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Le Chatelier’s Principle

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Presentation on theme: "Le Chatelier’s Principle"— Presentation transcript:

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

2 Concentration Fe3+ (aq) + SCN- (aq) FeSCN2+ add Fe(NO3)3 add reactant Q < K add NaSCN add reactant add C2O42- remove Fe2+ Q > K at equilibrium change K = [FeSCN2+] Q = [FeSCN2+] [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 K = [NO2]2 = (3.0)2 Q = = 10.3 (1.74) [N2O4] (0.87) Q > K decrease volume decrease n increase n increase volume Δn = 0 no effect of pressure

5 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 M = 0.87 mol/L PV = nRT at 298 K P(1.0 L)= (4.87) (.08206) (298) P (1.0 L) = (3.87) (.08206) (298) P = 119 atm P = 95 atm 2 PNO =(3/3.87) (95) =73 atm 2 PNO = (3/4.87) (119) =73atm 2 4 PN O = (.87/3.87) (95) =22atm KP unchanged Kp = (73)2 / 22 = 242

6 Temperature changes K heat heat + N2O4 (g) 2 NO2 (g) +heat ΔH = 58.0 kJ treat heat reactant product 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 H2 (g) + I2 (g)  2HI (g) at 453oC, calculate K at equilibrium, [H2] = M = 0.50 – x x = 0.393 K = [HI]2eq = (2x)2 = (0.786)2 = 54.3 [H2]eq [I2]eq (0.50–x) (0.50–x) (0.107)2 [H2] (M) [I2] (M) [HI] (M) Initial 0.50 0.50 0.00 Change - x - x +2x 0.50 – x 2x Equilibrium 0.50 – x

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


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