As soon as we introduce the notion of reversible reactions, we might think: what if I do want to convert reactants into products? What if I want to maximize the amount of products I get, can I do anything?

I hope you remember that we’ve dealt with a similar question before. Back when we discussed factors that affect reaction rate, we noted that we can easily answer the question if we know the inputs/outputs and the mechanism which links them. Refer to this note: Finding Factors Affecting X.

Let’s consider a chemical reaction:

$aA+bB cC+dD(1)$The inputs are concentrations $[A]_{0},[B]_{0},[C]_{0},[D]_{0}$. The outputs are concentrations we have at equilibrium: $[A]_{eq},[B]_{eq},[C]_{eq},[D]_{eq}$. The mechanism by which they’re connected is the chemical reaction and equilibrium constant:

$K=[A]_{eq}[B]_{eq}[C]_{eq}[D]_{eq} (2)$Let’s imagine, for a moment, that we want to maximize the amount of product $[C]_{eq}$ that we get.

in the lecture, we’re somewhat sloppy with the terminology. We say we wish to maximize the yield of the reaction and then we behave as if we’re trying to increase $[C]_{eq}$. I personally will avoid using the word “yield” as it usually refers to how much we get on practice compared to theoretical expectations, and we almost have no way of predicting yield besides by just performing the synthesis in practice. What we talk here is always about the theoretical amount we could get with a 100% yield.

## Overview

Given the expression for the equilibrium constant:

$K=[A]_{eq}[B]_{eq}[C]_{eq}[D]_{eq} (2)$How might we increase $[C]_{eq}$ given that the value of $K$ is constant and should remain constant? We can enumerate all possible options

- If we somehow end up with lower $[D]_{eq}$, all other things constant, $[C]_{eq}$ has to be greater to keep $K$ constant.
- If we somehow end up with greater $[A]_{eq}$, all other things constant, $[C]_{eq}$ has to be greater to keep $K$ constant.
- If we somehow end up with greater $[B]_{eq}$, all other things constant, $[C]_{eq}$ has to be greater to keep $K$ constant.

Let’s aggregate these 3 options into two categories: **decreasing the amount of other products** and **increasing the amount of reactants**. Before we proceed to look at them in detail, let’s also be bold and challenge our constraint: do we really have to accept unchanging $K$? Could we change it somehow? Let’s call it our 3rd option: **changing** $K$.

Before we proceed, you should agree each of the three statements above is true. It’s a direct mathematical consequence of (2). As always, let me know if you don’t see why it has to be true.

## Option 1. Aim to decrease amount of products

How might we end up with lower $[D]_{eq}$? Let’s consider the case when $[C]_{0}=[D]_{0}=0$, i.e. we don’t have any products in the beginning. We won’t consider the case when it’s not true simply because we’re interested in how much $[C]$ and $[D]$ can we get from the reaction we’re about to set. By stoichiometry, it has to be true that:

$c1 [C]_{eq}=d1 [D]_{eq}(3)$So if we aim to end up with greater $[C]_{eq}$, it seems like we’re destined to have greater $[D]_{eq}$ as well. That’s true, we will produce big amounts of $[D]_{eq}$. But nothing stops us from *somehow* collecting $[D]_{eq}$ and taking it out of the system. How does it look like in practice? Well, imagine if $[D]$ has some unique physical properties: maybe it precipitates out of the solution? Or it’s a gas that evolves and leaves the mixture? Or maybe it simply has a different boiling point than $C$? In fact, in practice, whenever we try to esterify carboxylic acids:

where $CHX_{3}COOH$ is the acetic acid, $CX_{5}HX_{11}OH$ is isoamyl alcohol, $CHX_{3}COOCX_{2}HX_{5}$ is isoamyl acetate, and $HX_{2}O$ is, well, you know what it is. Dihydrogen Monooxide. The boiling point of isoamyl acetate is ~142°C, so what happens in practice is we heat the reaction mixture to ~90°C so that the forming $HX_{2}O$ evaporates, while the desired product isoamyl alcohol stays in the mixture.

Btw, isoamyl acetate has a nice fruity pear-like smell. Esters typically have noticeable and nice smell, and for that reason are heavily used in the perfume industry.

## Option 2. Aim to increase amount of reactants

How might we end up with greater $[A]_{eq}$ or $[B]_{eq}$? Well, this is simpler, as we can just dump more $[A]_{0}$ or $[B]_{0}$. The only interesting question for us is should we simply increase one of them or do we have to increase both. It’s actually a constrained optimization problem from calculus, and you can show that you get maximal ratio of $[C]$ to the amount of the reactant that is in shortage if $[A]≈[B]$. So you have to increase both.

## Option 3. Change K

For a given reaction at a given temperature $K$ is fixed. So if you want to increase the production of products or reactants, you have to resort to previous options. However, as we’ll see later in the course, turns out that the equilibrium constant $K$ is related to thermodynamics. Specifically:

$lnK=−RTΔG =−RTΔH +RΔS (5)$So, it turns out, that $K$ depends on temperature, and the constant of proportionality is directly related to $ΔH$ or the enthalpy change of a reaction (if you don’t know what enthalpy is, it’s the heat change during the reaction). $ΔH$ is defined as the difference in enthalpies of products minus reagents, so whenever $ΔH<0$, products are lower in energy (and thus more stable), so the difference was released into the atmosphere (and we call such reactions exothermic). In contrast, if $ΔH>0$, the products have more energy than reactants, so that new energy should come from somewhere and it comes from the atmosphere, i.e. reactions absorb heat. Such reactions are called endothermic.

You might notice that if $ΔH<0$, the coefficient before $T1 $ is positive $−ΔH>0$, so $lnK∝mT_{−1}+b$. You could see that for low values of $T$, let’s say $T=1$ (Kelvin, of course) $lnK≈b+1$, but as $T$ increases, $1/T$ keeps decreasing, so $lnK$ decreases to $≈b$. All of this is to say that $lnK$ becomes smaller as $T$ increases. Which means that the $K$ itself also becomes smaller. This is the big takeaway: **for exothermic reactions (when heat is released), increase of temperature decreases K, and so practically results in the shift of an equilibrium towards reactants**.

In contrast, if $ΔH>0$, the $lnK∝−aT_{−1}+b$, so again, with greater temperatures the first term becomes smaller, but because of a negative sign it decreases $b$ to a smaller extent, so $lnK$ increases in value with temperature. And the takeaway is for **endothermic reactions (when heat is absorbed), increase of temperature increases K, and so the equilibrium shifts towards reactants.**

Quite surprisingly, I shall say, these fairly complicated mathematical corollaries are perfectly captured by the simple mnemonic rule:

whenever you try to influence the system at equilibrium, the system will adapt so as to decrease the impact of the influence

which is, as I understand, the only thing you’re supposed to know at this point with regard to the dependence of $K$ on temperature. The logical argument is as follows:

- You try to influence the system by applying heat.
- The system wants to minimize that heat. So, it will proceed in the direction in which heat is absorbed (if the forward reaction is endothermic, then bingo, if the forward reaction is exothermic, then the reverse reaction is endothermic)

This rule also explains how the system responds to changes in concentrations, but it’s nice to see where this rule comes from.