# Chemistry - Why does the reaction quotient use the products (multiplications) of reactants and products, rather than their respective sums?

## Solution 1:

The derivation cited can help you understand it mathematically. But the answer also turns out to be satisfactory intuitively. Let me try to explain how.

The term $RT\ln Q_c$ is derived form the entropy part or $\Delta G=\Delta H-T\Delta S$, and in a slightly abstract way, represents entropy of mixing of the products and the reactants. This is why reactions which are spontaneous initially, as concentrations change, reach to the equilibrium state before complete conversion of reactants into products. If there was no entropy of mixing, a reaction which had negative free energy change(under any set of conditions) would continue to have a negative change (hence, spontaneous) until all of the reactants is converted to products, which will be the state of maximum entropy. But since there is entropy of mixing, the state of maximum entropy is reached before all the reactants are converted into products (because somewhere in between the entropy of mixing due to having a mixture of reactants and products outweighs the negative $\Delta G$ for the reactants to convert into the products) , and the concentrations at which this maximum entropy state can be reached is precisely what the equilibrium constant tells us.

To calculate the entropy, we need to realize the fact that certain components of the reaction may have a greater contribution in deciding the equilibrium constant than others because, at the maximum entropy state, the components are present according to their stoichiometric relations and hence one might be in a well-defined excess over other. The relative contributions of each species to the entropy is logarithmic, that is, the increase in entropy is a logarithmic function of concentration (related slightly in a abstract manner to $S=k_B\ln \Omega$). Therefore, the contribution to the free energy and consequently in deciding the $K_c$ is logarithmic in the concentrations. The stoichiometric coefficientts are multiplied to the logarithmic contributions of each(which appear as the power to which the concentrations are raised when brought inside the logarithm) and the relative contributions are simply added, but since the addition of logarithms means multiplications of its arguments, we get the familiar $RT\ln Q_c$ term which gives the mentioned definitions of reaction quotient and the equilibrium constant.

I have used reaction quotient and equilibrium constant interchangeably but the RQ represents an unfulfilled equilibrium constant and hence represents the same drive towards maximum entropy as $K_c$ (which is the state of maximum entropy itself).

## Solution 2:

Is there some relation to kinetics that might give me a deeper insight into the meaning of this quotient?

Yes! The way I handle equilibrium reactions treats them as a "competition" between a particular reaction and its exact reverse:

$$\ce{\alpha A + \beta B ->[k_1] \gamma C + \delta D} \tag{1}$$ $$\ce{\gamma C + \delta D ->[k_{-1}] \alpha A + \beta B} \tag{2}$$

Written as an equilibrium reaction, of course, it looks like this:

$$\ce{\alpha A + \beta B <=>[K_1] \gamma C + \delta D}\tag{3}$$

So, the question then becomes how to relate the two kinetic reactions (1) and (2) to the equilibrium reaction (3). Assuming both Reactions (1) and (2) are elementary reactions as they're written, the two rate expressions become:

$$r_1 = k_1 \left[A\right]^\alpha \left[B\right]^\beta \tag{4}$$ $$r_{-1} = k_{-1}\left[C\right]^\gamma \left[D\right]^\delta \tag{5}$$

By definition, at equilibrium the rates of the forward and reverse reactions are equal. So, combining (4) and (5):

$$k_1 \left[A\right]^\alpha \left[B\right]^\beta = r_1 = r_{-1} = k_{-1}\left[C\right]^\gamma \left[D\right]^\delta \tag{6}$$

Now, dividing (6) through by $k_{-1}$ and by $\left[A\right]^\alpha\left[B\right]^\beta$ gives (almost) the familiar equilibrium constant expression:

$${k_1 \over k_{-1}} = {\left[C\right]^\gamma\left[D\right]^\delta \over \left[A\right]^\alpha\left[B\right]^\beta } \tag{7}$$

If we define $K_1 \equiv {k_1 \over k_{-1}}$ (as we should!), then (7) is exactly the standard form of an equilibrium constant expression.

It is (I think?) straightforward to see from here why the reaction quotient $Q_C$ is also defined this way, and why if $Q_1 < K_1$ the reaction will proceed forward toward equilibrium, while if $Q_1 > K_1$ the reaction will proceed in the reverse toward equilibrium.