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II.1 EQUILIBRIUM

REVERSIBLE REACTIO2S

Many reactions can go in reverse and have separate activation energies.

Reactants → Products or Reactants ← Products

Reactants form Products Products form Reactants

Example: i) N2O4(g) is heated in a CLOSED flask.

ii) 2NO2(g) molecules will build up and collide.

iii) This reaction may go forward and reverse (depends on the conditions). We can write both the forward and

reverse reactions on the same line using a double arrow. iv) As long as there is N2O4 present, the forward reaction will keep on happening and as long as there is NO2

present, the reverse reaction will keep on happening.

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DY2AMIC EQUILIBRIUM

Read p. 37 – 39 and Try #2

II.2 CHARACTERISTICS OF EQUILIBRIUM

How do we recognize a reaction is in equilibrium? 1.

2.

3.

4.

5.

What does equilibrium look like on a graph? 22O4 ↔ 2 2O2

! Equilibrium DOES NOT mean concentration of reactants and products are equal. ! Equilibrium DOES mean rates of forward and reverse reactions are the same.

Example 1: Given the reaction: 2O2(g) + CO(g) !!!! 2O(g) + CO2(g)

a. If one mole of NO2 and one mole of CO are mixed in a 1.0 litre container, the rate of the forward reaction

will initially be (fast/slow)

b. While the forward rate > the reverse rate, the [NO] and the [CO2] will be (increasing/decreasing)

c. After the initial mixing, the rate of the forward reaction will be (increasing/decreasing)

d. The rate of the reverse reaction will be (increasing/decreasing)

e. Once equilibrium is established, what can be said about the rates of the forward and reverse reaction?

f. NO2 is a dark brown colour. All of the other gases are colourless. Describe what will happen to the colour of

the gas mixture from when the NO2 and the CO are mixed until equilibrium is established.

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Hebden: Do Set 7 #3 – 6, 11, 12 (p. 38–43)

II. 3 PREDICTI2G THE SPO2TA2EITY OF A REACTIO2 A SPONTANEOUS reaction is a reaction that occurs by itself, without outside assistance.

What causes a reaction to: a) proceed forward b) not proceed (proceed in reverse) c) form an equilibrium

E2THALPY

Endothermic C + heat " D Exothermic A " B + heat

3 In exothermic reactions, the have minimum enthalpy, it will favour the , and tend

to

3 we should expect exothermic reactions to

3 In endothermic reactions , the reactants have minimum enthalpy, therefore, it will favour the ,

and tend to

3 we should expect endothermic reactions to (without outside assistance like increasing temperature)

BUT some endothermic reactions do occur spontaneously. For example, cold packs – mixing the chemicals

absorbs heat (making it very cold)

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E2TROPY

ENTROPY is the amount of randomness in a system.

Probability is high that events occurring in life will lead to

How do we decide which side of a reaction has the most entropy?

PHASE

Draw the KMT model of the phases. Which phase has the most entropy? Least entropy? Why?

gas solution liquid solid

<---------------------------------------------------------------------->

So you can look at the subscripts for the phases to tell which side (reactant/product) has maximum entropy.

Example 2: Label the side that has maximum entropy. Is the entropy increasing or decreasing?

C2H2Cl4(l) ! C2H2(g) + 2 Cl2(g)

But what if both reactant and product are the same phase?

2UMBER OF MOLECULES

The side having the

Example 3: Label the side that has maximum entropy.

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

Example 4:

a. Given the equation: Cl2(g) ! Cl2(aq) ∆H = –25 kJ

Enthalpy favours the (reactant/product) and entropy favours the .

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b. If the reaction CO(g) + 3 H2(g) ! CH4(g) + H2O(g) + 49.3 kJ was proceeding to the right, the enthalpy

would be (increasing/decreasing) and the entropy is

If we consider both enthalpy and entropy, how do we decide which side is favoured?

In nature, there is a tendency toward enthalpy and entropy.

