### Key Concepts and Summary

Entropy (*S*) is a state function that can be related to the number of microstates for a system (the number of ways the system can be arranged) and to the ratio of reversible heat to kelvin temperature. It may be interpreted as a measure of the dispersal or distribution of matter and/or energy in a system, and it is often described as representing the “disorder” of the system.

For a given substance, entropy depends on phase with *S*_{solid} < *S*_{liquid} < *S*_{gas}. For different substances in the same physical state at a given temperature, entropy is typically greater for heavier atoms or more complex molecules. Entropy increases when a system is heated and when solutions form. Using these guidelines, the sign of entropy changes for some chemical reactions and physical changes may be reliably predicted.

### Key Equations

$ΔS=\frac{q_{rev}}{T}$ |

$S=klnW$ |

$ΔS=kln\frac{W_f}{W_i}$ |

In Figure 1 on this page, all possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b).

In Figure 1 on this page all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, Δ*S*, for the system when it is converted from distribution (b) to distribution (d).

## Solution

There are four initial microstates and four final microstates.

How does the process described in the previous item relate to the system shown in Figure 1 on this page?

Consider a system similar to the one in Figure 1 on this page, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to $\frac{1}{8}$ What does this comparison tell us about even larger systems?

## Solution

The probability for all the particles to be on one side is $\frac{1}{32}$. This probability is noticeably lower than the $\frac{1}{8}$ result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.

Consider the system shown in Figure 1 on this page. What is the change in entropy for the process where the energy is initially associated only with particle A, but in the final state the energy is distributed between two different particles?

Consider the system shown in Figure 1 on this page. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?

## Solution

There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states.

Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set.

(a) H_{2}(*g*), HBrO_{4}(*g*), HBr(*g*)

(b) H_{2}O(*l*), H_{2}O(*g*), H_{2}O(*s*)

(c) He(*g*), Cl_{2}(*g*), P_{4}(*g*)

At room temperature, the entropy of the halogens increases from I_{2} to Br_{2} to Cl_{2}. Explain.

## Solution

The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I_{2} is a solid, Br_{2} is a liquid, and Cl_{2} is a gas.

Consider two processes: sublimation of I_{2}(*s*) and melting of I_{2}(*s*) (Note: the latter process can occur at the same temperature but somewhat higher pressure).

${\text{I}}_{2}(s)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{I}}_{2}(g)$

${\text{I}}_{2}(s)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{I}}_{2}(l)$

Is Δ*S* positive or negative in these processes? In which of the processes will the magnitude of the entropy change be greater?

Indicate which substance in the given pairs has the higher entropy value. Explain your choices.

(a) C_{2}H_{5}OH(*l*) or C_{3}H_{7}OH(*l*)

(b) C_{2}H_{5}OH(*l*) or C_{2}H_{5}OH(*g*)

(c) 2H(*g*) or H(*g*)

## Solution

(a) C_{3}H_{7}OH(*l*) as it is a larger molecule (more complex and more massive), and so more microstates describing its motions are available at any given temperature. (b) C_{2}H_{5}OH(*g*) as it is in the gaseous state. (c) 2H(*g*), since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).

Predict the sign of the entropy change for the following processes.

(a) An ice cube is warmed to near its melting point.

(b) Exhaled breath forms fog on a cold morning.

(c) Snow melts.

Predict the sign of the entropy change for the following processes. Give a reason for your prediction.

(a) ${\text{Pb}}^{\mathrm{2+}}(aq)+{\text{S}}^{\mathrm{2-}}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{PbS}(s)$

(b) $2\text{Fe}(s)+\frac{3}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{Fe}}_{2}{\text{O}}_{2}(s)$

(c) $2{\text{C}}_{6}{\text{H}}_{14}(l)+19{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}14{\text{H}}_{2}\text{O}(g)+12{\text{CO}}_{2}(g)$

## Solution

(a) Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution. (b) Negative. There is a net loss of three moles of gas from reactants to products. (c) Positive. There is a net increase of seven moles of gas from reactants to products.

Write the balanced chemical equation for the combustion of methane, CH_{4}(*g*), to give carbon dioxide and water vapor. Explain why it is difficult to predict whether Δ*S* is positive or negative for this chemical reaction.

Write the balanced chemical equation for the combustion of benzene, C_{6}H_{6}(*l*), to give carbon dioxide and water vapor. Would you expect Δ*S* to be positive or negative in this process?

## Solution

${\text{C}}_{6}{\text{H}}_{6}(l)+7.5{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6{\text{3H}}_{2}\text{O(}g)+{\text{6CO}}_{2}(g)$

There are 7.5 moles of gas initially, and 3 + 6 = 9 moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and Δ*S* is positive.