1. Equilibrium calculations To view as a slide show, Click on the “Slideshow” on the menu bar, and then on “View Show” Click the mouse or press Page Down to go to the next slide Press the Escape key to leave the presentation Press Page Up to go back

2. Problem type #1 Given some information about the conditions before equilibrium and after equilibrium, find the equilibrium constant.

3. Suppose a flask was made to be 0.500 M in NH 3 initially, and when equilibrium was reached, the [NH 3 ] had dropped to 0.106 M. Find the value of K for: N 2 + 3H 2 2NH 3

4. N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial 0 0 0.500 M Change At equilibrium 0.106 M

5. N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial 0 0 0.500 M Change 0.106 – 0.500 M = -0.394 M At equilibrium 0.106 M

6. N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial 0 0 0.500 M Change + 0.394/2 M 0.106 – 0.500 M = -0.394 M At equilibrium 0.106 M

7. N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial 0 0 0.500 M Change + 0.394/2 M 0.106 – 0.500 M = -0.394 M At equilibrium 0 +.197 M =.197 M 0.106 M

8. N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial 0 0 0.500 M Change + 0.394/2 M + 3(.394/2) M 0.106 – 0.500 M = -0.394 M At equilibrium 0 +.197 M =.197 M 0.106 M

9. N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial 0 0 0.500 M Change + 0.394/2 M + 3(.394/2) M 0.106 – 0.500 M = -0.394 M At equilibrium 0 +.197 M =.197 M 0 +.591 M 0.106 M

10. N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial 0 0 0.500 M Change + 0.394/2 M + 3(.394/2) M 0. 106 – 0.500M = -0.394 M At equilibrium 0 +.197 M =.197 M 0 +.591 M =.591 M 0.106 M

11. N 2 + 3H 2 2NH 3 Let 2x be the amount of NH 3 that reacts Use stoichiometry of reaction! N 2 H 2 NH 3 Initial 0 0 0.500 M Change +x +3x -2x At equilibrium

12. N 2 + 3H 2 2NH 3 Let 2x be the amount of NH 3 that reacts 2x = 0.500 – 0.106 = 0.394 N 2 H 2 NH 3 Initial 0 0 0.500 M Change +x +3x -2x At equilibrium 0+x0+3x0.500 – 2x = 0.106 M

13. N 2 + 3H 2 2NH 3 Let 2x be the amount of NH 3 that reacts 2x = 0.500 – 0.106 = 0.394 N 2 H 2 NH 3 Initial 0 0 0.500 M Change +x +3x -2x At equilibrium0+x = 0.394/2 = 0.197 M 0+3x = 0.197 x 3 = 0.591 M 0.500 – 2x = 0.106 M

15. Problem type #2a Given information about conditions before equilibrium and the equilibrium constant, find the concentrations when equilibrium is reached.

16. For the reaction N 2 + O 2 2NO at 1700 o C, K = 3.52x10 -4. If 0.0100 mole NO is placed in a 1 – L flask at 1700 o, what will [N 2 ] be at equilibrium? N2N2 O2O2 2NO Initial 0 00.0100 M Change At equilibrium ? ? ?

17. For the reaction N 2 + O 2 2NO at 1700 o C, K = 3.52x10 -4. If 0.0100 mole NO is placed in a 1 – L flask at 1700 o, what will [N 2 ] be at equilibrium? N2N2 O2O2 2NO Initial 0 00.0100 Change +x -2x At equilibrium

18. For the reaction N 2 + O 2 2NO at 1700 o C, K = 3.52x10 -4. If 0.0100 mole NO is placed in a 1 – L flask at 1700 o, what will [N 2 ] be at equilibrium? N2N2 O2O2 2NO Initial 0 00.0100 Change +x -2x At equilibrium x x0.0100-2x

19. N2N2 O2O2 2NO At equilibrium x x0.0100-2x

21. 0.0100 - 2x = (1.88 x 10 -2 )x 0.0100 = 2.02 x x = 4.95 x 10 -3 M = [N 2 ] (also = [O 2 ]) Note that because K was small, most of the NO became N 2 and O 2 Final [NO] = 0.0100 – 2(4.95 x 10 -3 ) =1.00 x 10 -4 M

22. Problem type #2b Given information about conditions before equilibrium and the equilibrium constant, find the concentrations when equilibrium is reached.....But the math doesn’t work out as nicely

23. For the reaction F 2 2F at 1000 o C, K = 2.7 x10 -3. If 1.0 mole F 2 is placed in a 1 – L flask at 1000 o, what will [F] be at equilibrium? F 2 2F Initial 1.0 M 0 M Change At equilibrium ? ?

24. For the reaction F 2 2F at 1000 o C, K = 2.7 x10 -3. If 1.0 mole F 2 is placed in a 1 – L flask at 1000 o, what will [F] be at equilibrium? F 2 2F Initial 1.0 M 0 M Change - x + 2x At equilibrium

25. For the reaction F 2 2F at 1000 o C, K = 2.7 x10 -3. If 1.0 mole F 2 is placed in a 1 – L flask at 1000 o, what will [F] be at equilibrium? F 2 2F Initial 1.0 M 0 M Change - x + 2x At equilibrium 1.0 – x 2x

26. F 2 2F At equilibrium 1.0 – x 2x

27. 4x 2 = 2.7 x 10 -3 (1.0 – x) = 4x 2 = 2.7 x 10 -3 – 2.7 x 10 -3 x

28. 4x 2 = 2.7 x 10 -3 (1.0 – x) = 4x 2 = 2.7 x 10 -3 – 2.7 x 10 -3 x This is a quadratic equation Rearrange to the form ax 2 + bx + c = 0 4x 2 + 2.7x10 -3 x – 2.7 x 10 -3 = 0 a = 4 b = 2.7 x 10 -3 c = -2.7 x 10 -3

29. 4x 2 + 2.7x10 -3 x – 2.7 x 10 -3 = 0 a = 4 b = 2.7 x 10 -3 c = -2.7 x 10 -3

30. 4x 2 + 2.7x10 -3 x – 2.7 x 10 -3 = 0 a = 4 b = 2.7 x 10 -3 c = -2.7 x 10 -3

32. We found x = 0.0256 M F 2 2F Initial 1.0 M 0 M Change - x + 2x At equilibrium 1.0 – x = 0.974 M 2x = 0.051 M

33. Don’t memorize....

34. Don’t memorize UNDERSTAND

35. Problem type 2c Maybe next time.......