# 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, $$\ce{A <=> 2B + 2C}),$$ 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?

• Hint: Write the expression for Q for this reaction, then substitute in 20mM for the concentrations of A, B, and C. Then solve for $\Delta G$. You'll see that it's positive. Apr 21 '20 at 7:43
• You are using Le Chatelier's principle the wrong way. It doesnt state that if product is more than the reaction moves to reactant side. It states that if the reaction has attained equilibrium then change in concentrations or conditions will change the reaction direction accordingly. Apr 21 '20 at 9:08

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&>K \rightarrow \text{reactants favored} \\ K&>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&>1 \rightarrow \text{reactants favored} \\ 1&>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 $$Q<1$$ the free energy difference is negative and products will be favored, and vice-versa when $$Q>1$$.

Inserting the definition of the reaction quotient

\begin{align} \Delta G &= 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) \\ &= 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 $$c<\pu{1 M}$$ then $$\Delta G<0$$ and products are favored.

One of my professors in a graduate Reaction Engineering course gave us a conceptual justification for a similar question.

Consider two reactions,

1. $$\ce{A + B <=> C + D}$$

2. $$\ce{A + B <=> 2C + 2D}$$

For reaction 1, it requires 2 mols in the case of both the forward (1 mol A, 1 mol B) and reverse cases (1 mol C, 1 mol D) to drive the reaction forward. Conversely in reaction 2, while it requires 2 mols for the forward reaction, it requires 4 mols for the reverse.

When one thinks of a reaction, the fundamental mechanism will involve the assumption of some level of simultaneous collision or interaction (for example in the liquid of gaseous phases). From a probability perspective, it is significantly more likely for 2 molecules to collide perfectly to initiate a reaction, than it is for 4 molecules to do the same.

This is usually an assumption that one makes when modeling a reaction from first principles. For example, for the same reaction $$\ce{A + B <=> 2C + 2D}$$, we could reasonably assume that this reaction system would likely favor the forward over the reverse reaction, i.e: $$\ce{A + B -> 2C + 2D}$$-or that $$K_{forward}$$ >> $$K_{reverse}$$, just based on the intuition provided in the previous paragraph.

This is not a one-size fits all explanation; chemical reaction systems are very complicated, but I hope this conceptual perspective helps.