Entropy and Heat - Experimental Basis of the Second Law of Thermodynamics | Chemistry Optional Notes for UPSC PDF Download

Introduction

• Experiments show that the magnitude of ΔSvap is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔS_vap. For instance, ΔS_vap for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution.
• As illustrated in Figure  13.3.1, the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of  NaCl and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. 
• At the same time, however, each dissolved Na⁺ ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl⁻ ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an  NaCl solution, disruption of the crystalline  NaCl structure and the hydrogen-bonded interactions in water is quantitatively more important, so  ΔS_soln > 0.

Figure  13.3.1: The Effect of Solution Formation on Entropy

Dissolving  NaCl in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive.

Solved Example

Example: Predict which substance in each pair has the higher entropy and justify your answer.
(a) 1 mol of NH3(g) or 1 mol of He(g), both at 25°C
(b) 1 mol of Pb(s) at 25°C or 1 mol of Pb(l) at 800°C
Given: amounts of substances and temperature
Asked for: higher entropy
Strategy: From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy.
Ans: (a) Both substances are gases at 25°C, but one consists of He atoms and the other consists of NH3 molecules. With four atoms instead of one, the NH3 molecules have more motions available, leading to a greater number of microstates. Hence we predict that the NH3 sample will have the higher entropy.
(b) The nature of the atomic species is the same in both cases, but the phase is different: one sample is a solid, and one is a liquid. Based on the greater freedom of motion available to atoms in a liquid, we predict that the liquid sample will have the higher entropy.

Summary

A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases.