Consider the two complexation equilibria in aqueous solution, between the cobalt (II) ion `Co^(2+)` (aq) and ethylenediamine (en) on the one hand and ammonia `NH_(3)` on the other.
`[CO(H_(2)O)_(6)]^(2+)+6NH_(3)hArr[Co(NH_(3))_(6)]^(2+)+6H_(2)O` ...(1)
`[Co(H_(2)O_(6))]^(2+)+3enhArr[Co(en)_(3)]^(2+)+6H_(2)O` ..(2)
Electronicaly, the ammonia and en ligands are very similar, since both bond through N and since the liwis base strengths of their nitrogen atoms are similar. This means that `DeltaH^(@)` must be very similar for the two reactions, since six Co-N bonds are formed in each case. Interestingly however, the equilibrium constant is 100,000 times larger for the second reaction than it is for the first. This is the so called chelate effect: "the enhanced affinity of chelating ligands for a metal ion compared to similar non-chelating (monodentate) ligands for the same metal". The chelate effect is entropy-driven.
Q. qWhat may be main reason for reaction (2) having relatively such a large equilibrium constant?

Options

A. Relatively small decrease in entropy in reaction 2
B. Relatively large decrease in entropy in reaction 2
C. Relatively small increase in entropy in reaction 2
D. relatively large increase in entropy in reaction 2

Answer

D

Explanation

The big difference between these two reactions is that of entropy change. The first reaction has a `DeltaS^(@)` value close to zero, because there is the same number of molecules on both sides of te equation. The second one has a positive `DeltaS^(@)` because four molecules come together but seven molecules are produced are produced. This helps to make reaction 2 more favourable.