2. Solv ing Equilibrium Problems - Chem 35.5 · 2007. 11. 21. · 2. Solv ing Equilibrium Problems ¥ T wo basic flavors of chemical equilibrium problems 1. Equilibrium quantities

2. Solving Equilibrium Problems

• Two basic flavors of chemical equilibrium problems

1. Equilibrium quantities are given (concentrations or partial pressures) and we solve for Kc. This is easy plug and chug.

2. Kc and initial quantities are given and we solve for the

equilibrium concentrations.

In either case we can use a book-keeping techniqueInitial, Change Equilibrium Method

Using ICE Book-keeping

1. Alway write a balanced chemical equation.

2. Make an organized table with a balanced equation at the top followed by a row of initial concentrations of reactants and products, another row of changes in concentrations of reactants and products and a final row of equilibrium concentrations.

3. Use as the unknown variable x for one substance and use the stoichiometry to determine relative concentrations of other reactants and products. Think, before choosing the x-variable.

6. Substitute the equilibrium concentrations into the equilibrium expression from the balanced equation.

7. Solve the equation for x

8. Answer the questions, remember significant figures.

Determining equilibrium concentrations from Kc

PROBLEM:

In a study concerning the conversion of methane to other

fuels, a chemical engineer mixes gaseous CH4 and H2O in a

0.32 L flask at 1200 K. At equilibrium, the flask contains 0.26

mol of CO, 0.091 mol of H2, and 0.041 mol of CH4. What is

the [H2O] at equilibrium? A Handbook states that Kc = 0.26

for this reaction is:

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

SOLUTION:

PLAN: Use the balanced equation to write the Kc expression, and then

substitute values for each component.

= 0.13 M

0.26mol

0.32 L

= 0.81 M

0.091 mol

0.32 L

= 0.28 M

[H2O]eq =[CO]eq[H2]eq

3

[CH4]eq Kc

= 0.53 M=(0.81)(0.28]3

(0.13)(0.26]

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

initial

change

equilibrium

?

? ??

? ?

0.041 mol/0.32 L ?

?

?

Kc =[CO][H2]3

[CH4][H2O]

PLAN:

Calculating Kc from concentration data

PROBLEM: In a study of hydrogen halide decomposition, a researcher fills an evacuated 2.00 L flask with 0.200 mol of HI gas and

allows the reaction to proceed at 453oC. At equilibrium, it is found that the concentration of [HI] = 0.078 M. What is the equilibrium constant Kc?

2HI(g) H2(g) + I2(g)

Find the molar concentration of the starting material (in this case [HI] and then use algebra and the balanced chemical equation to determine the amount of reactants and products at equilibrium.

The problem said that: [HI]eq = 0.078 M

[HI]eq = 0.078 M = 0.100 - 2x ; x = 0.011 M

Kc =[H2] [I2]

[HI]2=

[0.011][0.011]

[0.078]2= 0.020 = Kc

concentration (M) 2HI(g) H2(g) + I2(g)

initial

change

equilibrium

0.100

-2x xx

0 0

0.100 - 2x x x

SOLUTION:

[HI] = 0.200 mol

2.00 L= 0.100 M

2HI(g) H2(g) + I2(g)

Kc =[H2] [I2]

[HI]2

Determining equilibrium concentrations from initial

concentrations and Kc

PROBLEM:

Fuel engineers use the extent of the change from CO and H2O to

CO2 and H2 to regulate the proportions of synthetic fuel mixtures. If

0.250 mol of CO and 0.250 mol of H2O are placed in a 125 mL flask

at 900K, what is the composition of the mixture at equilibrium? At

900K, Kc is 1.56 for this reaction.

CO(g) + H2O(g) CO2(g) + H2(g)

PLAN: 1)Balance equation, 2)write equilibrium expression, 3)set up ICE table, 4)find the concentrations of all species at initial conditions or equilibrium in the problem 5)use algebra to determine equilibrium concentrations and then

substitute into a Kc expression.

2.00 - x = 0.89 M

[CO] = [H2O] = 0.89 M

[CO2] = [H2] = 1.11 M

(the negative result is ignored)

SOLUTION: Initial concentrations must be calculated as M, we have

from the data given [CO] = [H2O] = 0.250/0.125L = 2.00M.

CO(g) + H2O(g) CO2(g) + H2(g)concentration

initial

change

equilibrium

2.00 2.00 0 0

-x -x x x

2.00 -x 2.00 -x x x

_________________________________________________

Kc =[CO2][H2][CO][H2O]

=x2

(2.00− x)(2.00− x)=

x2

(2.00− x)2= 1.56

x

2.00− x=√

1.56 = ±1.25

x = 1.25(2.00− x) = 2.50− 1.25x

2.25x = 2.50x = 1.11M

Br2 (g) 2Br (g)

At 1280˚C the equilibrium constant (Kc) for the reaction

is 1.1 x 10-3. If the initial concentrations are [Br2] =

0.063 M and [Br] = 0.012 M, calculate the

concentrations of these species at equilibrium.

