Chemistry 12: Dynamic Equilibrium Date: 1 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) N 2 O 4(g) is heated in a CLOSED flask. ii) 2NO 2(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 N 2 O 4 present, the forward reaction will keep on happening and as long as there is NO 2 present, the reverse reaction will keep on happening.
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II.1 EQUILIBRIUM REVERSIBLE REACTIO2S · Chemistry 12: Dynamic Equilibrium Date: 1 II.1 EQUILIBRIUM REVERSIBLE REACTIO2S Many reactions can go in reverse and have separate activation
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Chemistry 12: Dynamic Equilibrium Date:
1
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.
Chemistry 12: Dynamic Equilibrium Date:
2
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.
Chemistry 12: Dynamic Equilibrium Date:
3
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)
Chemistry 12: Dynamic Equilibrium Date:
4
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?
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]
Chemistry 12: Dynamic Equilibrium Date:
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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
Chemistry 12: Dynamic Equilibrium Date:
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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)
Chemistry 12: Dynamic Equilibrium Date:
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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)
Chemistry 12: Dynamic Equilibrium Date:
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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)
Chemistry 12: Dynamic Equilibrium Date:
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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)