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

As you may know, the reaction quotient $Q_c$ is defined by the equation $$Q_c = \frac{[C]^\gamma [D]^\delta}{[A]^\alpha [B]^\beta}$$ for the chemical reaction $$\alpha A + \beta B \rightarrow \gamma C + \delta D.$$

This is something I've been struggling to understand at an intuitive level. Why is $Q_c$ the ratio of the product of the reaction's product concentrations and the product of the reaction's reactant concentrations (sorry for the confusing terminology; I can't think of a more precise way to phrase that), rather than the ratio of the sums of concentrations, like this: $$Q_c' = \frac{\gamma[C]+\delta[D]}{\alpha[A]+\beta[B]}$$

This second, incorrect equation makes more sense to me because concentrations of solutes add up. Is the ratio defined as such so that it is directly proportional to the concentrations of each product and inversely proportional to the concentrations of each reactant? Is there some relation to kinetics that might give me a deeper insight into the meaning of this quotient?

• See here for a mathematical derivation of the relationship between the change in free energy in a reaction and the concentrations of the species (or partial pressures, in that case). It turns out that you add and subtract logarithms of the concentrations, and using a property of logarithms it is possible to condense all concentrations into a ratio that we call the reaction quotient. Nov 13, 2013 at 21:55
• @NicolauSakerNeto - if you could summarize/explain that derivation it would make a good answer. Nov 13, 2013 at 22:52
• @BenNorris Physical chemistry isn't my forte, so I would rather someone else with a better background describe it, if possible. Nov 14, 2013 at 2:09
• Related question: Why is the equilibrium constant the way it is? May 3, 2016 at 13:13

## 2 Answers

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).

• Thanks for the answer, this is significantly more thorough than anything I could have managed. Nov 14, 2013 at 10:18
• @NicolauSakerNeto I'm pretty sure equilibrium is not actually a maximum-entropy state, but is in fact the state where the entropic and enthalpic contributions to the free energy change are in balance. Apr 24, 2016 at 23:32
• @Brian Really? Then what would happen in a simple closed universe with only one equilibrating reaction? Would it reach equilibrium or would there be some other universal maximum entropy state? This question has a related answer, and it rests on the fact that the second law of thermodynamics is equivalent to saying "a system with a fixed internal energy self-adjusts its free parameters in such a way that the entropy is maximized," though no mathematical proof of this was given (outside the scope). Apr 25, 2016 at 2:02
• @NicolauSakerNeto And this is why, to this day, thermodynamics consistently mystifies me, since it seems impossible for equilibrium to be a maximum-entropy state unless $\Delta S_\mathrm{rxn} = 0$ for every reaction (which, itself, seems paradoxical). (Otherwise, in a closed system it seems like every equilibrium would proceed to completion in the direction of increasing $S$.) There must be some factor in there that I don't understand or am not considering, because many people agree that I'm wrong. :-) Apr 25, 2016 at 2:06
• @Brian After mulling over it for a while, I think we may have been talking about different things. We need to differentiate between the entropy of the system and the universe. I think you're right, the entropy of a chemical reaction at equilibrium need not have the highest entropy possible for the system, but the universe containing a reaction at equilibrium and its surroundings will be at the maximum of entropy. The enthalpy change of the system in Gibbs' free energy equation is a surrogate for the entropy change of the surroundings. Apr 25, 2016 at 2:44

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.

• This doesn't answer why they are exponents. Aug 3, 2020 at 6:22
• @AdityaRoychowdhury The exponents enter at the "Assuming both Reactions (1) and (2) are elementary reactions as they're written..." part, due to the use of the principle of mass action in defining the rates of elementary reactions. Aug 10, 2020 at 17:01