AP Chemistry 9.1-9.2: Entropy and Absolute Entropy

Entropy ($S$)

Entropy is a state function that measures the dispersal of matter and energy in a system. Greater dispersal corresponds to a greater number of possible microscopic arrangements.

What is a microstate?

A microstate is one specific arrangement of the locations and energies of all particles in a system. The number of possible microstates is represented by $W$.

Boltzmann's entropy equation

$S=k\ln W$, where $k=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$ is the Boltzmann constant and $W$ is the number of microstates.

How is entropy change related to the initial and final numbers of microstates?

$\Delta S=S_f-S_i=k\ln W_f-k\ln W_i=k\ln\left(\frac{W_f}{W_i}\right)$.

How does a change in the number of microstates affect entropy?

If $W_f>W_i$, then $\Delta S>0$. If $W_f<W_i$, then $\Delta S<0$; if the numbers are equal, $\Delta S=0$.

Matter dispersal model of entropy

A system has higher entropy when its particles are distributed among more locations or arrangements. This is often described qualitatively as greater disorder, but the fundamental idea is the number of accessible microstates.

For $N$ distinguishable particles distributed between two boxes, how many microstates are possible?

There are $2^N$ possible microstates, because each particle can independently be in either box.

Why is an approximately even distribution of particles between equal-volume regions the most probable distribution?

The approximately even distribution corresponds to the greatest number of individual microstates. Since probability is proportional to the number of microstates, it is also the highest-entropy distribution.

What is the relationship between distributions and microstates?

A distribution describes a macroscopic particle arrangement, such as the number of particles in each box. Multiple distinct microstates can correspond to the same distribution.

Why does a gas expanding into a vacuum have $\Delta S>0$?

Expansion increases the volume available to the gas particles, thereby increasing the number of possible particle locations and microstates. Thus, $W_f>W_i$ and $\Delta S>0$.

Why does heat spontaneously flow from a hot object to a cold object?

Energy becomes more evenly dispersed between the objects, producing more possible energy distributions and a greater total number of microstates. Therefore, the entropy of the universe increases.

Reversible process

A reversible process is an idealized process occurring so slowly that the system remains essentially at equilibrium and can be reversed by an infinitesimal change. Real processes are irreversible, but reversible paths are used to calculate entropy changes.

How is entropy change defined using reversible heat?

$\Delta S=\dfrac{q_{\mathrm{rev}}}{T}$ for a reversible process at temperature $T$ in kelvins. For a real irreversible process, the entropy change is equal to that calculated from any reversible path connecting the same initial and final states.

Why is entropy a state function?

The entropy change depends only on the initial and final states, not on the particular path used to connect them. This permits calculation of an irreversible process using a theoretical reversible path.

Entropy ordering of the phases of a substance

For the same substance under comparable conditions, $S_{\mathrm{gas}}>S_{\mathrm{liquid}}>S_{\mathrm{solid}}$ because particles have progressively more freedom of motion and available locations.

What are the entropy signs for melting, freezing, vaporization, condensation, sublimation, and deposition?

Melting and vaporization have $\Delta S>0$; freezing and condensation have $\Delta S<0$. Sublimation has $\Delta S>0$, while deposition has $\Delta S<0$.

How does increasing temperature affect the entropy of a substance?

Entropy generally increases as temperature rises because particles have greater average kinetic energy and a broader distribution of accessible energies. Thus, more microstates become available.

How do particle size and molecular complexity generally affect entropy?

At a given temperature, heavier atoms generally have greater entropy than lighter atoms. Molecules with more atoms generally have more vibrational modes and therefore more accessible microstates.

Why does mixing different types of particles generally increase entropy?

Mixing creates additional possible arrangements, orientations, and interactions among nonidentical particles. The increased dispersal of matter usually gives $\Delta S>0$.

What is the usual entropy change when an ionic solid dissolves in water?

Dissolution usually gives $\Delta S>0$ because the ordered solid produces mobile ions that can disperse throughout the solution and interact with solvent molecules. The sign should still be evaluated from the specific process if competing effects are important.

How does forming a precipitate from aqueous ions usually affect entropy?

Combining dispersed aqueous ions into an ordered solid usually decreases the number of accessible arrangements, so $\Delta S$ is generally negative.

How can changes in gaseous particle amounts help predict the sign of $\Delta S$?

A net increase in the amount of gaseous species generally gives $\Delta S>0$, while a net decrease generally gives $\Delta S<0$. This is a qualitative guideline based on the much greater dispersal possible in the gas phase.

Predict the sign of $\Delta S$ for $\mathrm{CaCO_3(s)\rightarrow CaO(s)+CO_2(g)}$.

The sign is positive because a gaseous product is formed from solids, increasing the dispersal of matter and the number of accessible microstates.

Predict the sign of $\Delta S$ for $\mathrm{Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)}$.

The sign is negative because mobile, dispersed aqueous ions form a more ordered solid precipitate.

