4.3 Equilibria

4.3 Equilibria

RECAP: (for exothermic reaction)

LE CHATELIER – oppose the motion!	Where does equilibrium move and why?
Increase Temperature	Toward reactants, therefore less products; Move to the endothermic side. Higher kinetic energy so more chance of successful collision LOW temp = high yield = but slow process…
Increase Pressure	Toward side with less molecules of gas (only affects gases). Particles are pushed together, which increases chances of successful collision. HIGH pressure = high yield = expensive!
Introduce Catalyst	NO EFFECT ON EQUILIBRIUM POSITION (will affect rate)

 

At equilibrium the amount of reactants and products is the SAME.

Dynamic Equilibrium – a reaction that occurs in both ways at the same time (conditions; in a closed system at constant temperature)

Many industrial reactions are reversible; we use this sign for equilibria:

E.G. Both these experiments are good economically

1)            Contact process – making sulphuric acid

2SO₂(g) + O₂(g)  2SO₃(g)

USES = fertilisers, dyes, medicines, batteries

• Haber process – making ammonia

N₂(g) + 3H₂(g)  2NH₃(g)

USES = fertilisers, producing nitrogen-based compounds

EXPERIMENT: Hydrogen-Iodine Reaction                               (REVERSIBLE)

There is a relationship between the concentration of initial reactants/products and the equilibrium concentrations which are produced from them

E.G.                                                                        H₂(g) + I₂(g)  2HI(g)

Initial concentration:                      H₂ = 1.0moldm^-3                I₂ = 1.0moldm^-3

Equilibrium concentration:          H₂ = 0.228moldm^-3           I₂ = 0.228moldm^-3

From this we can see that the ratio has remained the same, i.e. 1:1

K_p / K_c

What is K_p / K_c?

K _p / K_c is the ratio of product concentration to reactant concentration, and is commonly known as the equilibrium

constant. For example, in the hydrogen-iodine reaction K_c will be;

*Note: products are ² because in a balanced equation, there is a 2 in front – see below

E.G. 4X  2Y + 3Z

*We can calculate K_p using partial pressures (see below)

 

As long as the equilibrium is HOMOGENOUS (all reactants/products in the same state) then we can use this general rule for finding K_c;

If the equilibrium is HETEROGENOUS (where reactants/products are in different states) then you must LEAVE OUT any concentrations that are solid.

For K_p, HOMOGENOUS equilibriums can be calculated using;

If the equilibrium is HETEROGENOUS then you only take into account the gases.

*Note: we dont use square brackets for equilibrium partial pressures

 

EXPERIMENT:                    Fe²⁺(aq) + Ag⁺(aq)  Fe³⁺(aq) + Ag(s)

• Add 500cm³ of 0.1moldm^-3 silver nitrate solution to 500cm³ of 0.1 moldm^-3 of iron (II) sulfate solution
• Leave mixture in stoppered flask at 298K, it will reach equilibrium
• Take samples and titrate

 

CALCULATION:

Calculating partial pressures;

Minty Fruits Taste Minty

EXAMPLE:

When 3.0 moles of PCl₅ is heated in a closed system, the equilibrium mixture has 1.75 moles of Cl. If total pressure of the mixture is 714kPa, what is the partial pressure of PCl₅?

Step 1) Find moles at equilibrium of all reactants and products;

We know 1.75 moles of Cl₂, therefore we must also have 1.75 moles of PCl₃ and so (3 – 1.75) will leave us with the moles at equilibrium for PCl₅ which is 1.25 moles. Adding these together we get 1.25+1.75+1.75 = 4.75 total moles at equilibrium.

Step 2) Find the mole fraction;

Mole fraction of a gas in mixture =  =  = 1.66

Step 3) Find partial pressure;

Partial pressure of gas = 714 x  = 187.9kPa

 

Equilibrium and Entropy are related

∆S_total = R lnK

When the total entropy, ∆S_total, increases, the equilibrium constant, K, will also increase.

If;            K = 10^-10 = reaction will not occur

K = 10^-5 = mostly reactants

K = 1 = balanced products and reactants

K = 10⁵ = mostly products

K = 10¹⁰ = reaction complete