1. Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Gaseous Chemical Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 12

2. Chemistry 1011 Slot 52 12.1 The N 2 O 4  NO 2 Equilibrium YOU ARE EXPECTED TO BE ABLE TO: Distinguish equilibrium from steady state situations. Recognize chemical equilibrium as a dynamic process taking place in a closed system. Identify the changes taking place at the molecular level in a chemical equilibrium process. Identify the equilibrium constant as a temperature dependent constant related to the equilibrium partial pressures of reactants and products.

3. Chemistry 1011 Slot 53 Review - Partial Pressure Total gas pressure is proportional to the number of moles present; it is independent of the identity of the material The partial pressure of a gas in a gas mixture is the fraction of the total pressure that is due to that gas The partial pressure of gas X 2 in a mixture is equal to the mole fraction of X 2 x Total Pressure

4. Chemistry 1011 Slot 54 Partial Pressure Example Air is approximately 80% N 2 and 20%O 2 Total air pressure is approximately 100kPa P N 2 + P O 2 = P TOTAL P N 2 = 80% x 100kPa = 80kPa P O 2 = 20% x 100kPa = 20kPa Partial pressures can be used to express the concentrations of reactants and products in chemical equilibria

5. Chemistry 1011 Slot 55 The N 2 O 4  NO 2 Equilibrium At any given temperature, a sample of NO 2 will exist as an equilibrium mixture of NO 2 and N 2 O 4 N 2 O 4(g) 2NO 2(g) The forward and reverse reactions are taking place at the same rate The concentrations of the species remain constant

6. Chemistry 1011 Slot 56

7. 7 Getting to Equilibrium Equilibrium can be approached starting from either “reactants” or “products” In the N 2 O 4  NO 2 example, we can theoretically start from either 100% reactants or 100% products The equilibrium position will be the same

8. Chemistry 1011 Slot 58 The Equilibrium Constant Equilibrium Measurements for the N 2 O 4  NO 2 System at 100 o C Original P (atm) Equilibrium P (atm) Expt 1 N 2 O 4 1.00 0.22 NO 2 0.00 1.56 Expt 2 N 2 O 4 0.00 0.07 NO 2 1.00 0.86 Expt 3 N 2 O 4 1.00 0.42 NO 2 1.00 2.16

9. Chemistry 1011 Slot 59 The Equilibrium Constant The relationship (P NO 2 ) 2 / P N 2 O 2 is a constant (P is the equilibrium partial pressure of the species) Experiment 1 (P NO 2 ) 2 / P N 2 O 2 = (1.56) 2 /0.22 = 11 Experiment 2 (P NO 2 ) 2 / P N 2 O 2 = (0.86) 2 /0.07 = 11 Experiment 3 (P NO 2 ) 2 / P N 2 O 2 = (2.16) 2 /0.42 = 11 The constant is called the equilibrium constant K for the system

10. Chemistry 1011 Slot 510 Characteristics of Equilibrium A dynamic equilibrium can only exist in a closed system – neither reactants nor products can enter or leave the system At equilibrium, the concentrations of reactants and products remain constant At equilibrium, the forward and reverse reactions are taking place at equal and opposite rates Equilibrium can be approached from either side of the reaction equation

11. Chemistry 1011 Slot 511 Steady State vs Equilibrium When a process takes place in an open system, a steady state may be set up In a steady state, the concentration of a product may be constant, but this will be because it is being formed, and is leaving the system at the same rate Consider the evaporation of water in closed and open systems:

12. Chemistry 1011 Slot 512 The Evaporation of Water Water in an open beaker –Water will evaporate at a rate dependent on its temperature. –The concentration of water vapour in the top of the beaker will be constant, but water molecules are entering the space and leaving it at the same rate –This is an open system – it is a steady state –Eventually, all of the water will evaporate

13. Chemistry 1011 Slot 513 The Evaporation of Water Water in a closed container –Water will evaporate at a rate dependent on its temperature –As the concentration of water vapour rises, water molecules will condense –Eventually a dynamic equilibrium will be established at that temperature –The rate of evaporation will equal the rate of condensation H 2 O (l) H 2 O (g)