Entropy change for a system and its surroundings

$\Delta S_{\mathrm{univ}}=\Delta S_{\mathrm{sys}}+\Delta S_{\mathrm{surr}}$. The universe includes both the system and its surroundings.

Second law of thermodynamics

Every spontaneous process produces an increase in the entropy of the universe: $\Delta S_{\mathrm{univ}}>0$.

What do the three possible values of $\Delta S_{\mathrm{univ}}$ indicate?

$\Delta S_{\mathrm{univ}}>0$ indicates a spontaneous forward process; $\Delta S_{\mathrm{univ}}<0$ indicates a nonspontaneous forward process that is spontaneous in the reverse direction; $\Delta S_{\mathrm{univ}}=0$ indicates equilibrium.

How is the surroundings' entropy change commonly approximated when the surroundings are very large?

$\Delta S_{\mathrm{surr}}\approx\dfrac{q_{\mathrm{surr}}}{T_{\mathrm{surr}}}$, so $\Delta S_{\mathrm{univ}}\approx\Delta S_{\mathrm{sys}}+\dfrac{q_{\mathrm{surr}}}{T_{\mathrm{surr}}}$.

Why can a process with $\Delta S_{\mathrm{sys}}<0$ still be spontaneous?

Spontaneity depends on the total entropy change, $\Delta S_{\mathrm{univ}}$, not on the system alone. A sufficiently large positive entropy change in the surroundings can make $\Delta S_{\mathrm{univ}}>0$.

Why does heat flow spontaneously from hot to cold in terms of entropy?

For heat $q$ transferred from a hotter system to a colder surrounding, $\Delta S_{\mathrm{sys}}=-q/T_{\mathrm{hot}}$ and $\Delta S_{\mathrm{surr}}=q/T_{\mathrm{cold}}$. Because $T_{\mathrm{hot}}>T_{\mathrm{cold}}$, the positive surroundings term is larger, giving $\Delta S_{\mathrm{univ}}>0$.

Third law of thermodynamics

The entropy of a pure, perfect crystal at absolute zero, $0\ \mathrm{K}$, is zero.

Why does a perfect crystal at $0\ \mathrm{K}$ have zero entropy?

It has only one possible microstate, so $W=1$ and $S=k\ln(1)=0$.

Standard entropy, $S^\circ$

Standard entropy is the absolute molar entropy of a substance under standard conditions, commonly $298.15\ \mathrm{K}$ and a pressure of $1\ \mathrm{bar}$. Its units are typically $\mathrm{J\,mol^{-1}\,K^{-1}}$.

Formula for the standard entropy change of a chemical reaction

$\Delta S^\circ=\sum \nu S^\circ(\text{products})-\sum \nu S^\circ(\text{reactants})$, where $\nu$ values are the stoichiometric coefficients from the balanced equation.

How should standard molar entropies be used in a reaction calculation?

Multiply each substance's tabulated $S^\circ$ by its balanced-equation coefficient, sum the products, sum the reactants, and subtract: products minus reactants. Keep units consistent, usually $\mathrm{J\,K^{-1}}$ for the reaction as written.

Calculate $\Delta S^\circ$ for $\mathrm{H_2O(g)\rightarrow H_2O(l)}$ given $S^\circ[\mathrm{H_2O(g)}]=188.8$ and $S^\circ[\mathrm{H_2O(l)}]=70.0\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

$\Delta S^\circ=70.0-188.8=-118.8\ \mathrm{J\,mol^{-1}\,K^{-1}}$. The negative value is consistent with condensation.

Calculate $\Delta S^\circ$ for $\mathrm{H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)}$ using $S^\circ(\mathrm{H_2})=130.7$, $S^\circ(\mathrm{C_2H_4})=219.3$, and $S^\circ(\mathrm{C_2H_6})=229.2\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

$\Delta S^\circ=229.2-(130.7+219.3)=-120.8\ \mathrm{J\,mol^{-1}\,K^{-1}}$ for the reaction as written.

Calculate $\Delta S^\circ$ for $\mathrm{2CH_3OH(l)+3O_2(g)\rightarrow2CO_2(g)+4H_2O(l)}$ using the listed standard entropies.

$\Delta S^\circ=[2(213.8)+4(70.0)]-[2(126.8)+3(205.2)]=-161.6\ \mathrm{J\,K^{-1}}$ for the reaction as written.

What entropy changes are expected when liquid water is warmed without changing phase?

The entropy change is positive because increasing temperature broadens the distribution of accessible particle energies, even though the substance remains liquid.

Why can the spontaneity of melting change with temperature?

Melting requires heat absorption, so $\Delta S_{\mathrm{surr}}=q_{\mathrm{surr}}/T$ is negative and its magnitude decreases as temperature increases. Therefore, a process that is nonspontaneous at a lower temperature can become spontaneous at a higher temperature when $\Delta S_{\mathrm{univ}}$ changes from negative to positive.