If enthalpy favours reactants and entropy favours reactants then

If enthalpy favours products and entropy favours products then

If enthalpy favours reactants and entropy favours products then

If enthalpy favours products and entropy favours reactants then

Example 5: Determine the spontaneity of the reaction: CaCO3 (s) + energy ! CaO(s) + CO2 (g)

Hebden: Do Set 8 #14, 15 (a, b, d, f, h), 16 (a, b, c, e) (p.48–49)

II.4 LE CHÂTELIER’S PRI2CIPLE

LE CHATELIER’S PRINCIPLE:

APPLICATIO2 OF LE CHÂTELIER’S PRI2CIPLE

TEMPERATURE CHA2GES

N2O4(g) + heat ! 2NO2(g)

Temperature changes affect all equilibrium (s, l, aq, g)

LCP: If we ADD HEATto the equilibrium, the reaction will shift in way that tends to

N2O4(g) + heat ! 2NO2(g)

LCP: If we REMOVE HEAT from the equilibrium, the reaction will shift in way that tends to

N2O4(g) + heat ! 2NO2(g)

CO2CE2TRATIO2 A2D PARTIAL PRESSURE CHA2GES

In a mixture of gases, partial pressure is the pressure exerted by ONE gas. e.g. In a mixture containing NO and

CO2 gases: total pressure = partial pressure of NO + partial pressure of CO2.

If you increase PNO, it’s like increasing the concentration of NO.

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CO2(g) + NO(g) ! CO(g) + NO2 (g).

LCP: If we I2CREASE THE CO2CE2TRATIO2 of a certain substance in an equilibrium

mixture, the reaction will shift in way that tends to

If you increase PNO, it’s like increasing the concentration of NO.

CO2(g) + NO(g) ! CO(g) + NO2 (g).

LCP: If we DECREASE THE CO2CE2TRATIO2 of a certain substance in an equilibrium

mixture, the reaction will shift in way that tends to

If you decrease PNO

CO2(g) + NO(g) ! CO(g) + NO2 (g).

Example 6: What is the effect of adding FeCl3(aq) to this equilibrium:

Fe(NO3)3(aq) + KSCN(aq) ! FeSCN2+(aq) + KNO3(aq)

Be careful with spectator ions.

Complete ionic equation:

Net ionic equation:

Also, changing solids or liquids have no effect on the equilibrium.

Example 7: Changing which species will have an effect on the equilibrium:

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

VOLUME OF THE CO2TAI2ER A2D TOTAL PRESSURE CHA2GES

Consider: N2O4(g) + heat ! 2 NO2(g)

Compare the reactant side and product side, which side has more pressure?

vs.

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If you decreased the volume of the balloon, which side will it affect more? (Which side will feel more concentrated?)

LCP: If we I2CREASE THE TOTAL PRESSURE (decrease volume) on an equilibrium

system involving gases, the reaction will shift in way that tends to

LCP: If we DECREASE THE TOTAL PRESSURE (increase volume) on an equilibrium

system involving gases, the reaction will shift in way that tends to

Example 8: 2C2H6(g) + 7O2(g) ! 4CO2(g) + 6H2O(g)

a) Increasing the total pressure on this system, will cause a shift to the side with moles of gas,

which in this case is the side.

b) Decreasing the total pressure on this system, will cause a shift to the side with moles of gas,

which in this case is the side.

c) Increasing the total volume on this system (the same as the total pressure) will cause a

shift to the side with moles of gas, which in this case is the side.

CATALYSTS

Consider:

LCP: ADDI2G A CATALYST speed up forward and reverse reactions; therefore, the

GRAPH OF EQUILIBRIUM CHA2GES

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TEMPERATURE N2O4(g) + heat ! 2NO2(g)

Suppose we increase the temperature at 2 min., according to LCP the equilibrium will shift to the .

