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C. Y. Yeung (CHW, 2009)

Predict how the Eqm Shift: Le Chatelier’s Principle. Predict how the Eqm Shift: Le Chatelier’s Principle. p.01. … the eqm position will shift in a way to OPPOSE the effect of CHANGE. temp. / conc. / pressure. e.g. If [reactant]  ,. the eqm position will shift FW.

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C. Y. Yeung (CHW, 2009)

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

  2. Consider: N2(g) + 3H2(g) 2NH3(g) p.02 Eqm will shift FW to reduce the amount of N2(g). Both the rate of FWR & BWR would be faster. BUT: Keq is unchanged! Eqm will shift FW to increase the amount of NH3(g). Both the rate of FWR & BWR would be slower. BUT: Keq is unchanged! (Given: at 600K, Kc = 2.0 mol-2 dm6 , DH < 0) CHANGE 1: Adding N2(g) at constant vol. … CHANGE 2: Removing NH3(g) at constant vol. …

  3. Continue ….. N2(g) + 3H2(g) 2NH3(g) p.03 (Given: at 600K, Kc = 2.0 mol-2 dm6 , DH < 0) 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: Keq is unchanged! Both [reactant] and [product] remain unchanged, so there will be no change in the system. Keq is unchanged too! CHANGE 3: Increasing the pressure by reducing vol CHANGE 4: Adding a Noble Gas at constant vol. …

  4. Question : How about adding a Noble Gas at constant pressure …? p.04 N2(g) + 3H2(g) 2NH3(g) Conc. of all species decrease. Eqm will shift BW so as to increase [Reactant]. Both the rate of FWR & BWR would be slower. Keq is unchanged! [Reactant] decrease more. Noble Gas is added but pressure remains the same, the volume must expand.

  5. p.05 2  P P HNO3(g) NO(g) Kp = (a) 3  P P NO2(g) H2O(g) p. 118 Check Point 16-10B Q.1 (iii) increases (ii) increases (b) (i) decreases ( shift FW to reduce partial pressure of H2O) (c) Value of Kp remains unchanged. ( change of conc. / pressure would not change the value of Keq)

  6. p.06 (0.364) Kp = = 0.900 atm-1 (a) (0.636)2 2(0.636) = 1.272 2(0.364) = 0.728 2NO2(g) N2O4(g) at start (atm) at eqm (atm) 1.272 – 2x 0.728 + x x = 0.144 P = 1.272 – 2(0.144) = 0.984 atm  0.728 + x NO2 Kp = 0.900 = P = 0.728 + 0.144 = 0.872 atm (1.272 – 2x)2 N2O4 p. 118 Check Point 16-10B Q.2 (b) Vol. compressed to one-half, partial pressures double

  7. How about increasing temp.? p.07 N2(g) + 3H2(g) 2NH3(g) DH ln Keq = const. – RT (DH < 0) As FWR is exothermic, increasing temp. would shift the eqm BW so as to absorb excessive heat. **** BUT: Keq would CHANGE!!!  By finding the values of Kc at diff. temp., DH of FWR (of a reversible rxn) could be determined. (ref.: p. 120)

  8. p.08 If FWR is exothermic, that means the Ea of FWR is lower than that of BWR. Ea of FWR DH Why change in Temp. changes the value of Keq? ** Ea of BWR = Ea of FWR + DH

  9. p.09 But the BWR (with higher Ea) will have a higher percentage increase in rate. lower % of increase higher % of increase Note: When temp. increases, both the rate of FWR and BWR increase ….  k1 and k-1 increase for different extent!

  10. (DH < 0) p.10 N2(g) + 3H2(g) 2NH3(g) 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. Keq 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. ( noble gas has no effect on eqm position) (d) The eqm position to the right (forward). ( NO2 decreases more in partial pressure) (e) The eqm position to the left (backward). ( shift BW [exothermic!] to release more heat)

  12. p.12      Factors affecting the Rate of Rxn, Eqm Position and Keq …          

  13. p.13 0 2.0 4.0 at start (mol) H2(g) + I2(g) 2HI(g) 0.5 2x at eqm (mol) 2.0 – x 4.0 – x [HI(g)]2 1 (2x/5)2 Kc = x = 1.87 Kc = 50 = [H2(g)][I2(g)] [(4-x)/5][(2-x)/5] 0.5 [H2(g)] = (4 – 1.87)/5 = 0.426 mol dm-3  [I2(g)] = (2 – 1.87)/5 = 0.026 mol dm-3 [HI(g)] = 2(1.87)/5 = 0.748 mol dm-3 2 2008 HKASL Paper 2 Q.4(a) [7M] (a) (i)

  14. p.14 0.5 1 0.5 1 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).

  15. p.15 20 drops H2SO4(l) 1cm3 sample 1cm3 sample (V2 – V1) = amount of H2SO4 Expt. 10 Eqm Constant of Esterification 0.25 mol glacial CH3COOH + 0.25 mol CH3CH2CH2OH 0.25 mol glacial CH3COOH + 0.25 mol CH3CH2CH2OH + H2SO4(l) + 25 cm3 D.I. H2O + phenolphthalein: Titrate against std. NaOH V1 + 25 cm3 D.I. H2O + phenolphthalein: Titrate against std. NaOH V2

  16. p.16 reflux 15 mins 1cm3 sample reflux30 mins. 1cm3 sample if V3 = V4 , eqm is reached! Expt. 10 Eqm Constant of Esterification (con’t) 0.25 mol glacial CH3COOH + 0.25 mol CH3CH2CH2OH + H2SO4(l) (remained) cool in an ice-bath Titrate against std. NaOH V4 Titrate against std. NaOH V3

  17. p.17 CH3COOH + CH3CH2CH2OH CH3COOCH2CH2CH3 + H2O at start (M) - a at eqm (M) a2 Kc = y2 Expt. 10 Eqm Constant of Esterification (con’t) x (calculated from V1) x a a y (calculated from V4 and V2-V1) y

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

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