\documentclass[12pt,a4paper]{article}
\usepackage{chemfig}
\begin{document}
\title{Entropy}
\author{Philip Thomas K.}
\maketitle
\section{Introduction}
Entropy is the amount of disorder in a system/surrounding. If a system has a high entropy, there is a lot of disorder. If a system has a low entropy, there is less disorder, meaning that there is more order.
\newpage
\section{Theory}
The concept of entropy is based around the second law of thermodynamics:
\begin{figure}[h]
\centering
\fbox{\parbox{\linewidth}{\centering For a process to be spontaneous, the amount of entropy in the universe must increase. It is the conversion of ordered motion and energy to disordered motion and energy.}}
\end{figure}
\medskip
$S$ represents the amount of entropy, and the amount of it in the universe increases. The units of entropy, $S$ are Jmol\textsuperscript{$-1$}K\textsuperscript{$-1$}.
\medskip
In nature, systems and surroundings will move spontaneously from order to disorder. This is because there are many more ways to reach a disordered state than an ordered state, therefore the probability of this occuring becomes very large. For example, a diatomic molecule such as H\textsubscript{$2$} will have two quanta (limited set amounts) of vibrational energy.
\begin{center}
\begin{tabular}{ | l | l | l |}
\hline
Energy Level & H\textsubscript{1} & H\textsubscript{2} \\ \hline
5e & & \\ \hline
4e & & \\ \hline
3e & & \\ \hline
2e & $\bullet$ & $\bullet$ \\ \hline
1e & & \\ \hline
\end{tabular}
\end{center}
If energy is transferred from H\textsubscript{1} to H\textsubscript{2}, such that H\textsubscript{1} will have an energy level of 1e and that H\textsubscript{2} will have an energy level of 3e, we can say that this system will have a total of 5 combinations of each particle having different energy levels that add up to a total of 4e. Increasing amounts of possible energy levels (increasing heat) and increasing amount of particles increases the amount of possible combinations. The larger the amount of combinations, the higher the disorder. Put simply, $\uparrow$Heat, and $\uparrow$No. of particles will result in $\uparrow$$S$. This poly-atomic molecule disorder can be calculated using the following equation:
\begin{equation}
W = \frac{(N + q - 1)!}{q!(N - 1)!}
\end{equation}
Where $W$ is the number of combinations, $N$ is the number of atoms, and $q$ is the number of quanta of energy that can be calculated with the following simple equation:
\begin{equation}
q = q_1 + q_2 + q_3 \dots
\end{equation}
It is important to note that the lowest vibrational energy is 1e (zero point energy) , not 0e, as particles at 0K violate Heisenberg's uncertainty principle. In all chemical reactions there are entropy changes, both in the reaction system, as well as the surrounding system. From this we can say that the total entropy change is:
\begin{equation}
\Delta S\textsubscript{total} = \Delta S\textsubscript{system} + \Delta S\textsubscript{surroundings}
\end{equation}
For example, the chemical equation below:
\smallskip
\begin{center}
\schemestart 2Na + Cl\textsubscript{2}\arrow{->}2NaCl \schemestop
\end{center}
This is a reaction that has a negative value for $\Delta S\textsubscript{system}$, however, this reaction produces heat, so the value of $\Delta S\textsubscript{surroundings}$ is positive to a degree that $\Delta S\textsubscript{total}$ is positive. $\Delta S\textsubscript{surroundings}$ is calculated by how much it changes the temperature of the surroundings during the reaction. It can be calculated using the following equation:
\begin{equation}
\Delta S\textsubscript{surroundings} = - \frac{\Delta H}{T}
\end{equation}
Where $\Delta H$ is the enthalpy change of the reaction and $T$ is the temperature of the surroundings in Kelvin. Using these two equations, we can calculate the equation for the Gibb's Free Energy, which is an equation that can determine the spontaneity of a reaction at a certain temperature. Substituting equation (4) into equation (3), we get the following equation:
\begin{equation}
\Delta S\textsubscript{total} = \Delta S\textsubscript{system} - \frac{\Delta H}{T}
\end{equation}
Multiplying the entire expression by $T$, we get:
\begin{equation}
T \Delta S\textsubscript{total} = T \Delta S\textsubscript{system} - \Delta H
\end{equation}
Multiplying the entire expression by -1, we get:
\begin{equation}
-T \Delta S\textsubscript{total} = \Delta H - T \Delta S\textsubscript{system}
\end{equation}
And since the Gibb's free energy is described as such:
\begin{equation}
\Delta G = - T \Delta S\textsubscript{total}
\end{equation}
We can substitute equation (8) into equation (7) to get:
\begin{equation}
\Delta G = \Delta H - T \Delta S\textsubscript{system}
\end{equation}
