Sol . As deltaG° = - 2.303 × R × T × log(Keq)
where deltaG° = standard free energy of reaction = -6.0 Kcal / mol
R = gas constant = 1.987 × 10-3 Kcal / K mol
T = Temperature = 25°C = 298 K
Keq = Equilibrium constant
So , log(Keq) = - deltaG° / (2.303 × R × T )
= - ( -6.0) / ( 2.303 × 1.987 × 10-3 × 298 )
= 4.3999
and , Keq = 104.3999 = 25113.081
Now , deltaG = deltaG° - 2.303 × R × T × log(Q)
where deltaG = free energy of reaction = - 14.3 Kcal / mol
Q = Reaction quotient
So , - 14.3 = - 6.0 - 2.303 × 1.987 × 10-3 × 298 × log(Q)
- 8.3 = - 2.303 × 1.987 × 10-3 × 298 × log(Q)
So , log(Q) = 6.0865
and , Q = 106.0865 = 1220393.821
Therefore , Q / Keq = 1220393.821 / 25113.081
= 48.596
The free energy (AG) for a reaction under a set of initial conditions/concentrations is -14.3 kcal/mol....
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