PRE-EQUILIBRIA

This article is written to give Grade 11 and above students an idea about how a reaction mechanism is used to write rate law with special emphasis on pre-equilibrium

REVIEW OF SELECTIVE BASICS

Chemical equilibrium or simply ‘equilibrium’ is the stage in a chemical reaction when the rate of the forward reaction is equal to the rate of the reverse reaction.
The rate law of a chemical reaction, for example, A + 2B $\rightarrow$ 3C + 2D, shows how the rate of a reaction depends on the concentrations of different reactants. It is of the form: Rate = k[A]^m[B]ⁿ, where m and n are experimentally determined exponents or powers referring to the order of reaction. They are different from the coefficients of chemical species in the balanced chemical equation. ‘k’ is the rate constant.
The reaction mechanism is a series of elementary steps which show how the reactants change into products. For example:

The chemical reaction A + B $\rightarrow$ C + D can be considered or hypothesized to take place in several steps. Each step is called an elementary reaction and the sum of all elementary steps should give the overall reaction, as illustrated below:

Step 1: A $\rightarrow$ C + F ………… (slow)

Step 2: B + F $\rightarrow$ D …………. (fast)

Add steps to get an overall reaction: A + B $\rightarrow$ C + D {*Note: F, the intermediate cancels out as it is produced in step 1 and used in step 2}

The rate of reaction depends on the slow step. In the example above, the rate law for the overall reaction will have the molar concentration of A, written as [A], and the exponent of [A] will reflect the number of molecules of A taking part in the slow step (molecularity). So, the rate law for A + B $\rightarrow$ C + D will be Rate = k. [A]

Remember, the mechanism proposed and the rate law derived from it must comply with experimentally determined rate law.

In the above example, had the change from A+B to C+D be proposed by the different steps (written below), the rate law would have been different.

Alternative Step 1: 2 A $\rightarrow$ 2C + F1 ……… (slow)

Alternative step 2: 2B + F1 $\rightarrow$ 2D + A …….(fast)

Alternative rate law, Rate law: Rate = k. [A]²

The rate law for each elementary step is written through its ‘molecularity’, i.e. the number of molecules taking part on the reactant side. For example, the rate law for the first elementary step written above is, Rate = k₁.[A], and that of the second elementary step is, Rate = k₂. [B].[F]. However, the overall rate of the reaction depends on the slower step. The faster step(s) is/are not taken into account. In this case, k = k₁.

THE CONDITION OF PRE-EQUILIBRIUM

To understand the condition of pre-equilibrium, let’s consider the following example:

Overall reaction: BrO₃^–_(aq) + H⁺_(aq) + 5 Br^–_(aq) $\rightarrow$ 3 Br_2(l) + 3 H₂O_(l)

If it is given that the above reaction proceeds through the following mechanism:

From the above elementary steps, it is clear that H₂BrO₃⁺ (formed in step 2) is consumed slowly in step 3. If the rate of its consumption in step 3 is insignificant as compared to its formation and decomposition. This condition is called a pre-equilibrium condition. In such conditions, a ‘bottleneck’ is created where an intermediate has to wait to get consumed in the slow step. So, when the intermediate is accumulating, it can turn back into the reactants of the faster step. An equilibrium of forward and reverse reaction is established in the faster elementary step. However, the overall reaction has not reached equilibrium yet. Hence, the term pre-equilibrium is used to describe such mechanisms.

How to calculate the rate law of reactions with pre-equilibrium condition

In reactions with pre-equilibrium condition, the faster steps that come before the slow step also play a role in determining the overall reaction rate because of the reverse reactions they suffer due to the accumulation of an intermediate.

From the slower step and its molecularity, the rate of overall reaction can be written as:

Rate = k₃.[H₂BrO₃⁺].[Br^–] ………………(i)

Since intermediates and rate constants of elementary steps cannot figure in the overall rate law, k₃ and [H₂BrO₃⁺] need to be written in terms of k (overall rate constant) and reactants of the overall reaction.

For this we have to make an assumption for the faster steps preceding the slow step. This is called ‘steady-state’ assumption, that is, the first and second steps are in equilibrium (rate of forward reaction = rate of reverse reaction)

Therefore, for the step 2:

Rate of forward reaction, Rate_2f= k₂.[HBrO₃].[H⁺]

Rate of reverse reaction, Rate_2r=k_-2.[H₂BrO₃⁺]

Equating the above two due to steady-state assumption,

Rate_2f = Rate_2r

=> k₂.[HBrO₃].[H⁺] = K_-2.[H₂BrO₃⁺]

=> $\left [ H_{2}BrO_{3}^{+} \right ]=\frac{k_{2}}{k_{-2}}\cdot\left [ HBrO_{3} \right ]\cdot\left [ H^{+} \right ]$ ………….(ii)

Similarly, for step 1,

k₁.[BrO₃^–].[H⁺] = k_-1.[HBrO₃]

=> $\left [ HBrO_{3} \right ]=\frac{k_{1}}{k_{-1}}\cdot\left [ BrO_{3}^{-} \right ]\cdot\left [ H^{+} \right ]$

Substituting [HBrO₃] in (ii),

$\left [ H_{2}BrO_{3}^{+} \right ]=\frac{k_{1}\cdot k_{2}}{k_{-1}\cdot k_{-2}}\cdot\left [ BrO_{3}^{-} \right ]\cdot\left [ H^{+} \right ]^{2}$

Now substituting for [H₂BrO₃⁺] in the rate law for the slow step (i.e. in (i)) to get the overall rate law,

$Rate=\frac{k_{1}\cdot k_{2}\cdot k_{3}}{k_{-1}\cdot k_{-2}}\cdot\left [ BrO_{3} \right ]\cdot \left [ H^{+} \right ]^{2}\cdot [Br^{-}]$

The above rate law is complete where the overall rate constant, $k=\frac{k_{1}\cdot k_{2}\cdot k_{3}}{k_{-1}\cdot k_{-2}}$

Since, Br^– figures in both the step 3 (a slow step) and the overall chemical equation, the above rate law can also be written as:

$Rate=- \frac{d\left [ Br^{-} \right ]}{dt}=k\cdot\left [ BrO_{3}^{-} \right ]\cdot\left [ H^{+}]^{2} \right ] \cdot\left [ Br^{-} \right ]$