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Le Chatelier’s Principle

Le Chatelier’s Principle. equilibrium. balance. forward reaction. reverse reaction. disturb balance. changes in experimental conditions. equilibrium shifts. counteract disturbance. concentration. (gas phase). pressure. temperature. Concentration. Fe 3+ (aq). FeSCN 2+.

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Le Chatelier’s Principle

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  1. Le Chatelier’s Principle equilibrium balance forward reaction reverse reaction disturb balance changes in experimental conditions equilibrium shifts counteract disturbance concentration (gas phase) pressure temperature

  2. Concentration Fe3+ (aq) FeSCN2+ + SCN- (aq)  add Fe(NO3)3 Q K add reactant < add NaSCN add reactant add C2O42- Q K remove Fe2+ > at equilibrium change [FeSCN2+] [FeSCN2+] K = Q = [Fe3+] [SCN-] [Fe3+] [SCN-] ratef = kf [Fe3+] [SCN-]

  3. Pressure N2O4 (g) 2 NO2 (g)  at 25oC, K = 10.38 increase P by adding reactant or product K = [NO2]eq2 add N2O4 Q = [NO2]eq2 [N2O4]eq [N2O4]i Q K <

  4. Pressure N2O4 (g) 2 NO2 (g)  decrease volume [N2O4] = mol N2O4 increase [N2O4] V [NO2] = mol NO2 increase [NO2] V (6.0)2 = 11.9 Q = K = [NO2]2 = (3.0)2 = 10.3 (1.74) [N2O4] (0.87) Q K > decrease volume decrease n increase n increase volume Δn = 0 no effect of pressure

  5. PNO PNO PN O 2 2 2 4 Pressure N2O4 (g) 2 NO2 (g) add inert gas  1.00 M Ar [NO2] = 3.0 M = 3.0 mol/L increase P [N2O4] = 0.87 mol/L = 0.87 M P(1.0 L)= (4.87) (.08206) (298) PV = nRT at 298 K P (1.0 L) = (3.87) (.08206) (298) P = 119 atm P = 95 atm (95) =73 atm =(3/3.87) (119) (3/4.87) =73atm = = (.87/3.87) (95) =22atm KP unchanged Kp= (73)2 / 22 = 242

  6. Temperature changes K heat + N2O4 (g) +heat heat 2 NO2 (g) ΔH = 58.0 kJ  product reactant treat heat endothermic exothermic ΔH > 0 ΔH < 0 raising T adding heat as reactant ΔH > 0 lowering T removing heat as product ΔH < 0

  7. Calculations reaction table ICE table + I2 (g)  2HI (g) H2 (g) at 453oC, calculate K at equilibrium, [H2] = 0.107 M = 0.50 – x x = 0.393 = (0.786)2 = 54.3 = (2x)2 K = [HI]2eq [I2]eq [H2]eq (0.50–x) (0.50–x) (0.107)2 [I2] (M) [HI] (M) [H2] (M) Initial 0.50 0.50 0.00 Change - x - x +2x 2x 0.50 – x 0.50 – x Equilibrium

  8. Calculations  K = 54.3 + I2 (g)  2HI (g) H2 (g) K = x = .782 54.3 = (.224 + 2x)2 Q = (.224)2 = .195 < K x = .355 (.623 – x) (.414 – x) (.414) (.623) x = -b ± - 57.2x 50.3x2 + 13.96 = 0 b2 – 4ac 2a ax2 bx c [H2] (M) [HI] (M) [I2] (M) Initial .623 .414 .224 Change - x -x +2x Equilibrium .623 - x .414 - x .224 + 2x

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