6.6 Entropy

Entropy, denoted by the symbol S, is a thermodynamic property that measures the degree of disorder or randomness in a system. The more disordered a system is, the higher its entropy.

• A system moving from a more ordered state to a less ordered state experiences an increase in entropy.
• Conversely, a system moving from a less ordered state to a more ordered state experiences a decrease in entropy.

Entropy and States of Matter

The state of matter is a key factor in determining a substance's entropy.

• Solids: Particles are in a fixed, ordered arrangement with minimum energy and movement. They have the lowest entropy.
• Liquids: Particles have more freedom of movement than in solids but are still relatively close. They have intermediate entropy.
• Gases: Particles are far apart and move randomly and rapidly. They have the highest entropy.

The general trend is:

$S_{\text{gas}}>S_{\text{liquid}}>S_{\text{solid}}$

Example: Melting Ice

• In ice (solid), water molecules are in a fixed, ordered crystal lattice ($S_{\text{solid}}$ is low).
• As ice melts into water (liquid), the molecules gain energy and freedom of movement, becoming more disordered. Entropy increases ($S_{\text{liquid}}>S_{\text{solid}}$).
• When water boils into steam (gas), the molecules become highly disordered and spread out, leading to a further, significant increase in entropy ($S_{\text{gas}}>S_{\text{liquid}}$).

$H_2{O}_{(s)}\to H_2{O}_{(l)}\to H_2{O}_{(g)}$

Entropy increases $\to$

Entropy Change ($\Delta S$)

The change in entropy for a process is denoted as $\Delta S$.

• If the entropy of the system increases (more disorder), the change is positive: $\Delta S>0$.
• If the entropy of the system decreases (more order), the change is negative: $\Delta S<0$.

Effect of Temperature

• Increasing the temperature of a system increases the kinetic energy of its particles, leading to more random motion and thus higher entropy ($\Delta S>0$).
• Decreasing the temperature reduces particle motion, making the system more ordered and lowering its entropy ($\Delta S<0$).

Entropy Change During a Reversible Process

For a reversible process, the entropy change is related to the heat transferred and the absolute temperature:

$\Delta S=\frac{q_{\text{rev}}}{T}$

where:

• $q_{\text{rev}}$ = heat transferred reversibly (in Joules)
• $T$ = absolute temperature (in Kelvin)
• $\Delta S$ = entropy change (in J/K)

A positive $\Delta S$ indicates increased disorder; a negative $\Delta S$ indicates increased order.

Calculation of the Standard Entropy Change

The standard entropy change ($\Delta S^{\circ }$) for a chemical reaction can be calculated by subtracting the sum of the standard entropies of the reactants from the sum of the standard entropies of the products.

The general equation is:

$\Delta S_{rxn}^{\circ }=\Sigma S_{\text{products}}^{\circ }-\Sigma S_{\text{reactants}}^{\circ }$

Note: The coefficients from the balanced chemical equation must be included in the calculation.

As a general rule:

• If the number of moles of gas increases during a reaction, $\Delta S^{\circ }$ is typically positive.
• If the number of moles of gas decreases, $\Delta S^{\circ }$ is typically negative.

Worked Example

Calculate the standard entropy change, $\Delta S^{\circ }$, for the following reaction:

$A+2B\to C$

Given:

• $S^{\circ }(A)=40\text{ J/mol⋅K}$
• $S^{\circ }(B)=50\text{ J/mol⋅K}$
• $S^{\circ }(C)=100\text{ J/mol⋅K}$

Solution:

1. Write the formula:

$\Delta S^{\circ }=\Sigma S_{\text{products}}^{\circ }-\Sigma S_{\text{reactants}}^{\circ }$

2. Apply the formula:

$\Delta S^{\circ }=[1\times S^{\circ }(C)]-[1\times S^{\circ }(A)+2\times S^{\circ }(B)]$

3. Substitute and calculate:

$\Delta S^{\circ }=[1\times (100)]-[1\times (40)+2\times (50)]$ $\Delta S^{\circ }=100-[40+100]$ $\Delta S^{\circ }=100-140$ $\Delta S^{\circ }=-40\text{ J/mol⋅K}$

The result is negative, indicating a decrease in entropy for this reaction.

Predicting the Sign of $\Delta S$

Q: For which of the following reactions is $\Delta S^{\circ }$ positive or negative?

a) $C{O}_{2(g)}\to C{O}_{2(s)}$

A: Negative. The process involves a transition from a highly disordered gaseous state to a very ordered solid state (deposition). This significant increase in order results in a decrease in entropy ($\Delta S<0$).

b) $2H_{2(g)}+O_{2(g)}\to 2H_2{O}_{(l)}$

A: Negative. The reaction starts with 3 moles of gas (2 mol $H_2$ + 1 mol $O_2$) and ends with 2 moles of liquid. There is a decrease in the number of particles and a transition from the highly disordered gas phase to the more ordered liquid phase. Both factors contribute to a decrease in entropy ($\Delta S<0$).

c) $H_2{O}_{(l)}\to H_2{O}_{(g)}$

A: Positive. This is the process of vaporization (boiling). The molecules transition from the relatively ordered liquid state to the highly disordered gaseous state. This increase in randomness means the entropy increases ($\Delta S>0$).

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