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Chemical Equilibrium: “ Big K”

Chemical Equilibrium: “ Big K”. kinetics: rate constant “little k”. kinetics “little k” told us how fast a reaction proceeds and is used to indicate a possible mechanism. Eq. tells us to what extent a RXN proceeds to completion. react.  prod. @ Eq: rate forward = rate reverse.

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Chemical Equilibrium: “ Big K”

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  1. Chemical Equilibrium: “Big K” kinetics: rate constant “little k” kinetics “little k” told us how fast a reaction proceeds and is used to indicate a possible mechanism. Eq. tells us to what extent a RXN proceeds to completion. react.  prod. @ Eq: rate forward = rate reverse H2CO3(l)  H2O(l) + CO2(g)

  2. The Equilibrium Constant: K (Temperature Dependent) (mechanism independent) aA + bB  dD + eE Law of Mass Action: Values of Kc are constant for a RXN at a given temperature. Any equilibrium mixture of the above system at that temperature should give the SAME Kc value.

  3. We also have Kp which is sometimes used when dealing with gases with the “P” referring to the pressure of the gases. aA(g) + bB(g)  dD(g) + eE(g) Since PV = nRT (ideal gas law) so pressure is proportional to concentration. moles of GAS temp in K 0.0821L•atm mol•K

  4. Problem: PCl3(g) + Cl2(g) PCl5(g) In a 5.00 L vessel (@ 230oC) an equilibrium mixture is found to contain: 0.0185 mol PCl3, 0.0158 mol PCl5 and 0.0870 mol of Cl2. Question: determine Kc and Kp Solution: Kc = 49.08 one mol less gas on product side. Question: what does the value of K mean?

  5. Using I.C.E. (Initial, Change, Equilibrium) Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc? CO(g) + 2H2(g)  CH3OH(g) 0.3000 0 I. 0.1500 -x -2x +x C. x 0.3000 - 2x 0.1500 - x E. Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187 Therefore x = 0.0313 We can now solve for each of the other Eq. terms.

  6. Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc? CO(g) + 2H2(g)  CH3OH(g) 0.3000 0 I. 0.1500 -x -2x +x C. x 0.3000 - 2x 0.1500 - x E. Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187 Therefore x = 0.0313 We can now solve for each of the other Eq. terms. H2: 0.3000 - 2x = 0.2374 moles CH3OH: x = 0.0313 moles CO(g) + 2H2(g)  CH3OH(g) Therefore E. 0.1187 0.2374 0.0313

  7. Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc? CO(g) + 2H2(g)  CH3OH(g) 0.3000 0 I. 0.1500 -x -2x +x C. x 0.3000 - 2x 0.1500 - x E. CO(g) + 2H2(g)  CH3OH(g) Therefore E. 0.1187 0.2374 0.0313 Now find Kc: Kc = 10.52

  8. More I.C.E. Problem: @ 77oC, 2.00 mol of NOBr (nitrosyl bromide) placed in a 1.000L flask dissociates to the extent of 9.4%. Find Kc. 2NOBr(g)  2NO(g) + Br2(g) I. 2.00 0 0 -2x +2x +x C. x E. 2.00 - 2x 2x Now what? Since 9.4% dissociates, the Change in NOBr, -2x = 2.00(0.094) and.... |x| = 0.0940 so substituting the x value into the “E. term” gives: 2NOBr(g)  2NO(g) + Br2(g) E. 1.812 0.188 0.0940

  9. More I.C.E. Problem: @ 77oC, 2.00 mol of NOBr (nitrosyl bromide) placed in a 1.000L flask dissociates to the extent of 9.4%. Find Kc. 2NOBr(g)  2NO(g) + Br2(g) I. 2.00 0 0 -2x +2x +x C. x E. 2.00 - 2x 2x 2NOBr(g)  2NO(g) + Br2(g) E. 1.812 0.188 0.0940 = 1.01 x 10-3 What does the value of Kc tell us?

  10. Treatment of Pure Solids and Liquids (as solvents) in K expressions. S(s) + O2(g)  SO2(g) would expect Kc = [SO2] [S(s)][O2] solids are dropped out. but since M is meaningless for solids, and... Experimentally Kc is found to be 4.2 x 1052 @ 25oC and independent of S.

