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Most chemical reactions are reversible. Chemical equilibrium can only occur in a closed system.

Most chemical reactions are reversible. Chemical equilibrium can only occur in a closed system. Equilibrium is when the rates are equal AND the [ ]s remains constant. K c can be used to determine the equilibrium position. K c and Equilibrium Problems. (4 types). CAN YOU? / HAVE YOU?

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Most chemical reactions are reversible. Chemical equilibrium can only occur in a closed system.

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  1. Most chemical reactions are reversible. • Chemical equilibrium can only occur in a closed system. • Equilibrium is when the rates are equal AND the [ ]s remains constant. • Kc can be used to determine the equilibrium position.

  2. Kc and Equilibrium Problems (4 types)

  3. CAN YOU? / HAVE YOU? • Make equilibrium constant (K) calculations with various missing values. • Use ICE tables to successfully complete Equilibrium calculations. *Always make sure the equation is balanced, FIRST*

  4. 1. Plug and solve

  5. 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

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

  7. 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

  8. 2. Rearrange and solve

  9. 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].

  10. 3. Substitute concentrations then solve.

  11. 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]

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

  13. 3. Simple I.C.E. Tables

  14. H2(g) + F2(g) 2 HF(g) 1.00 mole of hydrogen and 1.00 mole of fluorine are sealed in a 1.00 L flask at 150.0°C and allowed to react. At equilibrium, 1.32 moles of HF are present. Calculate KC. 1. Write out the Eq Law – for the equation as written. 1.00 mol/L 0 1.00 mol/L [Initial] + 1.32 [Change] - 0.66 - 0.66 [Eqlbm] 0.34 1.32 0.34

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

  16. 2 SO3 (g) 2 SO2 (g) + O2 (g) Kc = [SO2]2[O2] 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] [SO3]2

  17. 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] Kc = 5.6 x 10-3 Kc = 0.00557 [SO3]2 [0.52]2

  18. 4. HARD I.C.E. Tables

  19. 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. N2(g) + O2(g) 2 NO(g) 6.0 mol/L 6.0 mol/L 0 [I] [C] - x - x + 2x [E] 2x 6.0 - x 6.0 - x

  20. 1. Substitute known values. 2. Get rid of the square by taking the square root of both sides.

  21. 3. 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

  22. 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

  23. 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

  24. 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. Kc = [HCl]2 [H2][Cl2] 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

  25. 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

  26. 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

  27. CAN YOU? / HAVE YOU? • Make equilibrium constant (K) calculations with various missing values. • Use ICE tables to successfully complete Equilibrium calculations. *Always make sure the equation is balanced, FIRST*

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