# Relation between chemical kinetics and chemical equilibrium

In my chemistry book, the law of chemical equilibrium is derived from the law of mass action:

For a reversible chemical reaction $$\ce{aA +bB\rightleftharpoons cC + dD}$$ where $a$, $b$, $c$ and $d$ are stoichiometric coefficients in the balanced equation, we have, from the law of mass action, the rate of the forward reaction $$r_\mathrm{f} = k_1 \ce{[A]^a[B]^b}\label{1}\tag{1}$$ where $[\ce{A}]$ represents the concentration of $\ce{A}$ and so on. The rate of the backward reaction is $$r_\mathrm{b} = k_2\ce{[C]^c[D]^d} \label{2}\tag{2}$$

At equilibrium, $r_\mathrm{f} = r_\mathrm{b}$, so, we have $$\frac{k_1}{k_2} = \frac{\ce{[C]^c[D]^d}}{\ce{[A]^a[B]^b}}$$ Replacing $\displaystyle \frac{k_1}{k_2}$ by $K_\mathrm{c}$, we have

$$K_\mathrm{c} = \ce{\frac{[C]^c[D]^d}{[A]^a[B]^b}}\tag{3}\label{3}$$ which is the law of chemical equlibrium. $K_\mathrm{c}$ is supposed to be constant, for any concentration of the reactants, at a particular temperature.

However, in chemical kinetics, they tell us that the rate equation, derived from the law of mass action is not always correct. That is, the exponents of the concentration terms in the rate equations may not be equal to their stoichiometric coefficients. If this is true then, the above two rate equations ($\eqref{1}$ and $\eqref{2}$) may not be correct for a particular reversible reaction.

Then, how can we apply the law of chemical equilibrium on every reaction, without experimentally verifying the rate equations? If the rate equations derived from the law of mass action are wrong, then the value of $K_\mathrm{c}$ obtained by the final equation will not be constant.

• You are essentially correct in your reasoning. – porphyrin May 28 '17 at 15:52
• Just because the kinetics are not mass action does not mean that the thermodynamic equilibrium constant is not given by the usual relationship. – Chet Miller May 29 '17 at 1:52
• A cursory reading of en.wikipedia.org/wiki/Chemical_equilibrium suggests that the section cited by the asker is the historical derivation (which is also easier to understand). There is a derivation based on the Gibbs / free enthalpy. – TAR86 Oct 13 '17 at 5:14
• The above deduction is only a simplified way which must be accompanied by the observation: "The equilibrium constant can only be obtained by the kinetic way if the chemical equations are elementary reactions." – Éderson D'Martin Costa Feb 14 '18 at 3:04

The big issue here is that there is confusion between the observed rate law, which is not the same thing as the actual rate law.

For example, there are pseudo zero order reactions, but it is impossible for a reaction to be truly zero order (what is reacting then?).

Consider the following conversion of $\ce{A}$ and $\ce{B}$ to $\ce{C}$, via a two step process involving intermediate $\ce{A'}$.

$$\ce{A + B \overset{k1}{->} A' + B \overset{k2}{->} C}$$

Suppose that the first step is slow ($k1 \ll k2$). This means that the observed rate law would be $k\ce{[A]}$ because as soon as $\ce{A'}$ is formed, the reaction to the product is fast, so the rate is essentially determined by the first step. But it doesn't make any sense that the rate law would be independent of $\ce{B}$. What if $\ce{[B]} = 0$?

If you write out the full rate law without making any assumptions, every species is involved.

Let's see if I can do this properly.

The rate of the reaction is $\ce{k2[A'][B]}$. I'll apply the steady state approximation to $\ce{A'}$ here: because $\ce{A'}$ is a fleeting intermediate that is quickly consumed, we assume that $$\ce{[A']} \approx 0 \Rightarrow \frac{d\ce{[A']}}{dt} \approx 0$$

$$\frac{d\ce{[A']}}{dt} = -k_{-1}\ce{[A']} - k_{2}\ce{[A'][B]} + k_{1}\ce{[A]} = 0$$

$k_{-1}$ is the rate of the reverse reaction to form $\ce{A'}$.

$$\ce{[A']} = \frac{k_{1}}{k_{-1}+k_{2}\ce{[B]}}\ce{[A]}$$

$$\mathrm{rate} = \frac{k_{1}k_{2}}{k_{-1}+k_{2}\ce{[B]}}\ce{[A][B]}$$

There are two cases to consider.

