium

2 (g)

equilibrium

Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4

14.1

The Equilibrium constant

N 2 O 4 (g) 2NO 2 (g) K = [NO 2 ] 2 [N 2 O 4 ] = 4.63 x 10 -3 aA + bB cC + dD K = [C] c [D] d [A] a [B] b

Law of Mass Action

What Equilibrium Constants Tell Us

In fact, much can be known from the value of equilibrium constants where the following points could be made: 1.

When the equilibrium constant is very small. Most reactants do not undergo a reaction and very little products are formed. Usually the concentration of products can be neglected as compared to reactants concentrations.

2.

When the equilibrium constant is very high. In this case, most reactants disappear and the reaction container contains mainly the products. In calculations, one can neglect the concentration of remaining reactants as compared to concentrations of products.

3.

A situation where the equilibrium constant is moderate. Appreciable amounts of reactants are left and appreciable amounts of products are also formed in the reaction mixture. One cannot neglect the reactants or the products concentration.

•

Manipulating

K

Reverse reaction:

•

Adding reactions:

K K f net

 1

K r



K

1 

K

2 •

Multiplying reaction by some factor: #1 : C(s) + ½ O 2 (g)

⇌

#2: 2 C(s) + O 2 (g)

⇌

CO(g) 2 CO(g)

K

1

K

2  [

CO

]  [

O

2 ]   1 2 1 2  4 .

6  10 23

K

2  [

CO

] 2 [

O

2 ]  2 .

1  10 47

Ways of expressing equilibrium constants (K

c

and K

p

)

We can write the equilibrium constant in two forms: K c for reactions that occur in the solution or gaseous state. Their concentrations are calculated using molarities.

K p is used for reactions involving gases where pressures, in atm, are used to express concentrations.

For reactions involving gases, the partial pressures of reacting and produced gases are used instead of the respective concentrations. In this case, the equilibrium constant is referred to as K p where:

The equilibrium constant K

p

for the reaction 2NO 2 (g) 2NO (g) + O 2 (g) is 158 at 1000K. What is the equilibrium pressure of O 2 0.270 atm?

if the P NO = 0.400 atm and P NO

P

2 NO

P

O 2

K p

=

P

2 NO 2 =

P

O2 = K

p P

2 NO 2 2

P

NO

P

O2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm

14.2

Predicting the direction of a reaction

If Q c = K c or Q p equilibrium = K p , then the reaction is at If Q c is larger than K c or Q p is larger than K direction (will form reactants) p , then the reaction will proceed in the back (reverse) If Q c is smaller than K c or Q p is smaller than K p , then the reaction will proceed in the forward direction (will form products)

Solution Use the known concentrations to calculate Q. Compare Q with K c to answer questions (a) and (b). Use a table to answer part (c).

(a)Q is larger than K c , so the reaction is not at equilibrium.

(b)Because Q is larger than K c , the concentration of the product, NO, is larger than its equilibrium concentration and the concentrations of the reactants, N 2 and O 2 , are smaller than their equilibrium concentrations. Therefore some of the product, NO, will be consumed and more of the reactants, N 2 and O 2 , will be formed.

In the reaction: 2NO(g) + Cl 2 (g)

D

2NOCl(g) K c is 6.5*10 -4 at 35 o C and 2*10 -2 NO, 8.3*10 proceed?

-3 mole of Cl 2 mole of , and 6.8 moles of NOCl are mixed in a 2.0 L flask. In which direction will the equilibrium First, find the concentrations: [Cl 2 (g)]} = 6.5*10 -3 /2.0 = 3.25*10 -3 [NO(g)] = 2*10 -2 /2.0 = 1*10 -2 M [NOCl(g)] = 6.8/2.0 = 3.4 M M

Q c = [NOCl(g)] 2 Q c = (3.4) 2 /{[NO(g)] 2 [Cl /(1*10 -2 ) 2 (3.25*10 -3 ) 2 (g)]} Q c = 3.6*10 7 Q c >>> K c Therefore, the reaction has much more products than should be and thus the reaction will proceed in the back direction (towards reactants)

Relation between K

c

and K

p

K p and K c are not necessarily equal except in the case where the number of moles of gases in the reactants and products sides is equal. This can be rationalized using the following treatment of the hypothetical reaction:

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K

c

and K

p

.

CO (g) + Cl 2 (g) COCl 2 (g)

D

n = 1

K c

= [COCl 2 ] [CO][Cl 2 ] = 0.14

0.012 x 0.054

= 220 – 2 = -1

K p

= K

c (RT)

D

n

R = 0.0821

T = 273 + 74 = 347 K

K p

= 220 x (0.0821 x 347) -1 = 7.7

14.2