# How is it that the equilibrium constant does not depend on the mechanism?

For a reaction of the form

$$\ce{aA + bB <=> cC + dD}$$

the equilibrium constant is

$$K_c=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$$

regardless of the mechanism of the reaction. Why is this the case?

I've seen derivations that use a reaction in one elementary step to demonstrate this, but this obviously doesn't work in general. The sample reaction used was $$\ce{N2O4 <=> 2NO2}$$.

## 3 Answers

You should indeed be able to write every equilibrium reaction in this simple form. I guess that the reason for this might lie in the principle of microscopic reversibility (shameless self-plug: see this answer of mine). But for non-elementary reactions, i.e. those whose reaction mechanism consists of more than one step, you get a more complicated equilibrium constant that is the product of the equilibrium constants of all the intermediate reaction steps (if the reaction mechanism has no branches). An example: Consider the reaction

\begin{equation} \ce{H2(g) + I2(g) <=>> 2HI(g)} \ . \end{equation}

Its mechanism consists of three reaction steps:

1. $\ce{ I2 <<=> 2I } \qquad \qquad \qquad \qquad K_{1} = \frac{[\ce{I}]^{2}}{[\ce{I2}]}$
2. $\ce{I + H2 <<=> H2I} \ \ \quad \qquad \qquad K_{2} = \frac{[\ce{H2I}]}{[\ce{I}] [\ce{H2}]}$

3. $\ce{H2I + I <=>> 2HI} \quad \qquad \qquad K_{3} = \frac{[\ce{HI}]^{2}}{[\ce{I}] [\ce{H2I}]}$

Now, take a look at the equilibrium constant of the overall reaction

\begin{equation} K_{\mathrm{tot}} = \frac{[\ce{HI}]^{2}}{[\ce{I2}] [\ce{H2}]} \ . \end{equation}

Then make the following substitutions: From the third reaction step's equilibrium equation you get $[\ce{HI}]^{2} = K_{3} [\ce{I}] [\ce{H2I}]$. Substituting this into the equation for $K_{\mathrm{tot}}$

\begin{equation} K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] [\ce{H2I}]}{[\ce{I2}] [\ce{H2}]} \ . \end{equation}

From reaction step 2 you get $[\ce{H2I}] = K_{2} [\ce{I}] [\ce{H2}]$ which then gives

\begin{equation} K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} [\ce{I}]^{2}}{[\ce{I2}]} \ . \end{equation}

And finally, from reaction step 1 you get $[\ce{I}]^{2} = K_{1} [\ce{I2}]$. Substituting this into the equation for $K_{\mathrm{tot}}$ you see that the overall equilibrium constant reduces to the product of the equilibrium constants of the reaction steps

\begin{equation} K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} K_{1} [\ce{I2}]}{[\ce{I2}]} = K_{1} K_{2} K_{3} \ . \end{equation}

If you have a reaction that doesn't proceed along a chain of reaction steps but branches out the situation will become more complicated but can be treated in an analogous way as above.

• Because intermediates doesn't appear in the overall reaction does this mean that their concentrations are zero at equilibirum? Otherwise can we detect them at equilibrium? Is this experimentaly confirmed? – ado sar May 12 '20 at 19:42

Philipp's answer is a nice application of Hess's Law to equilibrium. Let's look at why this approach works.

The position of equilibrium (as given by the equilibrium constant $K$) is related to the standard Gibbs free energy change that occurs during the reaction:

$$\Delta_rG^\circ=-RT\ln{K}$$

The Gibbs free energy $G$ is a state function - a property that depends only on the current state of the system, not the pathway by which the system reached that state. The value of $\Delta_rG^\circ$ depends only on the values of $G^\circ$ (or $\Delta_f G^\circ$ if you prefer) of the reactants and the products, which depend only on the identity and amounts of the substances. $\Delta_fG^\circ$ is independent of the way in which any of the substances were prepared.

Likewise, $K$ is a state function. $K$ describes the compositions of the mixture which qualify as meeting the requirements for equilibrium. The value of $K$ is independent of the amounts of any specific substance or the pathway (mechanism) by which equilibrium is reached (see Philipp's answer). The individual equilibrium concentrations, however, in the system are not state functions. $[\ce{A}]_{eq}$ depends on the initial concentrations of all species in the mixture.

The rate of reaction is dependent on the mechanism. Different mechanisms may have different orders, different activation energies, etc. Reaction rate (and consequently the rate constant $k$) are not necessarily state functions.

This answer is the same as the one by Philipp, but instead of using a specific example, it makes the argument for any multi-step reaction where each step is an elementary reaction.

The overall reaction is a sum of elementary steps

The overall reaction happens because elementary steps happen consecutively. Along these multiple steps, some species are reactants or products (i.e. appear in the chemical equation of the overall reaction) and others are intermediates (i.e. made by one elementary reaction and consumed by another).

The overall chemical equation is equal to the sum of the elementary steps, with intermediates appearing both on the reactant and the product side (those can be removed to arrive at the net chemical equation of the overall reaction).

At equilibrium, all elementary steps are at equilibrium

The only way the reactant and product concentrations can be constant (i.e. at equilibrium) is when all the intermediates are at equilibrium as well. You arrive at the equilibrium constant of the overall reaction by multiplying the equilibrium constants of the elementary reactions.

The rate laws of the elementary steps have the stoichiometric coefficients as exponents

This is the definition of an elementary step. The rate law for the forward reaction is the product of the reactant concentrations raised to the power of their coefficients. The rate law for the reverse reaction is the product of the product concentrations raised to the power of their coefficients. (Many of these reactants and products are intermediates with respect to the overall reaction.) At equilibrium, the forward and reverse rates are equal, so product concentrations divided by reactant concentrations (raised to respective coefficients) for the elementary steps will equal the quotient of reverse and forward rate constants, and we will call this equilibrium constant.

The equilibrium constant expression of the overall reaction is the product of the equilibrium constant expression of the elementary steps

Because the equilibrium constant of the overall reaction is equal to the product of the equilibrium constants of all the elementary steps, we can derive the equilibrium constant expression of the overall reaction by writing the product of the individual equilibrium constant expressions.

Just like intermediates cancel out of the net chemical equation because they appear both as reactant and as product, concentrations of intermediates will cancel out of this derived equilibrium expression (they appear both in the denominator - as reactant of one step - and in the numerator - as product of a different step). What remains after canceling are concentrations of reactants and products of the overall reaction, raised to their respective coefficients in the net reaction.

So whatever the coefficients are in the chemical equation of the net overall reaction will appear as exponents in the equilibrium constant expression for that same net overall reaction.

For an example, look at the answer by Philipp.

• "At equilibrium, all elementary steps are at equilibrium". Does this mean that when we have a catalyst some quantitity of it is consumed as the steps that catalyst participate are all in equilibrium? – ado sar Feb 3 at 14:47
• @adosar The catalyst is not so much consumed as it is "busy", e.g. bound to one of the intermediates. – Karsten Theis Feb 4 at 1:58