Equilibrium

HL only

17.2 The Equilibrium Law

Easy if you are given equilibrium concentrations! You simply substitute the values in to an expression for Kc.

However, we are normally given starting amounts and the equilibrium amount or concentration of one substance at equilibrium.

Handy hint: if you are given equilibrium amounts (number of moles) then you must convert to concentration by dividing by the volume in dm3. This is not necessary if there are equal numbers of reactant moles and product moles (i.e. if Kc has no units).

Worked example 1

200 g of ethyl ethanoate and 7.0 g of water were refluxed together. At equilibrium the mixture contained 0.25 mol of ethanoic acid. Calculate the value for Kc for the hydrolysis of ethyl ethanoate.

CH3COOC2H5 + H2O CH3COOH + C2H5OH

We use the Postman Pat method.

CH3COOC2H5 + H2O CH3COOH + C2H5OH

Initial concs a /V b /V 0 0

Eqm concs a-x /V b-x /V x /V x /V

In this example we can ignore V. Why?

Calculating initial number of molesa = moles ethyl ethanoate = 200 / 88 = 2.27b = moles of water = 7 /18 = 0.39

Use info in question and Mr values.

2.27 0.39

Calculating equilibrium number of molesx = 0.25 (given in Q)

0.25 0.25

So a-x = 2.27-0.25 = 2.02& b-x = 0.39-0.25 = 0.14

2.02 0.14

CH3COOC2H5 + H2O CH3COOH + C2H5OH

Now write an expression for Kc.

Kc = [CH3COOH][C2H5OH]

[CH3COOC2H5][H2O]

Substitute in values

Kc= 0.25 x 0.25 / 2.02 x 0.14

= 0.22 (no units)

Worked example 2

2 moles of phosphorus(V) chloride vapour are heated to 500 K in a vessel of volume 20 dm3. The equilibrium mixture contains 1.2 moles of chlorine. Calculate Kc for the decomposition of phosphorus(V) chloride into chlorine and phosphorus(III) chloride.

PCl5 PCl3 + Cl2

We use the Postman Pat method.

Initial concs a /V 0 0

Eqm concs a-x /V x /V x /V

Calculating initial concentrationa /V = moles PCl5 / vol in dm3 = 2 /20 = 0.1 moldm-3

0.1

Calculating equilibrium concentrations x/V = 1.2/20 = 0.06 moldm-3 (from moles of Cl2 in Q)

0.06

So a-x/V = 0.1-0.06 = 0.04

0.04 0.06

PCl5 PCl3 + Cl2

Now write an expression for Kc.

Kc = [PCl3][Cl2]

[PCl5]

Substitute in values

Kc= 0.06 x 0.06 / 0.04

= 0.09 moldm-3

PCl5 PCl3 + Cl2

17.1 Liquid-Vapour Equilibrium

Consider an evacuated container containing a small amount of a volatile liquid at the bottom.

What is going on in that flask?

Molecules escape from the liquid and become vapour. They collide with the wall of the container and, therefore, exert a pressure (known as the VAPOUR PRESSURE). Some molecules will condense back into a liquid. After a while:

rate of vapourisation = rate of condensation

At this point the system is in a state of dynamic equilibrium.

Molecules on the liquid surface need a certain minimum kinetic energy before they can escape. Think Maxwell – Boltzmann distribution!

The amount of energy depends upon the intermolecular forces between the molecules.

Vapourisation is endothermic as it requires overcoming the intermolecular forces. The amount of energy required for this change of state is called the enthalpy of vapourisation.

H2O(l) H2O(g)

H = +40.7 kJmol-1 at 373 K & 101.3 kPa

The energy is mainly required to overcome the intermolecular forces.

When a substance boils its temperature does not increase (so no increase in KE), the energy is absorbed to overcome the attractive forces between the particles.

The stronger the forces, the higher the boiling point and the higher the enthalpy of vapourisation.

Compound Hvap / kJmol-1 b.p. / K Intermolecular forces

methane 9.0 109 Van der Waals only

methoxymethane 27.2 248 Van der Waals and dipole – dipole

ethanol 38.6 352 Van der Waals, dipole – dipole and hydrogen bonding

But what about the vapour pressure at a certain temperature?

For substances with higher boiling points the vapour pressure will be lower as fewer molecules will escape from the liquid phase.

For a given substance, there is an exponential increase in vapour pressure with temperature.

0

5

10

15

-45 -25 -5 15 35

Temperature

Vap

ou

r p

ress

ure

e.g for propane

A liquid boils when its vapour pressure is equal to the pressure on the surface of the liquid as this allows bubbles of vapour to form in the liquid.

The normal boiling temperature of a liquid is the temperature at which the vapour pressure is equal to standard atmospheric pressure (101.3 kPa).

A liquid will boil at a lower temperature if the external pressure is reduced (boiling water up a mountain, for example).

A liquid will boil at a higher temperature if the external pressure is increased (in a pressure cooker, for example).

The graph overleaf illustrates: