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Equilibrium calculationsPowerPoint Presentation

Equilibrium calculations

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Equilibrium calculations

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Equilibrium calculations

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Problem type #1

- Given some information about the conditions before equilibrium and after equilibrium, find the equilibrium constant.

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

N2 + 3H22NH3

N2 + 3H22NH3

N2 + 3H22NH3

N2 + 3H22NH3

N2 + 3H22NH3

N2 + 3H22NH3

N2 + 3H22NH3

N2 + 3H22NH3

- Let 2x be the amount of NH3 that reacts
- Use stoichiometry of reaction!

N2 + 3H22NH3

- Let 2x be the amount of NH3 that reacts
- 2x = 0.500 – 0.106 = 0.394

N2 + 3H22NH3

- Let 2x be the amount of NH3 that reacts
- 2x = 0.500 – 0.106 = 0.394

Problem type #2a

- Given information about conditions before equilibrium and the equilibrium constant, find the concentrations when equilibrium is reached.

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?

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?

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?

- 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

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

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?

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?

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?

- 4x2 = 2.7 x 10-3(1.0 – x) =
- 4x2 = 2.7 x 10-3 – 2.7 x 10-3 x

- 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

4x2 + 2.7x10-3x – 2.7 x 10-3 = 0

- a = 4 b = 2.7 x 10-3 c = -2.7 x 10-3

4x2 + 2.7x10-3x – 2.7 x 10-3 = 0

- a = 4 b = 2.7 x 10-3 c = -2.7 x 10-3

Don’t memorize

- UNDERSTAND

Problem type 2c

- Maybe next time . . . . . . .