# Question & Answer: The Delta G^0 for formation of a phosphodiester bond (shown below) is +25 kJ/mol. (Base)_n…..

The Delta G^0 for formation of a phosphodiester bond (shown below) is +25 kJ/mol. (Base)_n + Base rightarrow (Base)_n + 1 + H_2O Delta G^0 = +25 kJ/mol a) What is the equilibrium constant for this reaction at 37 degree C? b) How might this reaction be made favorable? (Show the math for full credit)

Given

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$\Delta$G of the reaction = +25KJ/mole = +25 * 103 J/mole

T = 370C + 273k = 310 K

we know

$\Delta$G = -RTlnKeq

+25 * 103 J/mole = – 8.314 J/K.mole * 310K * lnKeq

on solving equilibrium constant Keq = 6.2 * 10-5

b.

here equilibrium constant is less than 1. So it is reactant favourable. i.e, backward reaction occurs mostly.