Which is the Gibb's free energy equation. If ($\Delta G$) is negative, the reaction is spontaneous. However, even if a reaction is spontaneous, it might not take place since the kinetics of the reaction may be too slow. From this we can make a table on the spontaneity of a reaction based on the positivity of $\Delta S\textsubscript{system}$ and $\Delta H$:
\begin{center}
\begin{tabular}{ | l | p{5cm} | p{5cm} |}
\hline
Positivity & $\Delta H$ = -ve & $\Delta H$ = +ve \\ \hline
$\Delta S$ = -ve & Spontaneous at low temperatures & $\Delta G$ positive at all temperatures, process never spontaneous \\ \hline
$\Delta S$ = +ve & $\Delta G$ negative at all temperatures, process always spontaneous & Spontaneous at high temperatures \\ \hline
\end{tabular}
\end{center}
It is also possible to calculate the value of $\Delta S\textsubscript{system}$ from the standard values of entropy found in most chemical data books using the following equation:
\begin{equation}
\Delta S\textsubscript{system} = \sum S\textsubscript{products} - \sum S\textsubscript{reactants}
\end{equation}
This equation can be used for calculating precise value for $\Delta S\textsubscript{system}$, however there are good ways to predict the posistivity of $\Delta S\textsubscript{system}$ by considering the number of molecules formed from the reactants, as well as the phases of all the molecules in the reaction. For example:
\begin{center}
\schemestart N\textsubscript{2}O\textsubscript{4$(g)$} \arrow{<=>} 2NO\textsubscript{2$(g)$} \schemestop\par
\end{center}
The chemical equation shown above shows a reaction in which the forward reaction doubles the amount of particles, while the reverse reaction halves the amount of particles. It is more likely for the forward reaction to take place since the forward reaction produces more particles, and there are more possible configurations for those particles to be in with larger amounts of particles, as seen in equation (1). The forward reaction therefore increases the amount of entropy in the system, and the reverse reaction decreases the amount of entropy in the system.
\begin{center}
\schemestart H\textsubscript{2}O\textsubscript{$(s)$} \arrow{<=>} H\textsubscript{2}O\textsubscript{$(l)$} \schemestop\par
\end{center}
In the above reaction, where ice melts to water (the forward reaction), entropy increases since the energy of the particles has increased, creating more ways for the particles to be arranged at different energy levels.
\newpage
In reversible reactions, when the mixture reaches equilibrium, we say that the forward reaction is going at the same rate as the reverse reaction, meaning the mixture is in a dynamic equilibrium. This therefore means that the forward and backward reactions must both be spontaneous, meaning that $\Delta S\textsubscript{total}$ must be positive in both cases. The only way the forward and backward reaction can both be spontaneous, and not break the second law of thermodynamics, is if the value of $\Delta S\textsubscript{total}$ for both reactions is 0. This makes intuitive sense since there cannot be an actual increase in the amount of entropy if the amounts of all the molecules remain constant and the temperature remains constant. Therefore, we can conclude that when a system is in dynamic equilibrium:
\begin{equation}
\Delta S\textsubscript{total}[\textrm{forward reaction}] = \Delta S\textsubscript{total}[\textrm{backward reaction}] = 0
\end{equation}
The spontaneous direction of change in a reversible reaction is determined by its total entropy change. The direction and extent of a change in a reversible reaction is determined by the equilibrium constant for that change. The connection between these two statements can be expressed like so:
\begin{equation}
\Delta S\textsubscript{total} = R\textrm{ln}K
\end{equation}
Where $K$ can be $K_c$ or $K_p$, and where $R$ is the gas constant. The following table shows rough guidelines to the extent of the reaction which can be determined by the different values of $\Delta S\textsubscript{total}$:
\begin{center}
\begin{tabular}{ | l | p{5cm} | }
\hline
$\Delta S\textsubscript{total}$/Jmol\textsuperscript{-1}K\textsuperscript{-1} & Extent of Reaction \\ \hline
+40 to -40 & Mixture of reactants and products \\ \hline
\textgreater +200 & Reaction is complete, and all the reactants turned to products \\ \hline
\textless -200 & No evidence of products and the reaction does not take place \\ \hline
\end{tabular}
\end{center}
It is important to note that for exothermic reactions, an increase in temperature decreases the value of $\Delta S\textsubscript{total}$ since it reduces the positivity of the $\Delta S\textsubscript{surroundings}$ term due to the division by a larger number. An increase in temperature does not significantly affect the value of $\Delta S\textsubscript{system}$. This means that an increase in temperature decreases the spontaneity of the reaction, but it will not make it negative.
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