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K c and Equilibrium Problems

K c and Equilibrium Problems. (4 types). Solve 4 types of problems involving equilibrium constants. Additional KEY Terms ICE table. 1. Plug. and solve. *Always make sure the equation is balanced, FIRST*.

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K c and Equilibrium Problems

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  1. Kc and Equilibrium Problems (4 types)

  2. Solve 4 types of problems involving equilibrium constants. Additional KEY Terms ICE table

  3. 1. Plug and solve *Always make sure the equation is balanced, FIRST*

  4. At 225°C, a 2.0 L container holds 0.040 moles of N2, 0.15 moles of H2 and 0.50 moles of NH3. If the system is at equilibrium, calculate KC. N2(g) + 3 H2(g) 2 NH3(g) 1. Change all into concentrations - mol/L 0.040 mol N2 = 0.020 M N2 2.0 L = 0.075 M H2 = 0.25 M NH3 2. Write the equilibrium law for the reaction

  5. 3. Substitute the concentrations and calculate K. (Note: no units for K)

  6. 2 NO(g) N2(g) + O2(g) Kc = Kc = = 0.0725 [N2][O2] [1.40][1.40] The equilibrium concentrations of N2 and NO are 1.40 mol/L and 5.20 mol/L respectively. Calculate the KC. 1.40 M 1.40 M 5.20 M According to Eq stoichiometry – the same amount of O2 will be produced as N2. [NO]2 [5.20]2

  7. 2. Rearrange and solve

  8. N2(g) + O2(g) 2 NO(g) At 210°C, the Kc is 64.0 The equilibrium concentrations of N2 and O2 are 0.40 mol/L and 0.60 mol/L, respectively. Calculate the equilibrium concentration of NO. 1. Write out the Eq Law. 2. Rearrange for [NO].

  9. 3. Substitute concentrations then solve.

  10. SO2(g) S(g) + O2(g) Kc = [S][O2] [S][O2] = Kc At 10.0°C, the Kc is 215.0 The equilibrium concentrations of SO2 is 9.40 mol/L. Calculate the [S] and [O2] at equilibrium. x M x M 9.40 M Values should be the same amount for S and O2 – assign the unknown “x” [SO2] [SO2]

  11. [S][O2] = Kc [x][x] = 215 [9.40] √ √ [x]2 = 2021 x = 45.0 M [S]eq and [O2] eq = 45.0 M [SO2]

  12. 3. Simple I.C.E. Tables

  13. H2(g) + F2(g) 2 HF(g) 1.00 M of hydrogen and 1.00 M of fluorine are allowed to react at 150.0°C. At equilibrium, the [HF] is 1.32 M. Calculate KC. 1 1 + 1.32 [Change] - 0.66 - 0.66 1.00 M 0 1.00 M [Initial] [Eqlbm] 0.34 1.32 0.34 Tip #1 – the changes in concentration match the stoichiometry of the reaction.

  14. H2(g) + F2(g) 2 HF(g) [Eq] 0.34 1.32 0.34

  15. 2 SO3 (g) 2 SO2 (g) + O2 (g) Initially 2.0 mol of SO2, 1.0 mol of O2 are mixed in a 3.0 L reaction container. At equilibrium, 0.20 mol of O2 are found to remain. Calculate the Kc. 0 0.67 mol/L 0.33 mol/L [I] [C] + 0.52 - 0.52 - 0.26 0.52 0.15 0.067 [E] Tip #2 – make sure the +/- reflect the appropriate side being made or used up

  16. 2 SO3 (g) 2 SO2 (g) + O2 (g) Kc = Kc = [SO2]2[O2] [0.15]2[0.067] 0.52 0.15 0.067 [E] Tip #3 –Write Eq Lawfor equation as written. Kc = 5.6 x 10-3 Kc = 0.00557 [SO3]2 [0.52]2

  17. 4. HARD I.C.E. Tables

  18. 6.0 moles of N2 and O2 gases are placed in a 1.0 L container, what are all the concentrations at equilibrium? The Kc is 6.76. 1 N2(g) + O2(g) 2 NO(g) 1 6.0 mol/L 6.0 mol/L 0 [I] [C] - x - x + 2x [E] 2x 6.0 - x 6.0 - x Tip #4 – use x values to match the stoichiometry changes of the reaction.

  19. Substitute known values. Get rid of the square by taking the square root of both sides.

  20. Isolate, and solve for x. 6.0 mol/L 6.0 mol/L 0 [I] - x -3.4 - x -3.4 2(3.4) + 2x [C] [E] 6.0 - x 6.0 - x 2.6 6.8 2x 2.6

  21. Kc = [HI]2 [H2][I2] 2.0 moles of HI were placed in a 1.0 L flask at 430ºC. (Kc = 54.3) Calculate the equilibrium concentrations. H2 (g) + I2 (g) 2 HI (g) [I] 0 0 2.0 - 2x + x + x [C] [E] + x 2.0 - 2x + x

  22. Kc = [HI]2 54.3 = [2.0 - 2x]2 7.37 = 2.0 – 2x [+ x][+ x] x [H2][I2] √ √ 7.37 x = 2.0 – 2x 7.37 x + 2x = 2.0 [I] 0 0 2.0 9.37 x = 2.0 - 2x x = 0.21 + x + x - 0.42 0.21 0.21 [C] [E] + x 2.0 - 2x + x 0.21 1.58 0.21

  23. The Kc for the reaction 5.3 x 10 -2 at 0ºC. Initially, 2.5 mol each particle was injected into a 1 L reaction vessel. Find the Eq concentrations. H2 (g) + Cl2 (g) 2 HCl (g) [I] 2.5 mol/L 2.5 mol/L 2.5 mol/L + x + x -2x [C] 2.5 + x [E] 2.5 + x 2.5 - 2x Tip #5 – use the Kc valueto predictthe changes in the reaction.

  24. 5.3 x 10 -2 = [2.5 - 2x]2 5.3 x 10 -2 = [2.5 - 2x]2 [2.5 + x][2.5 + x] Kc = [HCl]2 [2.5+x]2 [H2][Cl2] √ √ √ 0.23 = 2.5 – 2x 2.5+ x

  25. x = 1.92 2.23 0.58 + 0.23x = 2.5 - 2x 2.23 x = 1.92 x = 0.86 mol/L [I] 2.5 mol/L 2.5 mol/L 2.5 mol/L [C] 2(0.86) + 0.86 + 0.86 [E] 3.4 mol/L 3.4 mol/L 0.8 mol/L

  26. CAN YOU / HAVE YOU? • Solve 4 types of problems involving equilibrium constants. Additional KEY Terms ICE table

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