How do cellular conditions change the Gibbs free energy of a reaction?
To understand that, you must have a basic understanding of Gibbs-Helmholtz equation. it states-
$\Delta G'=\Delta G^{'o} +RT\,ln\,(Q) $
Here, $\Delta G'$ = biochemical free energy change &amp; $\Delta G^{'o}$ = standard biochemical free energy change.
Suppose you have a reaction,
$x+FADH_2\rightleftharpoons xH_2+FAD $
So, $Q=\dfrac{[xH_2][FAD]}{[x][FADH_2]}$
Now let's assume that $\Delta G^{'o}&gt;0$.
Hence the spontaneity of the aforesaid reaction (in forward direction) will be dependent on the $\Delta G^{'o}$, which is a function of reaction quotient (Q) as well as temperature (T).
Now, if the body is in lower energy state, i.e. fasting; $\dfrac{[FAD]}{[FADH_2]}$ ratio will be high. So, in order to make $\Delta G'&lt;0$, the following condition must be satisfied $-$
$RT\,ln\,(Q)+\Delta G^{'o}&lt;0$,
or, $RT\,ln\,(Q)&lt; -\Delta G^{'o}$
or, $RT\,ln \dfrac{[xH_2][FAD]}{[x][FADH_2]}\,&lt; -\Delta G^{'o}$
With this reasoning, be it any metabolic pathway; we can predict its spontaneity.