The value of ΔG°' for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is 1.67 kJ/mol. If the concentration of glucose-6-phosphate at equilibrium is 2.35 mM, what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0 °C. The constant R = 8.3145 J/(mol·K)

Question

aaronq · Answer

First find eq. constant, K

$\sf \large \Delta G=-RTlnK\rightarrow K=e^{-\frac{\Delta G}{RT}}$

The equation is a rearragement: $G6P \rightleftharpoons F6P$ 

$\sf \large K=\dfrac{[F6P]}{[G6P]}\rightarrow [G6P]=K*[F6P]$


$\sf \large [G6P]= e^{-\frac{\Delta G}{RT}}*[F6P]$

aaronq · Answer

oh is solved for 6GP not F6P as i should have. just rearrange the equation