First, in most scenarios because the second step is fast, $k_{2}\ce{[B]} > k_{1}$. This means that $k_{-1}+k_{2}\ce{[B]} \approx k_{2}\ce{[B]}$, and this $\ce{[B]}$ will cancel the other one in the numerator. It's really important to note that this is a pseudo first order reaction with two component, but the real rate law contains both species.

The other regime is $k_{2}\ce{[B]} < k_{1}$. In other words, there is very little $\ce{B}$ present. Here, $k_{-1}+k_{2}\ce{[B]} \approx k_{-1}$, so the $\ce{[B]}$ in the rate law cannot be canceled out, and we see that the reaction is regained its true second order form, which was concealed by the specifics of the reaction before.

• Just a comment: "This means that the observed rate law would be..." I had a slight problem understanding things after here. But when I wrote the two reactions separately: $\ce{A \overset{k1}{->} A'}$ and then $\ce{A' + B \overset{k2}{->} C}$ your post made much more sense. – Gaurang Tandon Feb 21 '18 at 2:28
• "First, in most scenarios because the second step is fast, k2[B]>k1" I need help understanding this. The rate of second step is $r_2=k_2[B][A']$ but that of first step is $r_1=k_1[A]$. So, shouldn't always $r_1>r_2$ since you said that $[A']\approx0$? – Gaurang Tandon Feb 21 '18 at 2:33
• @GaurangTandon This has to be true. Imagine a case where the forward reaction weren't faster than the reverse reaction. Then you can't argue that the first step is slower and that the intermediate is fleeting. Note that approximately zero is not the same thing as zero. You might want to read up on some physical organic chemistry for some of these topics. – Zhe Feb 21 '18 at 13:57
• @Zhe Is there any way to do this without the steady-state approximation? – Apoorv Potnis Feb 22 '18 at 16:18
• @ApoorvPotnis Sure, but you might not be able to use this approach at all. I'd look at orthocresol's solution using chemical potentials instead. – Zhe Feb 22 '18 at 16:22

The derivation that you cite only holds true if both forward and backward reactions are elementary steps. In this case, the rate of the forward reaction is indeed given by

$$r = k[\ce{A}]^a[\ce{B}]^b$$

and so on, and hence the derivation is logically sound. If they are not elementary steps, then even though the conclusion is correct, the method of arriving there is not, for the reasons you have already pointed out. See also: hBy2Py's answer and Ben Norris's answer where the assumption of an elementary step is explicitly mentioned.

In general the equilibrium constant does not need to be derived in this manner; a more conventional method of arriving at it is by using (pressure here is taken to equal an arbitrary but specified standard pressure $p^\circ$)

$$\mu_i = \mu_i^\circ + RT\ln a_i$$

where $\mu_i$ is the chemical potential of species $i$, $\mu_i^\circ$ is the standard chemical potential of species $i$, and $a_i$ is the chemical activity of species $i$. We now have, for a reaction

$$\ce{a A + b B -> c C + d D},$$

the following equation:

\begin{align} \Delta_\mathrm r G^\circ &= c\mu_\ce{C}^\circ + d\mu_\ce{D}^\circ - a\mu_\ce{A}^\circ - b\mu_\ce{B}^\circ \\ &= c(\mu_\ce{C} - RT\ln a_\ce{C}) + d(\mu_\ce{D} - RT\ln a_\ce{D}) - a\mu_\ce{A} - RT\ln a_\ce{A} - b(\mu_\ce{B} - RT\ln a_\ce{B}) \\ &= c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B} - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right] \\ &= \Delta_\mathrm r G - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right] \end{align}

However, at equilibrium, $\Delta_\mathrm r G = 0$ and hence

$$\Delta_\mathrm r G^\circ = - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right]$$

The term in square brackets can then be identified as the equilibrium constant, $K$. As you can see there is no need for any recourse to kinetic rate laws.

