AnnouncementsHour Exam IIThursday Feb 5, 2009CTC 102 and 105Time: 6:00 - 8:00 PM (1.5 hrs long)Bring Blue Book, calculator and writing instrument.

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. We use ICE table to calculate either Kc from initial quantities or we calculate equilibrium concentrations given Kc.

---it’s the algebra that’s tricky

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

Let’s try it in an example problem!

PLAN:

Calculating Kc from concentration dataIn 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 453˚C. 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 therefore:

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

Kc = [H2] [I2][HI]2

=[0.011][0.011]

[0.078]2= 0.020

2HI(g) H2(g) + I2(g) Kc = [H2] [I2][HI]2

1. Write down what we solve: Kc and what we know.

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

initialchange

equilibrium

.200 mol/2.00 L-2x xx

0 0

0.100 - 2x x x

+

2. Set up the ICE table

PLAN:

Calculating equilibrium concentration using a simplfying assumption (no quadratic equation)

The reaction of nitrogen with oxygen giving nitrogen oxide contributes to air pollution whenever a fuel is burnt at high temperature. At 1500K the K = 1.0 X 10-5. Suppose a sample of air has [N2] = 0.80 M, and [O2] = 0.20M before a reaction occurs. Calculate the equilibrium concentrations at equilibrium.

Set up ICE table and begin filling it in.

N2(g) + O2(g) <==> 2NO(g) Kc = 1.0 X 10-5

= 0.020

1. Write down what we solve: Kc and what we know.

2. Set up the ICE table

Kc = [N2] [O2][NO]2 = 1.0 X 10-5

Molarity

initialchange

equilibrium

0.80-x +2x-x

0.20 0

0.80 - x 2x+

N2(g) + O2(g) <=> 2NO(g)

0.20 - x

=[.80 -x]

[2x]2Kc = [N2] [O2][NO]2 = 1.0 X 10-5

[.20 -x]

3. Use equilibrium expression and substitute values.

4. There are 3 cases: (1) perfect square, (2) simplifying assumption (both easy to solve) (3) quadratic equation (harder)

=[.80 -x]

[2x]2Kc = [N2] [O2][NO]2 = 1.0 X 10-5

[.20 -x]

3. Use equilibrium expression and substitute values.

4. We can use a simplifying assumption here: RULE: If 100 x K < [A]0 then .80 - x = .80100 X 10-5 = 10-3 < [A]0 .......it’s ok to rid x.

=Kc = [N2] [O2][NO]2 = 1.0 X 10-5

[.80][2x]2

[.20] [.16][2x]2=

x = (1.0 X 10-5 (.16) 1/2 = 6.3 X 10-4 4

)([N2] = .80 - x = 0.80 - 6.3 X 10-4 = 0.80M[O2] = -.20 - x = 0.20 - 6.3 X 10-4 = 0.20M[NO] = 2x = 6.3 X 10-4 X 2 = 1.3 X 10-3M

Determining equilibrium concentrations from Kc

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) Kc = 0.26

PLAN: Use the balanced equation to write the Kc expression, and then substitute values for each component.

[H2O]eq = [CO]eq[H2]eq3

[CH4]eq Kc= 0.53 M= (0.81)(0.28]3

(0.13)(0.26]

= 0.13 M

0.26mol0.32 L

= 0.81 M

0.091 mol0.32 L

= 0.28 M

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

initialchange

equilibrium

?

? ??

? ?

0.041 mol/0.32 L ?

?

?

Kc =[CO][H2]3

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

Determining equilibrium concentrations from initial concentrations and KcFuel 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.

Kc = 1.56

[CO] = [H2O] = 2.00 - x = 0.89 Mx = 1.11M = [CO2] = [H2]

It’s a perfect square and easy to solve!We toss out the negative result!

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)concentrationinitial

changeequilibrium

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 = 2.50

1. Concentration changes2. Pressure or Volume changes (gas)3. Temperature Changes4. Addition of a catalyst (not)

There are 4 Disturbances to consider

“If an external stress (disturbance) is applied to a system at equilibrium, the system will adjust to counter the stress according to the law of mass action.”

Qc = [C]c[D]d[A]a[B]b

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

Add more NH3

Equilibrium shifts left to offset stress

We can use if we wish Qc to help us decide what happens.

1. Changes in Concentration

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

RemoveNH3

Equilibrium shifts left to offset stress

Predicting the Effect of a Change in Concentration on the Equilibrium Position

To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O2;

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

What happens to(a) [H2O] if O2 is added? (b) [H2S] if O2 is added?

(c) [O2] if H2S is removed? (d) [H2S] if sulfur is added?

PLAN: We can write and compare Q with K when the system is disturbed to see in which direction the reaction will progress.

