Question

The value o f $\Delta G °$ for the phosphorylation o f glucose in glycolysis is $13.8 kJ/mol. $ The value of $K_c $ at $298 K $ is

• A. $2.5 \times l0^3 $
• B. $3.81 \times l0^{-3} $
• C. $5.0 \times l0^{-4} $
• D. $8.5 \times l0^{-3} $

Answer 1

Answer: (B)

$\Delta G° = 13.8 kJ/mol = 13.8 \times 10^3 J/mol $
Also,$ \Delta G^\circ = - RT \,lnK_c $
Hence,
In $ K_c = -13.8 \times 10^3 J/mol (8.314 J mol^{-1} K^{-1} \times 298 K) $
In $K_c = - 5.569 $
$K_c = e^{- 5.569} $
$ K_c = 3.81 \times 10^{-3} $