# 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*. **Example: Melting Ice** - In ice (solid), water molecules are in a fixed, ordered crystal lattice ($S_{solid}$ is low). - As ice melts into water (liquid), the molecules gain energy and freedom of movement, becoming more disordered. Entropy increases ($S_{liquid} > S_{solid}$). - When water boils into steam (gas), the molecules become highly disordered and spread out, leading to a further, significant increase in entropy ($S_{gas} > S_{liquid}$). $$ H_2O_{(s)} \rightarrow H_2O_{(l)} \rightarrow H_2O_{(g)} $$ *Entropy increases* $\rightarrow$ ### 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 ($\Delta S < 0$). ### Calculation of the 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: $$ \Delta S^{\circ}_{rxn} = \Sigma S^{\circ}_{\text{products}} - \Sigma S^{\circ}_{\text{reactants}} $$ *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, the entropy change ($\Delta S$) is typically positive. If the number of moles of gas decreases, it is typically negative. ### Worked Examples Calculate the standard entropy change, $\Delta S^{\circ}$, for the following reaction: $$ A + 2B \rightarrow C $$ **Given:** - Standard entropy of A, $S^{\circ}(A) = 40 \text{ J/mol·K}$ - Standard entropy of B, $S^{\circ}(B) = 50 \text{ J/mol·K}$ - Standard entropy of C, $S^{\circ}(C) = 100 \text{ J/mol·K}$ **Solution:** 1. **Write the formula:** $$ \Delta S^{\circ} = \Sigma S^{\circ}_{\text{products}} - \Sigma S^{\circ}_{\text{reactants}} $$ 2. **Apply the formula to the specific reaction:** $$ \Delta S^{\circ} = [1 \times S^{\circ}(C)] - [1 \times S^{\circ}(A) + 2 \times S^{\circ}(B)] $$ 3. **Substitute the given values 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. ### Possible Questions/Answers **Q: For which of the following reactions is $\Delta S^{\circ}$ positive or negative?** **a) $CO_{2(g)} \rightarrow CO_{2(s)}$** **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) $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$** **A:** **Negative**. The reaction starts with 3 moles of gas (2 moles of $H_2$ + 1 mole of $O_2$) 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) $H_2O_{(l)} \rightarrow H_2O_{(g)}$** **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 ($\Delta S > 0$). ### Maxwell's Demon and Entropy In the context of statistical mechanics, the Maxwell's Demon paradox was proposed to challenge the Second Law. However, resolution of the paradox shows that the entropy of the entire system (including the demon) must increase.