How is a reaction with more products than reactants spontaneous? If I have a reaction that has $\Delta G^\circ = 0,$ where there are more products than reactants (for example, $\ce{A &lt;=&gt; 2B + 2C}),$ then if I change the concentration of all the reactants and products (i.e. to 20 mM), then shouldn't the reaction favor the reactants, since you have more total product than reactants (in my example, you would have 40 mM of both $\ce{B}$ and $\ce{C}$ and then just 20 mM of $\ce{A}),$ and therefore produce a positive $\Delta G$ value? I get that if you use the equation $\Delta G = \Delta G^\circ + RT\ln Q,$ you would get a negative $\Delta G$ for my example, showing that it is a spontaneous reaction. 

However, can anyone explain this problem from a conceptual perspective, since it would seem that if you have more products than reactants with $Q,$ that would drive the reaction to favor the reactants, according to Le Chatelier's principle?
Note that whether the forward or reverse reaction is favorable depends entirely on the relative magnitudes of Q and K (the reaction quotient versus the equilibrium constant):

$$\begin{align} Q&amp;&gt;K \rightarrow \text{reactants favored}  \\ K&amp;&gt;Q \rightarrow \text{products favored}  \end{align}$$

Next, consider that when you declare a value for $\Delta G^\circ$ you can translate that into a value for K, using the following equation:

$$\Delta G^\circ = -RT\log K$$

For instance, if $\Delta G^\circ = 0$ then $K=1$.

For the example here, this means that

$$\begin{align} Q&amp;&gt;1 \rightarrow \text{reactants favored}  \\ 1&amp;&gt;Q \rightarrow \text{products favored}  \end{align}$$

Next, consider the standard state of all substances. For solutions this is typically $\pu{c^\circ = 1 M}$. Standard states are important because the reaction quotients and equilibrium constants, when described in terms of concentrations, contain ratios of the concentrations in solution to the standard concentrations, for instance

$$Q= \frac{\prod_{\text{products i}} (c_i/c^\circ)^{\nu_i}}{\prod_{\text{reactants i}} (c_i/c^\circ)^{\nu_i}}$$

If all substances are present at the standard concentration, $Q=1$. In the present example $K=1$ so the equilibrium state consists of all substances being at standard concentrations ($c_i=\pu{1 M}$).

Finally, consider the following way to relate Q, K and $\Delta G$: 

$$\Delta G = RT\log \left( \frac{Q}{K}\right)$$

If $K=1$ as in the present example then

$$\Delta G = RT\log Q $$ 

This just reinforces what was stated earlier for the special example presented here: when $Q&lt;1$ the free energy difference is negative and products will be favored, and vice-versa when $Q&gt;1$. 

Inserting the definition of the reaction quotient

$$\begin{align} \Delta G &amp;= RT\log \left( \frac{\prod_{\text{products i}} (c_i/c^\circ)^{\nu_i}}{\prod_{\text{reactants i}} (c_i/c^\circ)^{\nu_i}}\right) \\ &amp;= RT\log \left( \frac{c_B^2c_C^2}{c_Ac^{\circ\,3}}\right) \end{align}$$ 

If we assume all substances are present at the same concentration c, then

$$  \Delta G= 3RT\log \left( \frac{c}{c^{\circ}}\right) $$

Clearly if $c&lt;\pu{1 M}$ then $\Delta G&lt;0$ and products are favored.