CM4106 Chemical Equilibria & Thermodynamics

Lesson 1Introduction to Chemical Equilibria

A Chemistry Education Blog by Mr Tanhttp://chemistry-mr-tan-yong-yao.blogspot.sg/

Equilibrium Constant, K- describes the extent to which reaction proceeds (position

of chemical equilibrium)

Kc = [C]c[D]d

[A]a[B]b

aA + bB ⇌ cC + dD

equilibrium concentrations

Kp =(pC)c(pD)d

(pA)a(pB)b

equilibrium pressures

(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

Kp = Kc (RT)n

n = (moles of gaseous product) - (moles of gaseous reactant)

(II) Manipulating K

Examples:

SO2(g) + Cl2(g) ⇌ SO2Cl2(g) n = 1 – 2 = -1

CaCO3(s) ⇌ CaO(s) + CO2(g): n = 1 – 0 = 1

Kc = = 0.212 at 100C[NO2]2

[N2O4]

Kc’ = = 4.72 at 100C[N2O4]

[NO2]2

1. The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

N2O4(g) ⇌ 2 NO2 (g)

2 NO2 (g) ⇌ N2O4(g)

Kc = 1Kc’

(II) Manipulating K

Kc = = 0.212 at 100C[NO2]2

[N2O4]

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.

N2O4(g) ⇌ 2 NO2 (g)

Kc’ = (Kc)n

2 N2O4(g) ⇌ 4 NO2 (g)Kc = = (0.212)2 at 100C[NO2]4

[N2O4]2

(II) Manipulating K

(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.

Kc’ = = at 100C[B]2

[A]22 A (g) ⇌ 2 B (g)

2 B(g) ⇌ 3 C(g) Kc’’ = = at 100C[C]3

[B]2

2 A(g) ⇌ 3 C(g) Kc =[C]3

[A]2

[B]2

[A]2

[C]3

[B]2

= at 100C

= x

= Kc’ x KC’’

(III) Equilibrium Calculations

Step 1: Find I-C-E 2SO3(g) ⇌ 2 SO2(g) + O2(g)

Initial (atm)

Change (atm)

Equilibrium (atm)

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Step 2: Substitute equilibrium values into Kp

Kp =(pSO2)2(pO2)1

(pSO3)2

Kp = 0.338

Kp =(0.300)2(0.150)1

(0.200)2

0.500 atm 0 0

-0.300 atm +0.300 atm +0.150 atm

0.150 atm0.300 atm0.200 atm

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.

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)

Equilibrium (atm)

- y + y + y

1.66 - y y y

(III) Equilibrium Calculations

Kp = [PPCl3] [PCl2

]

[PPCl5]

= 0.497

0.497 =(y)(y)

(1.66 – y ) Solve for y,

y = 0.693 or -1.19 (rejected)

P (Cl2) = 0.693 atm

P (PCl3) = 0.693 atm

P (PCl5) = 0.967 atm

(III) Equilibrium CalculationsAt 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.

 

 Initial [ ]

Change [ ]

Equilibrium []

PCl5(g) ⇌ PCl3(g) + Cl2(g)

4.35 x 10-3 M 0.149 M 0

1.00 x 10-3 M

+ 1.00 x 10-3 M+ 1.00 x 10-3 M- 1.00 x 10-3 M

0.150 M3.35 x 10-3 M

Kc = [PCl3] [Cl2]

[PCl5]

Kc = [1.00 x 10-3] [0.150]

[3.34 x 10-3] Kc = 0.0449 (to 3.s.f.)

Take note of volume of system especially for Kc calculation

(III) Equilibrium Calculations

Q < K Q = K Q > K

Q < K

The [product] is too small and [reactant] is too large.

Reaction will proceed from left to right forming more products.

System is at equilibriumthere is no net movement

Q > K

The [product] is too large and [reactant] is too small.

Reaction will proceed from right to left forming more reactants

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(III) Equilibrium CalculationsAt 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

Since Q > K,

Reaction will proceed from right to left to achieve equilibrium by forming more reactants

Q = (PSO2)2(PO2) (PSO3)2

Q = (0.41)2(2.5) (0.16)2

Q = 16.4 > Kp

(IV) Le Châtelier’s PrincipleFactor Rate of

reactionRate

constant, k

Position of equilibrium

Equilibrium constant, Kc (or Kp)

Increase in reactant concentration

↑ No change Shifts to reduce reactant conc.

No effect

Increase in pressure (by decreasing volume)

↑ No change Shifts in direction w ↓ no. of moles

No effect

Increase in temperature ↑ ↑ Shifts in

direction of endothermic rxn

endothermic reaction

↑exothermic reaction

↓Adding a catalyst ↑ ↑ No change

Catalysts increase rate of forward and reverse reaction equally.

No effect