Review Slides of CM4106 Lesson 1 - Introduction to Chemical Equilibria

1. CM4106 Chemical Equilibria & Thermodynamics
Lesson 1
Introduction to Chemical Equilibria
A Chemistry Education Blog by Mr Tan
http://chemistry-mr-tan-yong-yao.blogspot.sg/

2. Equilibrium Constant, K
- describes the extent to which reaction proceeds (position
of chemical equilibrium)
aA + bB ⇌ cC + dD
[C]c[D]d (pC)c(pD)d
Kc = Kp =
[A]a[B]b (pA)a(pB)b
equilibrium equilibrium
concentrations pressures

3. (I) Writing K expressions
1. Write K as for reaction proceeding from left to right
2. Do not include solids and liquids in K expressions
3. Be careful not to leave out reacting ratios

4. (II) Manipulating K
Kp = Kc (RT) n
n = (moles of gaseous product) - (moles of gaseous reactant)
Examples:
SO2(g) + Cl2(g) ⇌ SO2Cl2(g) n = 1 – 2 = -1
CaCO3(s) ⇌ CaO(s) + CO2(g): n=1–0=1

5. (II) Manipulating K
1. The equilibrium constant of a reaction in the reverse
reaction is the reciprocal of the equilibrium constant of
the forward reaction.
[NO2]2
N2O4(g) ⇌ 2 NO2 (g) Kc = = 0.212 at 100 C
[N2O4]
[N2O4]
2 NO2 (g) ⇌ N2O4(g) Kc’ = 2 = 4.72 at 100 C
[NO2]
1
Kc =
Kc’

6. (II) Manipulating K
2. The equilibrium constant of a reaction that has been
multiplied by a number is the equilibrium constant raised
to a power that is equal to that number.
[NO2]2
N2O4(g) ⇌ 2 NO2 (g) Kc = = 0.212 at 100 C
[N2O4]
[NO2]4
2 N2O4(g) ⇌ 4 NO2 (g) Kc = [N O ]2 = (0.212)2 at 100 C
2 4
Kc’ = (Kc)n

7. (II) Manipulating K
3. The equilibrium constant for a net reaction made up of two
or more steps is the product of the equilibrium constants for
the individual steps.
[B]2
2 A (g) ⇌ 2 B (g) Kc’ = [A]2
= at 100 C
2 B(g) ⇌ 3 C(g) [C]3
Kc’’ = = at 100 C
[B]2
[C]3 = Kc’ KC’’
2 A(g) ⇌ 3 C(g) Kc = x
[A]2
[B]2 [C]3
= x
[A]2 [B]2
= at 100 C

8. (III) Equilibrium Calculations
Sulfur trioxide decomposes at high temperature in a sealed container:
2SO3(g) ⇌ 2SO2(g) + O2(g)
Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm.
At equilibrium, the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.
Step 1: Find I-C-E
2SO3(g) ⇌ 2 SO2(g) + O2(g)
Initial (atm) 0.500 atm 0 0
Change (atm) -0.300 atm +0.300 atm +0.150 atm
Equilibrium (atm) 0.200 atm 0.300 atm 0.150 atm
Step 2: Substitute equilibrium values into Kp
(pSO2)2(pO2)1
Kp =
(pSO3)2
(0.300)2(0.150)1
Kp = (0.200)2
Kp = 0.338 Page 6

9. (III) Equilibrium Calculations
For the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g), the equilibrium constant Kp has the
numerical value 0.497 at 500 K.
A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm.
What are the equilibrium pressures of PCl5, PCl3 and Cl2 at this temperature?
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Initial (atm) 1.66 0 0
Change (atm) -y +y +y
Equilibrium (atm) 1.66 - y y y
[PPCl3] [PCl2]
Kp = = 0.497
[PPCl5]
(y)(y) P (Cl2) = 0.693 atm
0.497 = (1.66 – y ) Solve for y, P (PCl3) = 0.693 atm
P (PCl5) = 0.967 atm
y = 0.693 or
-1.19 (rejected)

10. (III) Equilibrium Calculations
At a certain temperature a 2.00 L flask initially contained 0.298 mol PCl3(g)
and 8.70 x 10-3 mol PCl5(g). After the system had reached equilibrium,
2.00 x 10-3 mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes
according to the reaction:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of all species and the value of Kc.
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Initial [ ] 4.35 x 10-3 M 0.149 M 0
Change [ ] - 1.00 x 10-3 M + 1.00 x 10-3 M + 1.00 x 10-3 M
Equilibrium [] 3.35 x 10-3 M 0.150 M 1.00 x 10-3 M
[PCl3] [Cl2] Take note of volume of system especially
Kc = for Kc calculation
[PCl5]
[1.00 x 10-3] [0.150]
Kc = Kc = 0.0449 (to 3.s.f.)
[3.34 x 10-3]

11. (III) Equilibrium Calculations
Q<K Q=K Q>K
Q<K Q>K
The [product] is The [product]
too small and is too large
[reactant] is System is at and [reactant]
too large. equilibrium is too small.
there is no net
movement
Reaction will Reaction will
proceed from proceed from
left to right right to left
forming more forming more
products. reactants
Page 9

12. (III) Equilibrium Calculations
At 1000 K the value of Kp for the reaction 2SO3(g) ⇌ 2SO2(g) + O2(g)
is 0.338.
Predict the direction in which the reaction will proceed toward
equilibrium if the initial partial pressures are PSO3 = 0.16 atm,
PSO2 = 0.41 atm and PO2 = 2.5 atm.
Compare Q and Kp to predict direction reaction will proceed
towards eqm
(PSO2)2(PO2)
Q=
(PSO3)2
(0.41)2(2.5) Since Q > K,
Q=
(0.16)2 Reaction will proceed from right to left
to achieve equilibrium by forming
Q = 16.4 > Kp
more reactants

13. (IV) Le Châtelier’s Principle
Factor Rate of Rate Position of Equilibrium constant,
reaction constant, k equilibrium Kc (or Kp)
Increase in reactant No change Shifts to reduce No effect
concentration
↑ reactant conc.
Increase in pressure No change Shifts in direction No effect
(by decreasing
↑ w ↓ no. of moles
volume)
Increase in Shifts in direction endothermic reaction
↑ ↑
temperature of endothermic
rxn ↑
exothermic reaction
↓
Adding a catalyst No change No effect
↑ ↑ Catalysts
increase rate of
forward and
reverse reaction
equally.