Q =[H2O]2

[H2S]2[O2]

(b) When O2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H2S] decreases.

(c) When H2S is removed, Q increases and the reaction progresses to the left to come back to K. So [O2] decreases.

(d) Sulfur is not part of the Q (K) expression because it is a solid. Therefore, as long as some sulfur is present the reaction is unaffected. [H2S] is unchanged.

(a) When O2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H2O] increases.

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O2;

Predicting the Effect of a Change in Concentration on the Equilibrium Position

2. Pressure and Volume Effects In Gases

1N2O4 (g) 2NO2 (g)

There are two cases for changes in pressure and volume:

1. ∆n = 0 (mol of product gases - mol of reactant gases = 0)

2. ∆n ≠ 0

Cl2(g) + I2(g) 2ICl(g)any change in P or V has no affect on the equilibrium!

If pressure is increased (volume decreased) the system will move to the side with fewer number of moles.

How would you change the volume of each of the following reactions to increase the yield of products?

PLAN: When gases are present, a change in volume will affect the pressure and that change will alter the concentration of the gas. If the volume decreases (i.e.--pressure increases), the reaction will shift to fewer moles of gas and vice versa.

(a) CaCO3(s) CaO(s) + CO2(g)

(b) S(s) + 3F2(g) SF6(g)(c) Cl2(g) + I2(g) 2ICl(g)

SOLUTION:

(a) CaCO3(s) CaO(s) + CO2(g)

(b) S(s) + 3F2(g) SF6(g)(c) Cl2(g) + I2(g) 2ICl(g)

(a) CO2 is the only gas present. To increase its yield, increase the volume (decrease the pressure).

(b) There are more moles of gaseous reactants than products, so decrease the volume (increase the pressure) to shift the reaction to the right.

(c) There are an equal number of moles of gases on both sides of the reaction. Therefore, a change in volume will have no effect.

How would you change the volume or pressure of each of the following reactions to increase the yield of products?

Summary Changes in TemperatureChange Exothermic Rx

Increase temperature K decreasesDecrease temperature K increases

Endothermic Rx

K increasesK decreases

3. Changes in TemperatureThere are two cases. Think of heat as reactant or as product!

2. When ∆H > 0 = Endothermic => think heat is reactant

1. When ∆H < 0 => Exothermic => think heat is a product

heat + A (g) + B (g) C (g) + D (g)

A (g) + B (g) C (g) + D (g) + heat

Kc = [C]c[D]d

[A]a[B]b

What happens to K can be analyzed using mass-action!

ΔH = + 180.5 kJ

Is the amount of NO(g) formed from the given amounts of N2 and O2 greater at higher of lower temperatures?

N2 (g) + O2 (g) 2NO (g)

TRANSLATION: ΔH = 180.5 kJ =>ENDO = HEAT IS REACTANT!

Heat + N2 (g) + O2 (g) 2NO (g)

Increasing the temperature by adding more heat to this reaction will increase the amount of NO formed and decrease the reactant side. Kc will increase in value because NO2 increases!

PRODUCT-FAVORED SHIFT

Kc = [NO]2

[N2][O2]

How does an increase in temperature affect the concentration of the underlined substance and Kc for the following reactions?

(a) CaO(s) + H2O(l) Ca(OH)2(aq) ΔHo = -82 kJ

(b) CaCO3(s) CaO(s) + CO2(g) ΔHo = 178 kJ(c) SO2(g) S(s) + O2(g) ΔHo = 297 kJ

PLAN:

Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on Kc.

SOLUTION:

How does an increase in temperature affect the concentration of the underlined substance and Kc for the following reactions?

(a) CaO(s) + H2O(l) Ca(OH)2(aq) ΔHo = -82 kJ

(b) CaCO3(s) CaO(s) + CO2(g) ΔHo = 178 kJ

(c) SO2(g) S(s) + O2(g) ΔHo = 297 kJ

PLAN: Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on Kc.

(a) CaO(s) + H2O(l) Ca(OH)2(aq) + heatAn increase in temperature will shift the reaction to the left, decrease [Ca(OH)2], and decrease Kc.

(b) CaCO3(s) + heat CaO(s) + CO2(g)

The reaction will shift right, resulting in an increase in [CO2] and increase in Kc.

(c) SO2(g) + heat S(s) + O2(g)The reaction will shift right, resulting in an decrease in [SO2] and increase in Kc.

A catalyst lowers Ea for both forward and

reverse reactions and only achieves

equilibrium at a faster rate.

Adding a Catalyst to a system at equilibrium• does not shift the position of an equilibrium system• does not alter the value of Kc or Kp

UncatalyzedPathway

CatalyzedPathway