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HE III Wednesday, November 30 7:00 – 8:15

HE III Wednesday, November 30 7:00 – 8:15. 180 Bevier AQ5 Bhagi AQ6 Bhagi AQ8 Bhagi AQJ Blair. 160 English AQ2 Abadeer AQ3 Abadeer. 66 Library AQ1 Abadeer AQ4 Abadeer AQB Barrick. 101 Armory AQ7 Barrick AQ9 Barrick AQA Barrick. Conflict Exam 5:30 – 6:45 161 Noyes Lab.

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HE III Wednesday, November 30 7:00 – 8:15

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  1. HE III Wednesday, November 30 7:00 – 8:15 180 Bevier AQ5 Bhagi AQ6 Bhagi AQ8 Bhagi AQJ Blair 160 English AQ2 Abadeer AQ3 Abadeer 66 Library AQ1 Abadeer AQ4 Abadeer AQB Barrick 101 Armory AQ7 Barrick AQ9 Barrick AQA Barrick Conflict Exam 5:30 – 6:45 161 Noyes Lab

  2. Equilibrium Calculations K at a given temperature, it is constant K = [products]eqp For solution chemistry [reactants]eqr K = Pproductsp For gases Preactantsr or Kc = [products]eqp [reactants]eqr K = Kc (RT)n R = 0.08207

  3. Equilibrium Calculations it changes until it = K Q is not constant Q = [products]ip For solution chemistry [reactants]ir Q = Pprod ip For gases Preact ir if Q > K reverse reaction too much product initially if Q < K forward reaction too much reactant initially if Q = K system at equilibrium

  4. Equilibrium Calculations Write the expression for K for the reaction: CaCO3 (s)  CaO (s) + CO2 (g) K = 0.0045 at 800oC K = PCO2 1 mol of CaCO3, 1 mol of CaO and 1 atm CO2 Q = CaCO3 1 make more • forward CaO b) reverse 1 mol of CaCO3 and 1 atm CO2 0 Q =

  5. Equilibrium Calculations reaction table ICE table + I2 (g)  2HI (g) H2 (g) Q = 02/(0.5)2 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

  6. 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

  7. Equilibrium Calculations CH3OH (aq) CH2O (aq) + H2 (aq)  K = 3.7 x 10-10 at 298 K Starting with a 1.00 M solution of CH3OH, what are the equilibrium concentrations ? [CH2O] (M) [H2] (M) [CH3OH] (M) 0 Initial 1.00 0 -x +x +x Change Equilibrium 1.00 - x x x

  8. Equilibrium Calculations CH3OH (aq) CH2O (aq) + H2 (aq)  1.92 x 10-5 x 100 = 0.00182% 1.00 1.00 – 0.0000192 = 1.00 x = 1.92 x 10-5 K = 3.7 x 10-10 = x2 = x2 1.00 1.00 - x  1.00 very small 1.00 – x [CH2O] (M) [H2] (M) [CH3OH] (M) 0 Initial 1.00 0 -x +x +x Change Equilibrium 1.00 - x x x

  9. N2 O4 (g)  2 NO2 (g) Kc = 4.0 x 10-7 • If 0.1 mol of N2 O4 (g) is reacted in a 1.0 L , what • is the equilibrium concentration of NO2 (g) ? • 2.6 x 10-5M • 1.2 x 10-2M • 2.0 x 10-4M • 4.0 x 10-7M • 7.8 x 10-3M

  10. FeSCN2+ (aq)  Fe3+ (aq) + SCN- (aq) K = 8.3 x 10-5 Calculate the equilibrium concentrations when starting with 2.0 M FeSCN2+  8.3 x 10-5 = x2 x2 x = 0.0129 2.0 - x 2.0 0.0129 x 100% = 0.644% 2.0 5% approximation O.K.

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