  11. AgCl(s)  Ag+(aq) + Cl-(aq) Ksp = [Ag+(aq)][Cl-(aq)] = 1.8 x 10-10 @ 25oC This is an EQUILIBRIUM value independent of the amount of solid AgCl left sitting on the bottom of the container.

  12. PURE LIQUIDS (SOLVENTS) NH3(g) + H2O(l)  NH4+(aq) + OH-(aq) Note: by convention the water is ignored. HCOOH(aq) + H2O(l)  HCOO-(aq) + H3O+(aq)

  13. CaCO3(s)  CaO(s) + CO2(g) What is the Kc expression? Kc = [CO2] What is the Kp expression? Kp = PCO2

  14. Reversing Equations: Reactants become Products and Products become Reactants. What is the relationship between the two K values? 1. 2H2(g) + O2(g)  2H2O(g) 2. 2H2O(g)  2H2(g) + O2(g) Relationship: K2 = 1/K1 = K1-1

  15. Knet for summing RXN’s: If a RXN can be obtained from the sum of RXN’s, KRXN = K1K2 RXN 1: S(s) + O2 (g) SO2(g) RXN 2: SO2(g) + 1/2O2(g)  SO3(g) S(s) + 3/2O2(g)  SO3(g) Net RXN: KRXN = K1K2

  16. The RXN Quotient: Qc Consider a system that may not yet be @ Equilibrium. aA + bB  dD + eE @ Equilibrium If Qc = Kc ? is too small so RXN  ratio If Qc < Kc? is too large so RXN  If Qc > Kc? ratio

  17. Example: PCl5(g)  PCl3(g) + Cl2(g) @ 250oC Kc= 4.0 x 10-2 If: [Cl2] and [PCl3] = 0.30M and [PCl5] = 3.0M, is the system at Equilibrium? If not, which direction will it proceed? = 3.0 x 10-2 Qc = Find Qc and compare to Kc to decide. Qc < Kc (not @ Eq.) Which way must the RXN go to achieve Equilbrium? Remember ratio is prod./React more products makes the number bigger RXN goes

  18. Calculations Using Kc: (1st case....Perfect Square) Kc = 55.17 @ 699K H2(g) + I2(g)  2HI(g) Experiment: 1.00 mol of each H2 and I2 in a 0.500 L flask. Find [ ] of products and reactants @ Equilibrium. H2(g) + I2(g)  2HI(g) conc. in mol/L [ ] [ ] [ ] 0 2.00 2.00 I. 2x -x -x C. 2.00 - x 2x 2.00 -x E. = 55.17 “perfect square”

  19. Calculations Using Kc: (1st case....Perfect Square--continued) Kc = 55.17 @ 699K H2(g) + I2(g)  2HI(g) H2(g) + I2(g)  2HI(g) [ ] [ ] [ ] 2.00 - x 2x 2.00 -x E. = 55.17 “perfect square” 7.428(2.00 - x) = 2x 1.58 = x

  20. Calculations Using Kc: (1st case....Perfect Square) Kc = 55.17 @ 699K H2(g) + I2(g)  2HI(g) H2(g) + I2(g)  2HI(g) [ ] [ ] [ ] 2.00 - x 2x 2.00 -x E. 7.428(2.00 - x) = 2x 1.58 = x [H2] = 2.00 - 1.58 = 0.42M [I2] = 0.42M [HI] = 2(1.58) = 3.16M

  21. Kc Problems with Quadratic Equation ax2 + bx +c = 0 If equilibrium expression is not a perfect square must use quadratic equation.

  22. Problem: H2(g) + I2(g) 2HI(g) @ 458 oC Kc = 49.7 Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask. Find: conc. of the equilibrium mixture. H2(g) + I2(g) 2HI(g) [ ] [ ] [ ] 2.00 0 I. 1.00 -x 2x -x C. 2x 2.00 - x 1.00 - x E. 0.920x2 - 3.00x + 2.00 = 0

  23. Problem: H2(g) + I2(g) 2HI(g) @ 458 oC Kc = 49.7 Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask. Find: conc. of the equilibrium mixture. H2(g) + I2(g) 2HI(g) [ ] [ ] [ ] 2.00 0 I. 1.00 2x -x -x C. 2x 2.00 - x 1.00 - x E. 0.920x2 - 3.00x + 2.00 = 0 2 solutions for x: 1.63  0.70 x = 2.33 or 0.93 will give positive solution for Eq. Conc.

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