1. C. Y. Yeung (CHW, 2009) p.01 … the eqm position will shift in a way to OPPOSE the effect of CHANGE. Predict how the Eqm Shift: Le Chatelier’s Principle temp. / conc. / pressure so as to reduce [reactant]. e.g. If [reactant] , the eqm position will shift FW.

2. p.02 Consider: N 2 (g) + 3H 2 (g) 2NH 3 (g) (Given: at 600K, K c = 2.0 mol -2 dm 6,  H < 0) Eqm will shift FW to reduce the amount of N 2 (g). Both the rate of FWR & BWR would be faster. BUT: K eq is unchanged! CHANGE 1: Adding N 2 (g) at constant vol. … Eqm will shift FW to increase the amount of NH 3 (g). Both the rate of FWR & BWR would be slower. BUT: K eq is unchanged! CHANGE 2: Removing NH 3 (g) at constant vol. …

3. p.03 Eqm will shift FW to reduce the pressure by reducing the no. of mol of gaseous molecule. Both the rate of FWR & BWR would be faster. BUT: K eq is unchanged! CHANGE 3: Increasing the pressure by reducing vol Both [reactant] and [product] remain unchanged, so there will be no change in the system. K eq is unchanged too! CHANGE 4: Adding a Noble Gas at constant vol. … Continue ….. N 2 (g) + 3H 2 (g) 2NH 3 (g) (Given: at 600K, K c = 2.0 mol -2 dm 6,  H < 0)

4. p.04 Eqm will shift BW so as to increase [Reactant]. Both the rate of FWR & BWR would be slower. K eq is unchanged! Question :How about adding a Noble Gas at constant pressure …? Noble Gas is added but pressure remains the same, the volume must expand. Conc. of all species decrease. [Reactant] decrease more. N 2 (g) + 3H 2 (g) 2NH 3 (g)

5. p.05 p. 118 Check Point 16-10B Q.1 (b)(i)decreases (a) Kp =Kp =Kp =Kp = P HNO 3 (g) 2P NO(g)  P NO 2 (g) 3 P H 2 O(g)  (  shift FW to reduce partial pressure of H 2 O) (ii)increases (iii)increases (c)Value of K p remains unchanged. (  change of conc. / pressure would not change the value of K eq )

6. p.06 p. 118 Check Point 16-10B Q.2 (b)Vol. compressed to one-half, partial pressures double (a) Kp =Kp =Kp =Kp =(0.364) (0.636) 2 = 0.900 atm -1 2NO 2 (g) N 2 O 4 (g) 2(0.636) = 1.272 at start (atm) 2(0.364) = 0.728 at eqm (atm) 1.272 – 2x 0.728 + x (1.272 – 2x) 2 K p = 0.900 = 0.728 + x x = 0.144  P = 1.272 – 2(0.144) = 0.984 atm NO 2 P = 0.728 + 0.144 = 0.872 atm N2O4N2O4N2O4N2O4

7. p.07 How about increasing temp.? As FWR is exothermic, increasing temp. would shift the eqm BW so as to absorb excessive heat. N 2 (g) + 3H 2 (g) 2NH 3 (g) (  H < 0)  By finding the values of K c at diff. temp.,  H of FWR (of a reversible rxn) could be determined. (ref.: p. 120) ln K eq = const. – HHHH RT **** BUT: K eq would CHANGE!!!

8. p.08 Why change in Temp. changes the value of K eq ? If FWR is exothermic, that means the E a of FWR is lower than that of BWR. E a of FWR HHHH ** E a of BWR = E a of FWR +  H

9. p.09 Note: When temp. increases, both the rate of FWR and BWR increase …. But the BWR (with higher E a ) will have a higher percentage increase in rate. lower % of increase higher % of increase  k 1 and k -1 increase for different extent!

10. p.10 Therefore, when temp. increases, the eqm position would shift BW to absorb excessive heat, and also increase the conc. of reactant. Reactant become more predominant than before. N 2 (g) + 3H 2 (g) 2NH 3 (g) (  H < 0) K eq decreases! (ref. p. 119)

11. p.11 p. 120 Check Point 16-10C (a)The eqm position shifts to the left (backward). (b)The eqm position shifts to the left (backward). (c)The eqm position remains unchanged. (d)The eqm position to the right (forward). (e)The eqm position to the left (backward). (  noble gas has no effect on eqm position) (  NO 2 decreases more in partial pressure) (  shift BW [exothermic!] to release more heat)

12. p.12 Factors affecting the Rate of Rxn, Eqm Position and K eq … FactorsRate of RxnEqm PositionK eq Temp. Conc. Pressure Surface Area Catalyst              

13. p.13 2008 HKASL Paper 2 Q.4(a) [7M] (a)(i) H 2 (g) + I 2 (g) 2HI(g) 4.0 at start (mol) 02.0 at eqm (mol) 4.0 – x 2x 2.0 – x K c = 50 = [(4-x)/5][(2-x)/5] (2x/5) 2 x = 1.87  [H 2 (g)] = (4 – 1.87)/5 = 0.426 mol dm -3 [I 2 (g)] = (2 – 1.87)/5 = 0.026 mol dm -3 [HI(g)] = 2(1.87)/5 = 0.748 mol dm -3 Kc =Kc =Kc =Kc = [H 2 (g)][I 2 (g)] [HI(g)] 2 0.5 1 0.5 2

14. p.14 2008 HKASL Paper 2 Q.4(a) [7M] (a)(ii) (I)No change. There is no change in the no. of mol of gases in the rxn. No shifting of eqm position will result. (II)Increased. The eqm position will shift to the right to give a greater no. of mol of HI(g). 0.5 1 0.5 1

15. p.15 Expt. 10 Eqm Constant of Esterification 0.25 mol glacial CH 3 COOH + 0.25 mol CH 3 CH 2 CH 2 OH 1cm 3 sample + 25 cm 3 D.I. H 2 O + phenolphthalein: Titrate against std. NaOH  V 1 20 drops H 2 SO 4 (l) 0.25 mol glacial CH 3 COOH + 0.25 mol CH 3 CH 2 CH 2 OH + H 2 SO 4 (l) 1cm 3 sample + 25 cm 3 D.I. H 2 O + phenolphthalein: Titrate against std. NaOH  V 2 (V 2 – V 1 ) = amount of H 2 SO 4

16. p.16 Expt. 10 Eqm Constant of Esterification (con’t) 0.25 mol glacial CH 3 COOH + 0.25 mol CH 3 CH 2 CH 2 OH + H 2 SO 4 (l) (remained) reflux 30 mins. cool in an ice-bath 1cm 3 sample Titrate against std. NaOH  V 3 reflux 15 mins 1cm 3 sample Titrate against std. NaOH  V 4 if V 3 = V 4, eqm is reached!

17. p.17 Expt. 10 Eqm Constant of Esterification (con’t) CH 3 COOH + CH 3 CH 2 CH 2 OH CH 3 COOCH 2 CH 2 CH 3 + H 2 O x (calculated from V 1 ) x at start (M) at eqm (M) y (calculated from V 4 and V 2 -V 1 ) y - a aa Kc =Kc =Kc =Kc = y2y2y2y2 a2a2a2a2

18. Assignment Study the examples in p. 113-122, p.128 Q.8 - 13, p. 232 Q.13 [due date: 9/3(Wed)] p.18 Next …. Partition Equilibrium (p. 104 - 109) [Tue] Pre-Lab: Expt. 10 Eqm Constant of Esterification