The [N2O4] will and the [NO2] will

originally at

equilibrium

conc. are constant

CONCENTRATION N2O4(g) + heat ! 2NO2(g)

Suppose we inject more NO2 into the container at 2 min., so the [NO2] will suddenly increase. According to

LCP the equilibrium will shift to the .

originally at

equilibrium

conc. are constant

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PRESSURE N2O4(g) + heat ! 2NO2(g)

Suppose we decrease the volume of the container suddenly at 2 min., so the [N2O4] and [NO2] will suddenly

. According to LCP in order to decrease the pressure, the equilibrium will shift to the side

with moles of gas, that is to the .

originally at

equilibrium

conc. are constant

CATALYST N2O4(g) + heat ! 2NO2(g)

Adding a catalyst speeds up BOTH the forward and reverse reactions, so the concentrations of the species

originally at

equilibrium

conc. are constant

add catalyst

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Hebden: Do Set 9 #17, 19, 21, 23, 24, 26, 27 (p.54–55)

II.5 I2DUSTRIAL APPLICATIO2S OF EQUILIBRIUM PRI2CIPLES Hebden: Read “The Haber Process” on p. 56 and Do #29

N2(g) + 3H2(g) ! 2NH3(g) + 92 kJ

29. a) For highest yield of NH3(g), use pressure.

b) For highest yield of NH3(g), use temperature.

c) For fastest reaction rate, use temperature.

d) Problem:

Solution:

e) Speed up the reaction by

2OTE: “Shift” in equilibrium means when a new equilibrium is

reached, the concentration of reactants or products is more or less. It

does not say anything about the rate of the reaction.

f) Increasing the temperature will the rate of reaction. The equilibrium will shift

. This means that when the new equilibrium is reached, the amount of reactants will be

(more/less than) the amount of products.

Hebden: Do #30 (p.56)

II.6 EQUILIBRIUM EXPRESSIO2

We can quantify (attach a numerical value to) the equilibrium for most reactions by creating a RATIO of

PRODUCTS TO REACTANTS

For a reaction: aA + bB ! cC + dD

The EQUILIBRIUM EXPRESSION, Keq =

Example 9: Write the equilibrium expression for the reaction for :

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

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

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EQUILIBRIUM CO2STA2T

The RATIO obtained from the equilibrium expression is a unique to a particular reaction.

This ratio is called the EQUILIBRIUM CONSTANT or Keq

Example 10: The equilibrium constant expression for the reactions:

BaSO4(s) ! Ba2+(aq) + SO4

2-(aq) CH3COCH3(l) + Cl2(g) ! CH3COCH2Cl(l) + HCl(g)

Do Hebden: Set 10 # 31 (a, c, e, f, h, i), 32, 34, 35 (a, c, e, h) (p.60)

II.7 THE MEA2I2G OF KEQ

Large Keq means at equilibrium, there is lots of and very little

Small Keq means at equilibrium, there is lots of and very little

FACTORS AFFECTI2G KEQ

Temperature

Consider the original reaction: reactants ! products + heat

If we increase the temperature, reactants ! products + heat

The equilibrium will shift reactants ! products + heat

The concentration of solids and liquids are fairly constant. The do not expand or

contract much. Since they are constants, their values are already incorporated with Keq.

OMIT solids and pure liquids from the Keq expression.

Keq = [Products]

[Reactants]

Keq = [Products]

[Reactants]

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Write the Keq expression Keq =

So Keq (remains the same/becomes larger/becomes smaller)

Concentration

Consider the original reaction: reactants ! products

If we increase reactants, reactants ! products

The equilibrium will shift

Write the Keq expression Keq =

So Keq (remains the same/becomes larger/becomers/smaller)

Pressure

Consider the original reaction: N2(g) + 3H2(g) ! 2NH3(g)

If we decrease the volume (or increase the pressure), N2(g) + 3H2(g) ! 2NH3(g)

The equilibrium will shift

Write the Keq expression Keq =

So Keq (remains the same/becomes larger/becomes smaller)

Changes in concentration, pressure, surface area, and adding a catalyst

ONLY CHANGING affects Keq.

Example 11: 2 NO(g) + Cl2(g) ! 2 NOCl(g) + 76 kJ

What happens to the Keq in the above reaction if the temperature is decreased? Why?

