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Problem type #1

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

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Problem type #1

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

  2. Suppose a flask was made to be 0.500 M in NH3 initially, and when equilibrium was reached, the [NH3] had dropped to 0.106 M. Find the value of K for:N2 + 3H22NH3

  3. N2 + 3H22NH3

  4. N2 + 3H22NH3

  5. N2 + 3H22NH3 • Let 2x be the amount of NH3 that reacts • 2x = 0.500 – 0.106 = 0.394

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

  7. For the reaction N2 + O2 2NO at 1700oC, K = 3.52x10-4. If 0.0100 mole NO is placed in a 1 – L flask at 1700o, what will [N2] be at equilibrium?

  8. 0.0100 - 2x = (1.88 x 10-2)x • 0.0100 = 2.02 x • x = 4.95 x 10-3 M = [N2] (also = [O2]) Note that because K was small, most of the NO became N2 and O2 Final [NO] = 0.0100 – 2(4.95 x 10-3) =1.00 x 10-4 M

  9. 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

  10. For the reaction F2 2F at 1000oC, K = 2.7 x10-3. If 1.0 mole F2 is placed in a 1 – L flask at 1000o, what will [F] be at equilibrium?

  11. 4x2 = 2.7 x 10-3(1.0 – x) = • 4x2 = 2.7 x 10-3 – 2.7 x 10-3 x • This is a quadratic equation • Rearrange to the form ax2 + bx + c = 0 4x2 + 2.7x10-3x – 2.7 x 10-3 = 0 • a = 4 b = 2.7 x 10-3 c = -2.7 x 10-3

  12. 4x2 + 2.7x10-3x – 2.7 x 10-3 = 0 • a = 4 b = 2.7 x 10-3 c = -2.7 x 10-3

  13. We found x = 0.0256 M

  14. Don’t memorize • UNDERSTAND

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