# Why are reversible reactions spontaneous?

If a reversible chemical reaction is spontaneous in one direction, shouldn't it be non spontaneous $$(ΔG > 0)$$ in the other? In other words for the reaction:

$$\ce{A + B ⇌ C + D}$$

if

$$\ce{A + B -> C + D}$$

has $$ΔG < 0$$ and thus proceeds spontaneously, doesn't this imply that

$$\ce{C + D -> A + B}$$

would have $$ΔG > 0$$ and thus proceed non spontaneously?

A negative $$\Delta G^0$$ implying that a reaction will proceed spontaneously or a positive $$\Delta G^0$$ implying that a reaction will not proceed spontaneously is just a "rule of thumb" to give you a feel for what to expect. But all a negative $$\Delta G^0$$ really means is that the equilibrium constant for the reaction is large (and formation of the products is highly favored), and a positive $$\Delta G^0$$ means that the equilibrium constant for the reaction is small (and formation of the products is not highly favored). However, if you start out with only reactants present initially, irrespective of the sign of $$\Delta G^0$$, the reaction will proceed spontaneously. However, if $$\Delta G^0$$ is negative, at final equilibrium, the extent of the reaction will be greater.

Remember not to confuse the standard free energy change (measured/computed for the standard states and concentrations) with the real free energy change.

In order to obtain the $$\Delta G$$ for the reaction, you should apply:

$$\Delta G = \Delta G ^{\varnothing} + RT \ln Q$$

Then, a reaction which is nonspontaneous in one context may be spontaneous in another context where the concentrations of participating species are different.

• Probably nitpicking, but shouldn't the upper symbol at $ΔG$ be circle $\circ$ or a plimsoll ⦵? An empty set symbol ∅ looks somewhat confusing to me. – andselisk Jan 8 at 5:37
• Probably? \varnothing is the only one I ever remember... – Zhe Jan 8 at 14:10
• That's what the compose key is for:) – andselisk Jan 8 at 15:38