You have a reaction in which you have 1.0 M of substrate and 0.01 M enzyme .
Let's suppose that most of the enzyme forms a complex with substrate so that at equilibrium 90% of the enzyme molecules have bound substrate and 10% enzyme molecule is free. This gives
(90% of 0.01M) is
0.9 X 0.01 = .009 M (enz-sub complex)
If, on the other hand very little of the enzyme is bound to substrate so that 10% of the enzyme was bound to substrate and 90% was free the equation would be
10% of 0.01M is
0.1 X 0.01 = .001 (enz-sub complex)
If the Keq is a small number then the enzyme has a high affinity for substrate.
If the Keq is a large number then the enzyme has a relatively low affinity for substrate.
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