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 & $Delta G^{'o}$ = standard biochemical free energy change.

Suppose you have a reaction, $x+FADH_2rightleftharpoons xH_2+FAD $

So, $Q=dfrac{[xH_2][FAD]}{[x][FADH_2]}$

Now let's assume that $Delta G^{'o}>0$.

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 $Delta G'<0$, the following condition must be satisfied $-$

$RT,ln,(Q)+Delta G^{'o}<0$,

or, $RT,ln,(Q)< -Delta G^{'o}$

or, $RT,ln dfrac{[xH_2][FAD]}{[x][FADH_2]},< -Delta G^{'o}$

With this reasoning, be it any metabolic pathway; we can predict its spontaneity.