Entropy: A Measure of Disorder and Energy Dispersal.
What Is Entropy?
Entropy
Entropy, denoted as $S$, is a thermodynamic property that quantifies the dispersal or distribution of energy and matter in a system.
- When energy or matter becomes more spread out, the system's entropy increases.
- Conversely, when energy or matter becomes more concentrated, entropy decreases.
Entropy as Disorder
Entropy is often referred to as a measure of "disorder" or "randomness."
- Low entropy: Imagine a tidy room where everything is in its assigned place.
- High entropy: Now picture a messy room with items scattered randomly.
- While this analogy is useful, it’s important to understand that entropy is rigorously defined by the number of "microstates" available to a system.
- Microstates refer to the different ways energy can be distributed among particles.
Entropy is not solely about "messiness" but about the distribution of energy and matter within a system.
Entropy Comparisons: Solids, Liquids, and Gases
The state of matter significantly influences entropy. Under identical conditions:
- Gases exhibit the highest entropy because their particles can move freely in all directions, resulting in a vast number of possible arrangements.
- Liquids have lower entropy than gases, as their particles are more constrained in motion.
- Solids have the lowest entropy due to their structured, fixed arrangement.
This trend can be summarized as:
$$
S_{\text{gas}} > S_{\text{liquid}} > S_{\text{solid}}
$$
Entropy Changes During Phase Transitions
Entropy changes during phase transitions because the arrangement and motion of particles evolve:
- Melting (solid → liquid): Entropy increases ($\Delta S > 0$) as particles gain more freedom to move.
- Vaporization (liquid → gas): Entropy increases significantly ($\Delta S > 0$) because particles become highly dispersed.
- Condensation (gas → liquid): Entropy decreases ($\Delta S< 0$) as particles become more ordered.
Melting Ice
- When ice melts into water, the rigid hydrogen-bonded structure of ice collapses, allowing water molecules to move more freely.
- This increases the system's entropy.
Many students mistakenly believe that all exothermic processes decrease entropy. This is incorrect! Entropy depends on how energy and matter are distributed, not just on heat flow.
Calculating Standard Entropy Change ($\Delta S^\circ$)
- Entropy is a state function, meaning its value depends only on the initial and final states of the system, not on the path taken.
- The standard entropy change for a reaction, $\Delta S^\circ$, can be calculated using the standard entropy values ($S^\circ$) of the reactants and products, which are provided in the IB Chemistry data booklet.
$$
\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})
$$
- $S^\circ$ values are expressed in $\text{J K}^{-1} \ \text{mol}^{-1}$.
- Multiply each $S^\circ$ value by its stoichiometric coefficient from the balanced chemical equation.
Always use the coefficients from the balanced equation when calculating $\Delta S^\circ$.
Consider the reaction:
$$\text{N}_2\text{O}_4(g) \rightarrow 2\text{NO}_2(g)$$
From the data booklet:
$$S^\circ(\text{N}_2\text{O}_4) = 304 \, \text{J K}^{-1} \text{mol}^{-1}$$
$$S^\circ(\text{NO}_2) = 240 \, \text{J K}^{-1} \text{mol}^{-1}$$
$$\Delta S^\circ = [2 \times S^\circ(\text{NO}_2)] - [S^\circ(\text{N}_2\text{O}_4)]$$
$$\Delta S^\circ = [2 \times 240] - [304] = 480 - 304 = +176 \, \text{J K}^{-1} \text{mol}^{-1}$$
The positive $\Delta S^\circ$ indicates an increase in entropy, consistent with the formation of two gas molecules from one.
Predicting Entropy Changes
You can often predict whether a reaction increases or decreases entropy by considering:
- States of Matter:
- Reactions producing gases generally increase entropy.
- Reactions consuming gases generally decrease entropy.
- Number of Particles:
- If the total number of product molecules exceeds the total number of reactant molecules, entropy likely increases.
- If the number decreases, entropy likely decreases.
For the reaction:
$$
\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)
$$
- The reactants include 1 solid.
- The products include 1 solid and 1 gas.
- The formation of a gas increases the dispersal of matter, so $\Delta S > 0$.
Why Is the Entropy of a Perfect Crystal at 0 K Zero?
- At absolute zero (0 K), a perfect crystal has no thermal motion, and its particles are perfectly ordered.
- According to the Third Law of Thermodynamics, the entropy of a perfect crystal at 0 K is zero.
This is because there is only one possible arrangement (microstate) for the particles, meaning no randomness or dispersal of energy.
- Why does entropy increase when a liquid evaporates into a gas?
- Calculate $\Delta S^\circ$ for the reaction: $$
\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)
$$ Given:- $S^\circ(\text{H}_2) = 131 \, \text{J K}^{-1} \ \text{mol}^{-1}$,
- $S^\circ(\text{Cl}_2) = 223 \, \text{J K}^{-1} \ \text{mol}^{-1}$,
- $S^\circ(\text{HCl}) = 187 \, \text{J K}^{-1}\ \text{mol}^{-1}$.
