The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{~K}\) is zero. In Appendix \(4, \mathrm{~F}^{-}(a q), \mathrm{OH}^{-}(a q)\), and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

Short Answer

Expert verified
The negative standard entropy values for F^-, OH^-, and S^2- ions do not contradict the third law of thermodynamics because these values are relative quantities compared to a reference substance at a pressure of 1 bar and temperature of 298 K. The negative values indicate that these ions are more ordered compared to the reference substances, in their aqueous solution, due to ion-dipole interactions with water molecules. These values should not be misinterpreted as implying negative entropy at absolute zero.

Step by step solution

01

Understand Entropy

Entropy (S) is a measure of the disorder or randomness of a system. According to the third law of thermodynamics, the entropy of a perfect crystal at 0 K is assumed to be zero. This is because, in a perfect crystal, all particles are in a well-ordered state at absolute zero (0 K).
02

Comparing Entropy Values to Absolute Zero

The negative values of the standard entropy (S°) of F^-, OH^-, and S^2- indicate that their entropy values are less than that of a perfect crystal at 0 K. This might seem counterintuitive, given the third law of thermodynamics. However, it is important to note that these negative entropy values are obtained through comparisons and measurements relative to other substances, not by direct measurements of the entropy at absolute zero.
03

Understand Standard Entropy

Standard entropy (S°) refers to the entropy of a substance at a reference state, typically at a pressure of 1 bar and temperature of 298 K. The standard entropy values, like any other thermodynamic properties, are based on a reference standard. Just like how the standard enthalpy of formation for an element in its standard state is defined to be zero, the standard entropy values are relative quantities based on the reference.
04

Comparing to a Reference Substance

The negative standard entropy values for F^-, OH^-, and S^2- ions essentially indicate that these ions are more ordered than the reference substances to which they are compared. It should be noted that these negative values are for the entropy of the solvated ions in their aqueous solution, which means that we are comparing the entropy of these ions in water to their reference state. In that respect, the negative entropy values make sense as adding ions in an aqueous solution creates a more ordered environment around the ions due to the formation of ion-dipole interactions with water molecules. In conclusion, the negative values of standard entropy for F^-, OH^-, and S^2- ions do not contradict the third law of thermodynamics. These values signify that the disorder or randomness is less than that of the reference substances with which they are compared, and should not be misinterpreted as implying negative entropy at absolute zero.

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Most popular questions from this chapter

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

Consider the following reaction: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}\), \(P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)

Given the following data: $$ \begin{aligned} 2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow & 12 \mathrm{CO}_{2}(g)+& 6 \mathrm{H}_{2} \mathrm{O}(l) \\ & \Delta G^{\circ}=-6399 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta G^{\circ}\) for the reaction $$ 6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l) $$

Consider a weak acid, HX. If a \(0.10-M\) solution of HX has a \(\mathrm{pH}\) of \(5.83\) at \(25^{\circ} \mathrm{C}\), what is \(\Delta G^{\circ}\) for the acid's dissociation reaction at \(25^{\circ} \mathrm{C} ?\)

Consider the system $$ \mathrm{A}(g) \longrightarrow \mathrm{B}(g) $$ at \(25^{\circ} \mathrm{C}\). a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}\), calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if \(1.00 \mathrm{~mole}\) of \(\mathrm{A}(g)\) at \(1.00\) atm and \(1.00\) mole of \(\mathrm{B}(g)\) at \(1.00 \mathrm{~atm}\) are mixed at \(25^{\circ} \mathrm{C}\). c. Show by calculations that \(\Delta G=0\) at equilibrium.

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