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

N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial M Change At equilibrium M

N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial M Change – M = M At equilibrium M

N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial M Change /2 M – M = M At equilibrium M

N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial M Change /2 M – M = M At equilibrium M =.197 M M

N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial M Change /2 M + 3(.394/2) M – M = M At equilibrium M =.197 M M

N 2 + 3H 2 2NH 3 N 2 H 2 NH 3 Initial M Change /2 M + 3(.394/2) M – M = M At equilibrium M =.197 M M M

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

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 M Change +x +3x -2x At equilibrium

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

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

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

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

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

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

N2N2 O2O2 2NO At equilibrium x x x

x = (1.88 x )x = 2.02 x x = 4.95 x M = [N 2 ] (also = [O 2 ]) Note that because K was small, most of the NO became N 2 and O 2 Final [NO] = – 2(4.95 x ) =1.00 x 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 F 2 2F at 1000 o C, K = 2.7 x 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 ? ?

For the reaction F 2 2F at 1000 o C, K = 2.7 x 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

For the reaction F 2 2F at 1000 o C, K = 2.7 x 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

F 2 2F At equilibrium 1.0 – x 2x

4x 2 = 2.7 x (1.0 – x) = 4x 2 = 2.7 x – 2.7 x x

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

4x x10 -3 x – 2.7 x = 0 a = 4 b = 2.7 x c = -2.7 x 10 -3

4x x10 -3 x – 2.7 x = 0 a = 4 b = 2.7 x c = -2.7 x 10 -3

We found x = M F 2 2F Initial 1.0 M 0 M Change - x + 2x At equilibrium 1.0 – x = M 2x = M

Don’t memorize....

Don’t memorize UNDERSTAND

Problem type 2c Maybe next time