A wee-bit tougher ICE problem

Let x

be the

change

in

concent

ration

of Br2

[Br]2

[Br2]Kc =

Solve for x using quadratic equation

Br2 (g) 2Br (g)1. Write Balanced Equation

2. Write Equilibrium Expression

3. Set Up ICE TableBr2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

[0.063] [0.012]

-x +2x

0.063 - x 0.012 + 2x

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

[Br]2

[Br2]Kc = =

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x

4x2 + 0.0491x + 0.0000747 = 0

-b ± b2 – 4ac !2a

x =

x = -0.00178x = -0.0105

At equilibrium, [Br] = 0.012 + 2x = -0.009 M and 0.00844 M

At equilibrium, [Br2] = 0.062 – x = 0.06378 M

ax2 + bx + c =0

Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

[0.063] [0.012]

-x +2x

0.063 - x 0.012 + 2x

Two solutions

PROBLEM: Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It decomposes by the reaction,

COCl2(g) CO(g) + Cl2(g) Kc = 8.3 x 10-4 (at 360 oC)

Calculate [CO], [Cl2], and [COCl2] at equilibrium when the initial

amount of phosgene gas is 0.100 mol in a 10.0 L flask.

Given information in the problem

[COCl2] = 0.100 mol/10.0 L = 0.0100 M

Kc = 8.3 x 10-4 (at 360 oC)

COCl2(g) CO(g) + Cl2(g)

Kc = 8.3 x 10-4 =[CO] [Cl2]

[COCl2]

Initial (M)

Change (M)

Equilibrium (M)

0.0 0.0

-x +x

0.010 - x x

COCl2(g) CO(g) + Cl2(g)

0.010

+x

x

Kc = 8.3 x 10-4 =(x) (x)

(0.010 - x)

(continued) Kc = 8.3 x 10-4 =(x) (x)

(0.010 - x)

Solving for x yields x = 2.5 x 10-3 M and 0.0100 - x = 7.5 x 10-3 M.

Kc = 8.3 x 10-6 - 8.3 x 10-4 x = x2

Kc = 8.3 x 10-4 (0.010 - x) = x2

Kc = x2 + 8.3 x 10-4 x - 8.3 x 10-6 = 0

ax2 + bx + c =0

-b ± b2 – 4ac !2a

x =

a = 1

b = 8.3 x 10-4

c = - 8.3 x 10-6

[COCl2] [CO] + [Cl2]

(continued) Kc = 8.3 x 10-4 =(x) (x)

(0.010 - x)

Kc = 8.3 x 10-6 - 8.3 x 10-4 x = x2

Kc = 8.3 x 10-4 (0.010 - x) = x2

Kc = x2 + 8.3 x 10-4 x - 8.3 x 10-6 = 0

ax2 + bx + c =0

-b ± b2 – 4ac !2a

x =

a = 1

b = 8.3 x 10-4

c = - 8.3 x 10-6

4 a c = 4 x 1 x - 8.3 x 10-6

b2 = (8.3 x 10-4)2 = 6.89 X 10-7

c = - 8.3 x 10-6

c = - 8.3 x 10-6 Solving for x yields x = 2.5 x 10-3 M

and 0.0100 - x = 7.5 x 10-3 M.

Predicting reaction direction and calculating equilibrium concentrations

The research and development unit of a chemical company is

studying the reaction of CH4 and H2S, two components of natural gas.

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

In one experiment, 1.00 mol of CH4, 1.00 mol of CS2, 2.00 mol of H2S and 2.00 mol of H2 are mixed in a 250 mL vessel at 960 oC. At this temperature, Kc = 0.036.

(a) In which direction will the reaction proceed to reach equilibrium?

(b) If [CH4] = 5.56 M at equilibrium, what are the equilibrium

concentrations of the other three substances?

PLAN: Find the initial molar concentrations of all components and use these to

calculate Qc. Compare Qc to Kc, determine in which direction the reaction

will progress, and draw up expressions for the equilibrium concentrations.

SOLUTION:

Qc = [CS2][H2]4

[CH4][H2S]2=

[4.0][8.0]4

[4.0][8.0]2= 64.0

Qc of 64 is >> than Kc = 0.036

The reaction will progress to the left.

[CH4]initial = 1.00 mol/0.25 L = 4.00 M

[H2S]initial = 2.00 mol/0.25 L = 8.00 M

[H2]initial = 2.00 mol/0.25 L = 8.00 M

[CS2]initial = 1.00 mol/0.25 L = 4.00 M

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

SOLUTION:

At equilibrium [CH4] = 5.56 M, so 5.56 = 4.00 + x; thus,

x = 1.56 M

Therefore:[H2S] = 8.00 + 2x = 11.12 M

[CS2] = 4.00 - x = 2.44 M

[H2] = 8.00 - 4x = 1.76 M

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)concentrations

initial

change

equilibrium

4.00 8.00 4.00 8.00

+ x + 2x - 4x

4.00 + x 8.00 + 2x

- x

4.00 - x 8.00 - 4x

______________________________________________________