Chemistry - How can we define a non-spontaneous reaction?

Solution 1:

It sounds like your confusion arises from not making a distinction between $\Delta G$ and $\Delta G^\circ$ when describing a reaction as spontaneous or not.

The $\Delta G^\circ$ is the free energy change for the reaction at the defined "standard" conditions of 1 M solute concentrations and/or 1 bar gas partial pressures of both the reactants and products. When we make the general statement that a reaction is spontaneous or not, we are usually referring to whether this $\Delta G^\circ$ is greater or less than 0.

If $\Delta G^\circ < 0$, then the reaction is described as spontaneous, and it will proceed in the forward direction when starting from a mix of 1 M reactants and 1 M products until equilibrium is reached.

In contrast, $\Delta G$ is the free energy change for the reaction in whatever state is of interest. In your case, you have chosen a state with no product present at all. When we say that a reaction is spontaneous based on $\Delta G$, we mean whether it will go in the forward or reverse direction from this specific state of interest, not the standard state.

Obviously, if there is no product present, the reaction cannot go in the reverse direction (forming reactant from product), so all reactions will go in the forward direction if started in a state of $0$ product. That is what you have shown quantitatively using $Q=0$ so that $\Delta G <<0$.

So you are absolutely correct that every reaction in a state in which no product is present is spontaneous. That does not mean, however, that the reaction will necessarily happen anytime soon. Very often, the activation energy barrier is high enough that the reaction rate is extremely slow. Or it is possible that the equilibrium constant may favor reactants so much that even one molecule of product is sufficient to reach equilibrium and the reaction stops very quickly.

Solution 2:

In general it is necessary to consider any entropy changes in determining whether a system is at equilibrium or if a spontaneous change will occur.

As there must be an increase in entropy in actual processes then $dS_{system}+dS_{surr}=dS_{irrev} \ge 0$. By using the first law with the last expression and after several steps, we find that in an isothermal reaction (surrounding temperature $T$) in which no work is done other than $pV$ (gas expansion) work the condition becomes

$$dG + TdS_{irrev}=0$$

thus a process can take place spontaneously with entropy production only if the free energy decreases. If there is no change in free energy, no change can take place and the system is at equilibrium.

This is what you have determined in your question but with the implicit assumption that reaction will occur.

A non-spontaneous process is one that is already at equilibrium or that will require input of some work to make it change.

Solution 3:

I agree with Andrew that it depends on whether you define spontaneity based on $\Delta G^\circ <0$ or $\Delta G < 0$. Typically, freshman chemistry books use the former.

However, I've never liked equating spontaneity with the sign of $\Delta G^\circ$, prefering to instead use the sign of $\Delta G^\circ$ as an indicator of whether reactants or products are "favored".

The reason I don't like using the sign of $\Delta G^\circ$ (as opposed to that of $\Delta G$) as an indicator of spontaneity is it causes the very confusion you are experiencing. It also obscures the concept of chemical equilibrium.

I thus favor equating spontaneity with the sign of $\Delta G$ (which in turn depends not only on $K_{eq}$, but also upon where the reaction mixture is relative to it), which is the view you are articulating. For clarifying this latter view, I think the following diagram, which I've borrowed and modified from one of my earlier answers (Are all exothermic reactions spontaneous?), is helpful: enter image description here