If the equilibrium constant (K) = 100, this reaction favors the (reactants/products) and proceeds (very little/almost to completion). Equilibrium?

1 Answer
Martin M.
May 26, 2017

Let's take the general equilibrium reaction:
#wA+xBharryC+zD#

In which A, B, C and D are compounds and w,x,y,z are coefficients.
To determine the equilibrium constant, we write:

#K=([C]^yxx[D]^z)/([A]^wxx[B]^x)color(white)(aaaa)color(white)(color(blue)(1.)/color(blue)(2.)#

The equilibrium constant says something about the proceeding of the reaction: does it favour the forward reaction or the backwards reaction? The blue numbers represent the concentration of the two sides:

  1. "Combined concentration products"
  2. "Combined concentration reactants"

We have given that #K=100#, this means that side 1 must be a larger number than number 2. Only then will K be larger than 1. This means that the combined concentration of side 1 (products) is larger than the combined concentration of side 2.
Therefore the reaction is favoured to the right side (the product side). Since K=100 is pretty big, the reaction will probably go almost into completion.

By following this method, you can figure out that

K>>1 #color(white)(aa)# #color(blue)(|) #Product favored #color(white)(aaa)# #color(blue)(|)#Close to completion
K>1 #color(white)(aaa)# #color(blue)(|) #Product favored#color(white)(aaa)# #color(blue)(|)# Very little
K=1 #color(white)(aaa)# #color(blue)(|) #Equilibrium#color(white)(aaaaaaa)# #color(blue)(|)#-
K<1 #color(white)(aaa)# #color(blue)(|) #Reactant favored#color(white)(aaa)# #color(blue)(|)#Very little
K<<1 #color(white)(aa)# #color(blue)(|) #Reactant favored#color(white)(aaa)# #color(blue)(|)#Close to completion

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