Chapter 07 Chemical Equilibria

Chapter 8

Chemical Equilibria

Just as reactions vary in their speeds, they also vary in their extents. The topic of chemical equilibria will deal with how far a reaction can proceed. Students should familiarise themselves with both the qualitative and quantitative aspects of this chapter in order to secure marks in the examinations.

Definitions that students should memorise

A Reversible reaction

B A closed system

C Dynamic equilibrium

D Le Chatelier’s principle

The Equilibrium Law

Difference between Kc vs Kp

Students tend to get confused and mess up these two equilibrium constants. We will break down the various differences and make it more straightforward and easily understandable.

Similarities:	Differences:
1. Both are equilibrium constants, and a constant at constant temperature	1. The reasons why Kc and Kp expressions do not include solids are different. – For Kc, concentration of a solid is always constant at constant temperature. – For Kp, solids only add a negligible amount of pressure to the system.
2. Both Kc and Kp expressions do not include solids.	2. For all Kp expressions, liquids are not included. However, liquids are included for some Kc expressions. – For Kp, liquids only add a negligible amount of pressure to the system. – As for Kc expressions, concentration of water is not included if equilibrium is established in aqueous medium. In other words, H2O(l) is excluded if the system contains aqueous reactant, and H2O(l) is included if the system contains all (l) reactants

Test yourself

Construct the Kc or Kp expression for the following reactions and state its units

Hints:

• Do we include every single substance in the K_c or K_p expression?
• When do we include H₂O in the K_c or K_p expression?

• H₂O(g) + C(s) ⇌ H₂ (g) + CO(g)
• CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃CO₂C₂H₅ (l) + H₂O(l)
• 2CrO₄ – (aq) + 2H + (aq) ⇌ Cr₂O₇²– (aq) + H₂O(l)

Answers:

• Since C is a solid, we leave it out of the K_c or K_p expression.
• Since this reaction is in a non-aqueous reaction where H₂O is present in stoichiometric amounts, we include H₂O in the Kc or Kp expression.
• Since this reaction is in an aqueous medium, we do not include H₂O.

Questions involving changes in concentrations or partial pressures of reactants can quickly become confusing. But organising the information given in an initial-change-equilibrium (ICE) table helps make answering these questions more straightforward. Let’s take a look at how an ICE table can be utilised in the example below.

Worked example #1

At a very high temperature and a total pressure of 1atm, 20% of the steam dissociates into H₂ and O₂. Calculate the partial pressure that each of the 3 gases exert at equilibrium. Hence calculate the equilibrium constant, Kp, for the above reaction.

Amount/ mol	2H2O(g)	2H2(g)	O2(g)
Initial	a	0	0
Change	-0.20a	+0.20a	+0.10a
Equilibrium	0.80a	0.20a	0.10a

N_total = 0.80_a + 0.20_a + 0.10_a = 1.10_a mol
(Recall: The gaseous state) Partial pressure of gas x =

Answer = 5.68 × 10^-3 atm

When calculating Kc or Kp of the system, only use concentrations or partial pressure respectively, never use amounts. Although in cases when the mole ratio of the product to all of the reactants is 1:1 and the volume of the container hence cancels out in calculation, omitting the volume might result in mark deduction.

Answering questions using Le Chatelier’s Principle

• By Le Chatelier’s principle, (state the change of condition)
• Predict which reaction is favoured, and why
• Identify how the position of equilibrium (POE) will shift
• Describe the observation (if question asks)

Worked example #2

Ammonia is an important raw material in making fertilisers and explosives. Nitrogen is obtained from the fractional distillation of liquid air and hydrogen is abstained from the steam reforming of hydrocarbons. With both nitrogen and hydrogen, ammonia can be produced using the Haber process.

Temperature

• By Le Chatelier’s Principle, a low temperature will
• Favour the exothermic forward reaction to compensate for the lowered heat and
• POE will shift to the right
• Producing more ammonia However, at low temperatures, the rate of reaction is slow, and the system will take a long time to reach equilibrium. Hence an optimal temperature of 450℃ maximised both yield and rate.

Pressure

• By Le Chatelier’s Principle, a high pressure will
• Favour the forward reaction to decrease the total amount of gases and
• POE will shift to the right
• Producing more ammonia However, extremely high pressure leads to higher cost of production and requires equipment maintenance. Hence an optimal pressure of 250 atm is used.

Catalyst

• A catalyst of finely divided iron increases the rate of reaction so that quilibrium yield is reached faster.

As we know, if ΔG can determine the spontaneity of a reaction, how does ΔG⁸ relate to the position of equilibrium of a reaction?

We know that,

●  ΔG<0 indicates that the forward reaction is spontaneous and the forward reaction occurs at a faster rate,
●  ΔG=0 indicates that the system is in dynamic equilibrium and both the forward and backward rate of reactions are occurring at the same rate,
●  ΔG>0 indicates that the backward reaction is spontaneous and the backward reaction occurs at a faster rate.

ΔGɵ >0	ΔGɵ =0	ΔGɵ < 0
Minimum G lies close to reactants [products]<[reactants] POE lies to left Backward reaction favoured Kc small (<1)	Minimum G lies about equal reactants and products [products]=[reactants] Kc = 1	Minimum G lies close to reactants [products]<[reactants] POE lies to left Backward reaction favoured Kc small (<1)