Why is $K$ a constant? Well, the standard chemical potentials $\mu_i^\circ$ are thermodynamic constants, just like standard enthalpies of formation, standard entropies, etc. So $\Delta_\mathrm r G^\circ$ - a sum of constants - is itself a constant, and it follows that $K = \exp(-\Delta_\mathrm r G^\circ/RT)$ must be a constant.

• Note for high-schoolers like me: The chemical potential $\mu_i$ is nothing but the partial molar Gibbs energy of the species $i$. – Apoorv Potnis Feb 22 '18 at 15:50
• Just a question, where does the equation of chemical potential come from? How is it derived? – Shoubhik Raj Maiti Feb 26 '18 at 14:44
• @ShoubhikRajMaiti See this. – Apoorv Potnis Feb 27 '18 at 5:19

Then, how can we apply the law of chemical equilibrium on every reaction, without experimentally verifying the rate equations?

You can derive the equilibrium constant from the rate constants even if the rate law is different than the stoichiometry of the net reaction suggests. It just gets a bit more complicated than in your example.

You do need to know all the elementary reactions. At equilibrium, rates for each elementary reaction will be equal to the rate of the reverse reaction, giving one equation about rates for each elementary reaction.

If you multiply all these equations, you get an expression that contains all the rate constants and lots of concentration terms which will all cancel out unless they are reactants or products of the net equation. After cancelling and sorting rate constants on one side and concentration terms on the other side, you will arrive at the equilibrium constant expression.

Here is a simple example with A reacting to form B with the help of a catalyst E (E for enzyme):

$$\ce{A(aq) <=>[E] B(aq)}$$ (net reaction)

First elementary reaction:

$$\ce{A + E <=> I}$$

Second elementary reaction:

$$\ce{I <=> B + E}$$

First set of equal rates:

$$k_1 [A] [E] = k_{-1} [I]$$

Second set of equal rates:

$$k_2 [I] = k_{-2} [B] [E]$$

Multiply:

$$k_1 [A] [E] k_2 [I] = k_{-1} [I] k_{-2} [B] [E]$$

Cancel:

$$k_1 [A] k_2 = k_{-1} k_{-2} [B]$$

Sort:

$$\frac{k_1 k_2}{k_{-1} k_{-2}} = \frac{[B]}{[A]}$$

Compare to the expected equilibrium expression

$$K = \frac{[B]}{[A]}$$

For the specific example, this shows that while the enzyme influences the rates (and some steps depend on the enzyme concentration), it does not change the equilibrium (it speeds up both the forward and the reverse reaction). If you think of the common energy diagram, the enzyme (or in more general terms - the path of the reaction) influences the rates and activation energies, but not the position of the equilibrium, i.e. the Gibbs energy of reaction.

If the rate equations derived from the law of mass action are wrong, then the value of Kc obtained by the final equation will not be constant.

The equilibrium constant expression is correct even if the reaction has a mechanism more complex than a single elementary reaction. One more aspect is the unit of the equilibrium constant. Rate equations are based on concentrations (units mol/L) while equilibrium constants are based on activities (dimensionless). If the number of reactants is different from the number of products in the net equation, you have to consider this by dividing concentrations by the respective standard states when using rate constants to calculate K.

I’m stating this sentence from the book ‘Modern approach to chemical calculations’ by R C Mukerjee (p. 581 footnote)

“In chemical kinetics, reactions which occur in a single step are termed as simple or elementary reactions and those which take place in two or more steps are termed as complex reactions. For elementary reactions, the law of mass action and the rate law expressions are generally same.”

In other words, for elementary reactions, the exponent is basically the stoichiometric coefficient.

The reason why we don’t use stoichiometric coefficients as the exponents for complex reactions is as follows :

When we derive rate law of complex reactions, we combine the rate laws of the elementary reactions (actually, we combine the laws of mass action, but they are the same for an elementary reaction) by getting rid of many concentration terms (practically, we can only measure the concentration of reactants of the rate determining or slowest step, and the final product). So the stoichiometric coefficient of the reactants may or may not appear directly in the final product.

Complex reactions comprise of many elementary reactions and finding out the actual path of every reaction is not very easy. Even if we can find out every intermediate and every reaction involved, it is very tiresome to write the law of mass action for each reaction and get rid of the concentration terms we can’t measure.

So we use experimental data and define order of the reaction (with respect to the reactants) which appear as the exponents in the rate law.