Answer:
This reaction is favorable, and is likely regulated.
Explanation:
The equation to calculate delta G (dG) of a reaction is dG = dGo' + RTln [initial P]/[initial R]. You could use just dG = RTln [initial P]/[initial R] if (and that's a big IF) dGo' is zero, meaning that the reaction is at equilibrium when we have equal amounts of [P] and [R] (which is rarely the case). What really matters is the ratio of Q ([initial P]/[initial R]) to Keq ([P at equilibrium]/[R at equilibrium]), meaning how far off are we from equilibrium.
If Q=Keq, we are already at equilibrium (EQ).
If Q<Keq, we are not yet at EQ, having relatively more [R], or less [P] than under EQ conditions. This means the reaction will move forward to produce more P until EQ is achieved (dG is therefore NEGATIVE).
If Q>Keq, we are also off EQ, but we have relatively more [P], or less [R] than under EQ conditions. This means the reaction will move backwards to produce more R until EQ conditions are achieved (dG is therefore POSITIVE).
Try to understand these equations below (they say what I tried to describe in words)
dGo' = -RTlnKeq (under "standard conditions", i.e. we try to figure out how a reaction "behaves" if we start out with the same molar concentrations of R and P)
dG = dGo' + RTlnQ Q=[initial P]/[initial R] or
dG = -RTlnKeq + RTlnQ or
dG = RTlnQ - RTlnKeq or
dG = RTln Q/Keq
