The Equilibrium Law

Section 17.2 (AHL)

Vocabulary

Homogeneous equilibrium: all the reactants and products are in the same phase

Heterogeneous equilibrium: when there are two or more phases

Qc: reaction quotient, refers to a quotient obtained

by applying the equilibrium law to initial concentrations (instead of equilibrium concentrations)

Using Qc

The value of the reaction quotient can be used to determine how a reaction needs to shift in order to attain equilibrium

Qc=K

c equilibrium is established

Qc< K

c the system must shift to the right to

establish equilibrium Q

c> K

c the system must shift to the left

Example Calculation #1

The acid-catalyzed hydrolysis of ethyl ethanoate can be achieved by mixing the ester with dilute HCl

CH3COOC

2H

5(l) + H

2O

(l) CH

3COOH

(l) + C

2H

5OH

(l)

The products are ethanoic acid and ethanol

H+

Problem #1 Continued

If 1.00 mole of ethyl ethanoate is mixed with 1.00 mole of water and the reaction allowed to reach equilibrium at a particular temperature, then 0.30 moles of ethanoic acid is found in the equilibrium mixture

Calculate the value of Kc at this temperature

Make an ICE table

CH3COOC2H5 H2O CH3COOH C2H5OH

Initial Amount in moles

1.00 1.00 0.00 0.00

Change in amount +0.30

Equilibrium amount in moles

0.30

Continued

The rest of the table needs to be filled in to find K

c at that temperature

According to the stoichiometry, all the coefficients in the equation are all “1”, so if 0.30 mol of CH

3COOH is produced, then

0.30 mol ofvC2H

5OH is also produced

CH3COOC2H5 H2O CH3COOH C2H5OH

Initial Amount in moles

1.00 1.00 0.00 0.00

Change in amount +0.30 +0.30

Equilibrium amount in moles

0.30 0.30

Problem Continued

If 0.30 mole are present in the equilibrium mixture, then they must have been produced from the reaction of 0.30 moles of the ester and the water.

The amount of ester and water remaining at equilibrium must be 1.00-0.30 moles of each

CH3COOC2H5 H2O CH3COOH C2H5OH

Initial Amount in moles

1.00 1.00 0.00 0.00

Change in amount -0.30 -0.30 +0.30 +0.30

Equilibrium amount in moles

0.70 0.70 0.30 0.30

Problem Continued

We now have the number of moles, but not the volumes of each liquid.

You can assume 1 dm3 All molar values then become concentration

values(0.30)(0.30)

0.18(0.70)(0.70)cK

Example Problem #2 (ex. on page 194)

When a mixture initially containing 0.0200 mol dm-3 SO

2 and an equal concentration of

O2 is allowed to reach equilibrium in a closed

container of fixed volume at 1000K, it is found that 80.0% of the SO

2 is converted to

SO3. Calculate the value of the equilibrium

constant at that temperature.

Problem #2 Continued

Write the balanced equation first 2SO

2(g) + O

2(g) 2SO

3(g)

Make your table of ICE

SO2 O2 SO3

Initial Amount in moles

0.0200 0.0200 0.00

Change

Equilibrium amount in moles

Continued

Equilibrium [SO3] = (0.0200) (0.800) since

80% of the the SO2 turns into SO

3

[SO3] = 0.0160 mol dm-3

The stoichimetry says a 1:1 ratio between SO

3 and SO

2, so the [SO

2] is equal to 0.0200

minus 0.0160 which is 0.0040 mol dm-3

Continued

According to the stoichiometry, each sulfur trioxide molecule requires only ½ an oxygen molecule so:

[O2] = 0.0200 – (1/2)(0.0160) = 0.0120 mol

dm-3

SO2 O2 SO3

Initial Amount in moles

0.0200 0.0200 0.00

Change-0.0160 -0.0080 +0.0160

Equilibrium amount in moles

0.0040 0.0120 0.0160

Calculate Kc

=1333 mol-1dm3

1330 mol-1dm3 (to 3 sig figs)

2

2

(0.0160)

(0.0040) (0.0120)cK

Another Fun Problem #3

SO3(g)

+ NO(g)

NO2(g)

+ SO2(g)

Kc for this reaction is 6.78 at a specified

temperature If the initial concentrations of NO and SO

3

were both 0.03 mol dm-3, what would be the equilibrium concentration of each component?

Continued

We will have to use variables for the exponents to keep track of the concentrations and to solve

Let the change in concentration of the reactants both equal -x

Let the change in concentration of the products both equal +x

SO3 NO NO2 SO2

Initial Amount in moles

0.03 0.03 0.00 0.00

Change in amount -x -x +x +x

Equilibrium amount in moles

0.03-x 0.03-x x x

Continued

2 2

3

[ ][ ]

[ ][ ]c

NO SOK

SO NO

2

26.78

(0.03 )

x

x

6.780.03

x

x

Continued

2.60 (0.03 – x) = x

0.078 – 2.60x = x

0.078 = 3.60x

0.0217 = x (realize you are not finished when you calculate “x”)

2 .600 .03

x

x

SO3 NO NO2 SO2

Initial Amount in moles

0.03 0.03 0.00 0.00

Change in amount -x -x +x +x

Equilibrium amount in moles

0.03-x 0.03-x x x

SO3 NO NO2 SO2

Initial Amount in moles

0.03 0.03 0.00 0.00

Change in amount -0.0217 -0.0217 +0.0217 +0.0217

Equilibrium amount in moles

0.0083 0.0083 0.022 0.022

One more type of problem

N2 (g) 3H2 (g) 2NH3

Initial Amount in moles 1.00 3.00 0.00

Change in amount -0.062÷2= 0.031

-(3*(0.062÷2))=0.093 +0.062

Equilibrium amount in moles

0.969 2.91 0.062

Finish

Kc = 1.61 x 10-4

2

3

(0 .0 62 )

(0 .969 )(2 .91)cK