Hebden: Do Set 11 #36, 38, 39, 44, 45 (p.62)

II. 8 EQUILIBRIUM CALCULATIO2S

Chemical systems can either be

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AT EQUILIBRIUM or NOT AT EQUILIBRIUM

type 1

type 2

type 3

type 4

Method: 1) find eqm

concentrations 2) write Keq expression 3) plug in eqm

concentrations into Keq

Method: 1) set up an ICE box 2) fill in what you are

given. I = initial conc., C = change according to mole ratio, E = equilibrium conc.

3) write Keq expression 4) plug in eqm conc. into

Keq

Method: 1) set up an ICE box 2) fill in what you are

given. I, C, E. Let “x” = unknown.

3) write Keq expression 4) plug in eqm conc.

into Keq

Method: 1) plug in intial conc.

into Trial Keq expression

2) compare trial Keq with Keq:

Trial Keq < Keq = shift right

Trial Keq > Keq = shift left

Trial Keq = Keq = no shift

Type 1: System is AT EQUILBRIUM

Example 12: What is the Keq for the reaction, 2 HI(g) ! H2(g) + I2(g) if AT EQUILIBRIUM, there is 2.0

mol of HI, 3.0 mol of I2, and 3.5 mol of H2 in a 5.0 L closed flask? (Answer Keq = 2.6)

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2 NO(g) + O2(g) ! 2 NO2(g)

I

C

E

PCl5(g) ! PCl3(g) + Cl2(g)

I

C

E

Type 2: System is CHANGING TO EQUILIBRIUM Calculate Keq

Example 13: 2 NO(g) + O2(g) ! 2 NO2(g)

4.0 mol of NO2 is INTIALLY ADDED to a 2.0 L flask. After a while, equilibrium is reached. At equilibrium,

0.50 mol of NO is found. What is Keq?

Example 14: PCl5(g) ! PCl3(g) + Cl2(g)

8.0 moles of PCl3 and 6.0 moles of Cl2 are placed in a 2.0 L flask. After equilibrium, 0.50 moles of Cl2 are

found. What is Keq? (answer: Keq = 0.11)

Hebden: Do Set 12 #47 – 49 (p.70)

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2 A(g) + B(g) ! 2 C(g)

I

C

E

Type 3: System is CHANGING TO EQUILIBRIUM Calculate [species]

Example 15: 2 A(g) + B(g) !!!! 2 C(g)

An unknown amount of C was INITIALLY PLACED into a 3.0 L flask. When equilibrium was reached, the

concentration of A was 0.60 M. If Keq has a value of 34.0, how many MOLES of C were placed into the flask

originally?

Example 16: A2(g) + B2(g) ! 2AB(g) If 6.0 mol of A2 and 6.0 mol of B2 are placed in a 1.0 L bulb and allowed to come to equilibrium, what will be

the concentration of all the species at equilibrium? Keq = 2.5. (answer: [A2] = [B2] = 3.4 M, [AB] = 5.3 M)

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Trial Keq > Keq reaction will shift

Trial Keq < Keq reaction will shift

Trial Keq = Keq reaction will

Trial Keq > Keq reaction will shift

Trial Keq < Keq reaction will shift

Trial Keq = Keq reaction will

Type 4: PREDICTING REACITON SHIFT We can also use the Keq to predict which direction a reaction will go (left or right) for any initial concentrations of reactants and products.

We use REACTION QUOTIENT (Q) or TRIAL Keq. It has the same form as Keq EXCEPT:

Keq = [product] Trial Keq = [product]

[reactant] [reactant]

Example 19: 2SO2(g) + O2(g) !!!! 2SO3(g)

The Keq is 0.14 for the above reaction. If a 1.0 L flask is filled with 0.10 mol of SO3 and with 0.20 mol of SO2

and O2, is the reaction at equilibrium? If not, in which direction does it proceed?

Hebden: Do Set 13 #52, 53, 55 – 59, (60 – 63) [Type 3] and 50, 54 [Type 4] (p.71–72)