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K = [ NH 3 ] 2 = (0.031) 2 = 38000

The following concentrations at equilibrium were determined for a mixture of nitrogen gas, hydrogen gas, and ammonia at 125° C [ NH 3 ] = 0.031 M [ N 2 ] = 0.85 M [ H 2 ] = 0.0031 M Calculate K for the following reaction: N 2 + 3 H 2  2 NH 3.

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K = [ NH 3 ] 2 = (0.031) 2 = 38000

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  1. The following concentrations at equilibrium were determined for a mixture of nitrogen gas, hydrogen gas, and ammonia at 125° C [ NH3 ] = 0.031 M [ N2 ] = 0.85 M [ H2 ] = 0.0031 M Calculate K for the following reaction: N2 + 3 H2 2 NH3 K = [ NH3]2= (0.031)2 = 38000 [N2] [H2] 3 (0.85) (0.0031) 3

  2. The following concentrations at equilibrium were determined for a mixture of nitrogen gas, hydrogen gas, and ammonia at 125° C [ NH3 ] = 0.031 M [ N2 ] = 0.85 M [ H2 ] = 0.0031 M Calculate K for the following reaction: ½ N2 + 3/2 H2 NH3 K = [ NH3]= (0.031) = 200 [N2] 1/2 [H2] 3/2 (0.85) ½ (0.0031) 3/2

  3. For: N2 + 3 H2 2 NH3 K = [ NH3]2= (0.031)2 = 38000 [N2] [H2] 3 (0.85) (0.0031) 3 For: 1/2 N2 + 3/2 H2 NH3 K = [ NH3]= (0.031) = 200 [N2] 1/2 [H2] 3/2 (0.85) ½ (0.0031) 3/2 If the balanced chemical equation is multiplied by a constant n, the resulting K equals Kn 38000 ½ = 194.94  200

  4. The following concentrations at equilibrium were determined for a mixture of nitrogen gas, hydrogen gas, and ammonia at 125° C [ NH3 ] = 0.031 M [ N2 ] = 0.85 M [ H2 ] = 0.0031 M Calculate K for the following reaction: 2 NH3  N2 + 3 H2 K = [N2] [H2]3= (0.85) (0.0031)3 = 0.000026 [ NH3] 2 (0.031) 2

  5. For: N2 + 3 H2 2 NH3 K = [ NH3]2= (0.031)2 = 38000 [N2] [H2] (0.85) (0.0031) 3 For: 2 NH3  N2 + 3 H2 K = [N2] [H2]3= (0.85) (0.0031)3 = 0.000026 [ NH3] 2 (0.031) 2 If the balanced chemical equation is reversed , the resulting K equals 1/K 1 / 38000 = 0.000026

  6. If two balanced chemical reactions are combined into a single equation, the resulting K is the product of the two original K values 2 NOBr  2 NO + Br2K= 0.42 Br2 + Cl2  2 BrCl K= 7.2 2 NOBr + Cl2  2 NO + 2 BrCl K= (0.42) (7.2) = 3.0

  7. Manipulations of the Equilibrium Equation 1. If the balanced chemical equation is multiplied by a constant n, the resulting K equals K n 2. The K for the reverse reaction is equal to 1/ K 3. If two balanced chemical reactions are combined into a single equation, the resulting K is the product of the two original K values

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