The value of $Δ$ $G^{0}$ for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of $K_{c}$ at 298 K.

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Solution

$Δ G^{0} = - R T l n K$
$13.8 \times \frac{10^{3} J}{m o l} = - \frac{8.314 (J)}{(m o l K)} (298) (K) l n K_{c}$
$l n K_{c} = - \frac{13.8 \times 10^{3}}{8.314 \times 298}$
$K_{c} = 10^{- 5.57} = 10^{- 6 + 0.43}$
$K_{c} = 10^{- 6} \times 10^{0.43} = 2.69 \times 10^{- 6}$

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