7.3: Entropy and Heat - Experimental Basis of the Second Law of Thermodynamics
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Learning Objectives
- The Learning Objective of this Module is to understand the relationship between internal energy and entropy.
Thermodynamic Definition of Entropy
Experiments show that the magnitude of ΔS vap 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 \(\PageIndex{1}\), 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 \(\ce{NaCl}\) 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 \(\ce{Na^{+}}\) 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 \(\ce{NaCl}\) solution, disruption of the crystalline \(\ce{NaCl}\) structure and the hydrogen-bonded interactions in water is quantitatively more important, so \(ΔS_{soln} > 0\).
Dissolving \(\ce{NaCl}\) 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.
Example \(\PageIndex{1}\)
Predict which substance in each pair has the higher entropy and justify your answer.
- 1 mol of NH 3 (g) or 1 mol of He(g), both at 25°C
- 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.
Solution :
- Both substances are gases at 25°C, but one consists of He atoms and the other consists of NH 3 molecules. With four atoms instead of one, the NH 3 molecules have more motions available, leading to a greater number of microstates. Hence we predict that the NH 3 sample will have the higher entropy.
- 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.
Exercise \(\PageIndex{1}\)
Predict which substance in each pair has the higher entropy and justify your answer.
- 1 mol of He(g) at 10 K and 1 atm pressure or 1 mol of He(g) at 250°C and 0.2 atm
- a mixture of 3 mol of H 2 (g) and 1 mol of N 2 (g) at 25°C and 1 atm or a sample of 2 mol of NH 3 (g) at 25°C and 1 atm
- Answer a
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1 mol of He(g) at 250°C and 0.2 atm (higher temperature and lower pressure indicate greater volume and more microstates)
- Answer b
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a mixture of 3 mol of H 2 (g) and 1 mol of N 2 (g) at 25°C and 1 atm (more molecules of gas are present)