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

2. Chemistry 1011 Slot 52 12.3 Determination of K YOU ARE EXPECTED TO BE ABLE TO: Calculate the value of K from experimental data for the system at equilibrium Calculate the value of K from experimental data for the system at the start of a reaction and at equilibrium

3. Chemistry 1011 Slot 53 Determination of K from Equilibrium Data 1.Write a balanced equation for the equilibrium 2.Write an expression for the equilibrium constant 3.Input equilibrium partial pressures of components 4.Calculate K

4. Chemistry 1011 Slot 54 Determination of K from Equilibrium Data Ammonium chloride decomposes on heating At 400 o C, 12.0g of NH 4 Cl is present in a closed system. The partial pressures of NH 3 and HCl are 3.0atm and 5.0atm. Calculate K NH 4 Cl (s) NH 3(g) + HCl (g) K P = P NH 3 x P HCl = 3.0 x 5.0 = 15

5. Chemistry 1011 Slot 55 Determination of K from Initial and Equilibrium Data 1.Write a balanced equation for the equilibrium 2.Write an expression for the equilibrium constant 3.Distinguish equilibrium from initial partial pressures 4.Use stoichiometry to determine equilibrium partial pressures from initial partial pressure data 5.Input equilibrium partial pressures of components 6.Calculate K

6. Chemistry 1011 Slot 56 Determination of K from Initial and Equilibrium Data Consider the system: 2HI (g) H 2(g) + I 2(g) Initially, a system contains HI only, at a pressure of 1.00atm at 520 o C. At equilibrium, the partial pressure of H 2(g) is found to be 0.10atm Find 1.P I 2 at equilibrium 2.P HI at equilibrium 3.Kp

7. Chemistry 1011 Slot 57 Determination of K from Initial and Equilibrium Data 2HI (g) H 2(g) + I 2(g) Initially 1.00atm 0.00atm 0.00atm At equilibrium ? 0.10atm ? 2 moles of HI must react in order to produce 1 mole of H 2 At the same time, 1 mole of I 2 will be produced Partial pressure is proportional to the number of moles of gas  P -0.20atm +0.10atm +0.10atm At equilibrium 0.80atm 0.10atm 0.10atm Kp = P H 2 x P I 2 = 0.10 x 0.10 = 0.016 (P HI ) 2 (0.80) 2

8. Chemistry 1011 Slot 58 Equilibrium Reaction of NOBr 2.00atm of NOBr (g) are placed into a closed container at 350 o C NOBr (g) decomposes to NO (g) and Br 2(g) and an equilibrium is established At equilibrium, the partial pressure of Br 2 is 0.25atm 1.Write an equation for the equilibrium 2.Write an expression for Kp 3.Evaluate Kp for the reaction at 350 o C

9. Chemistry 1011 Slot 59 Equilibrium Reaction of NOBr 2 NOBr (g) 2NO (g) + Br 2(g) Initial 2.00atm 0.00atm 0.00atm Equilibrium ? ? 0.25atm Each mole of NOBr that decomposes will produce 1 mole of NO and ½ mole of Br 2  P -0.50atm +0.50atm +0.25atm Equilibrium 1.50atm 0.50atm 0.25atm Kp = (P NO ) 2 x P Br 2 = (0.50) 2 x (0.25) = 2.8 x 10 -2 (P NOBr ) 2 (1.50) 2