Question

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Answer 1

Free energy change of a chemical process is given by , $\Delta$G^o = -2.303RTlogKc

Given that free enrgy change of conversion from   G 1p to G 6p = - 7.1 kj/mol.

Given that free enrgy change of conversion from   G 6p to G 1p = - 2 kj/mol.

G 1p $\Leftrightarrow$ G 6p

so free energy change for the reversible process above = -7.1 - (-2.0) kj/mo

= -5.1 kj/mol

For the above process , Kc = [G 6p] / [G1p]

Temperature = 273 + 37^oC

= 310k

But  $\Delta$G^o = -2.303RTlogKc

for the above process ,given that ,  $\Delta$G^o = - 5.1 kj/mol. = -5100 j/mol

so logKc = 5100 j/ 2.303RT

= 5100 j / 2.303 x 8.3145 j/k.mol x 310 k

= 0.85917

Kc = 10^0.85917

=7.23

or concentration ratio ASKED ,   [G 6p] / [G1p] = 7.2