2. Equilibrium HL only

3. 17.2 The Equilibrium Law Easy if you are given equilibrium concentrations! You simply substitute the values in to an expression for K c. 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 dm 3. This is not necessary if there are equal numbers of reactant moles and product moles (i.e. if K c has no units).

4. 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 K c for the hydrolysis of ethyl ethanoate. CH 3 COOC 2 H 5 + H 2 O CH 3 COOH + C 2 H 5 OH We use the Postman Pat method.

5. CH 3 COOC 2 H 5 + H 2 O CH 3 COOH + C 2 H 5 OH Initial concs a /Vb /V00 Eqm concs a-x /Vb-x /V x /V In this example we can ignore V. Why? Calculating initial number of moles a = moles ethyl ethanoate = 200 / 88 = 2.27 b = moles of water = 7 /18 = 0.39 Use info in question and M r values. 2.270.39 Calculating equilibrium number of moles x = 0.25 (given in Q) 0.25 So a-x = 2.27-0.25 = 2.02 & b-x = 0.39-0.25 = 0.14 2.020.14

6. CH 3 COOC 2 H 5 + H 2 O CH 3 COOH + C 2 H 5 OH Now write an expression for K c. K c = [CH 3 COOH][C 2 H 5 OH] [CH 3 COOC 2 H 5 ][H 2 O] Substitute in values KcKc = 0.25 x 0.25 / 2.02 x 0.14 = 0.22 (no units)

7. Worked example 2 2 moles of phosphorus(V) chloride vapour are heated to 500 K in a vessel of volume 20 dm 3. The equilibrium mixture contains 1.2 moles of chlorine. Calculate K c for the decomposition of phosphorus(V) chloride into chlorine and phosphorus(III) chloride. PCl 5 PCl 3 + Cl 2 We use the Postman Pat method.

8. Initial concs a /V 00 Eqm concs a-x /V x /V Calculating initial concentration a /V = moles PCl 5 / vol in dm 3 = 2 /20 = 0.1 moldm -3 0.1 Calculating equilibrium concentrations x/V = 1.2/20 = 0.06 moldm -3 (from moles of Cl 2 in Q) 0.06 So a-x/V = 0.1-0.06 = 0.04 0.040.06 PCl 5 PCl 3 + Cl 2

9. Now write an expression for K c. K c = [PCl 3 ][Cl 2 ] [PCl 5 ] Substitute in values KcKc = 0.06 x 0.06 / 0.04 = 0.09 moldm -3 PCl 5 PCl 3 + Cl 2

10. 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

11. 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. H 2 O(l)  H 2 O(g)  H = +40.7 kJmol -1 at 373 K & 101.3 kPa

12. 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  H vap / kJmol -1 b.p. / KIntermolecular forces methane9.0109Van der Waals only methoxymethane27.2248Van der Waals and dipole – dipole ethanol38.6352Van der Waals, dipole – dipole and hydrogen bonding

13. 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. e.g for propane

14. 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: