Question

Ligands can be classified structurally, discuss this statement .Given that the relaxation time for the equilibrium H2O H+ + OH- is 37 × 10-6 at 298.15K. Calculate the rate coefficient for the forward and backward reaction.(Hint : the H2O is present in excess at 55.5 moldm-3).

Answer 1

1) Based on structure the classification of ligands is that from how many sides they attack or form bond with metal leading to different structures of metal ligand complex

a) Monodentate: single site (CN-, H2O NH3 etc)

b) Bidentate : two sites (ethylene diamine)

c) multidentate : more than two sites

2)

d[H2O] / dt = -K1[H2O] + K2[H+][HO-]

Let after relaxation ,

[H2O] = [H2O]eq + x

[H+] = [H+]eq -x

[HO-] = {HO-]eq -x

On differentiatin x

We will obtain

1/ relaxation time = (K1 + K2 { [H+]eq + {HO-]eq }

at equilibrium

K1[H2O]eq = K2 { [H+]eq + {HO-]eq

Therefore K1/K2 = [H+]eq[HO-]eq / [H2O]eq = Kw / 55.5 = K = 0.018 X 10^-14

Given , [H2O] =55.5

Also [H+] [OH-] = Kw

1/ relaxation time = K2 ( K + 2 X Kw^1/2)

On calculating

1/ 37 × 10-6 = K2 (0.018 X 10^-14 + 2X 10^-14) = K2 (2.018 X 10^-14 )

0.0133 X 10^20 = K2

K1 = (0.018 X 10^-14 X 0.0133 X 10^20 ) = 0.0002394 X 10^6 = 2.394 X 10^2