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Chapter 18: Problem 70

A carbon monoxide (CO) crystal is found to have entropy greater than zero at absolute zero of temperature. Give two possible explanations for this observation.

Short Answer

Expert verified
The carbon monoxide (CO) crystal's entropy is greater than zero at absolute zero due to the potential existence of a degenerate ground state and/or structural disorder.

Step by step solution

01

Concept of residual entropy

In some cases, substances can have what's known as 'residual entropy' at absolute zero (0 K). This most often occurs when a system possesses multiple microscopic arrangements (microstates) that are energetically equivalent. Hence, even at absolute zero, it's unclear which specific microstate the system is in, resulting in an entropy greater than zero.
02

Outline of Possible Explanation 1: Degenerate ground state

A possible explanation for CO crystal's non-zero entropy at absolute zero could be a degenerate ground state. A degenerate ground state is a state with minimum energy where there exist multiple possible equivalent microstates. In the case of a CO crystal, if the CO molecules can align in multiple possible ways with the same energy, there would be a non-zero entropy at absolute zero. This is because the system doesn't know which specific alignment to adopt as all such configurations are energetically equivalent.
03

Outline of Possible Explanation 2: Structural disorder

Another potential explanation might be structural disorder. Even if the energy states are not degenerate, if the CO molecules in the crystal do not adopt a fully ordered configuration at T=0 K, it can lead to non-zero entropy. This can occur because of some intrinsic asymmetry in the crystal structure, leading the system to have multiple equivalent (but structurally distinct) ways to arrange the CO molecules.

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Most popular questions from this chapter

Describe two ways that you could measure \(\Delta G^{\circ}\) of a reaction.

Use the following data to determine the normal boiling point, in kelvins, of mercury. What assumptions must you make in order to do the calculation? $$ \begin{aligned} \mathrm{Hg}(l): & \Delta H_{\mathrm{f}}^{\circ} &=0 \text { (by definition) } \\\ & S^{\circ} &=77.4 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \\ \mathrm{Hg}(g): & \Delta H_{\mathrm{f}}^{\circ} &=60.78 \mathrm{~kJ} / \mathrm{mol} \\ & S^{\circ} &=174.7 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \end{aligned} $$

Calculate \(\Delta G^{\circ}\) for the following reactions at \(25^{\circ} \mathrm{C}\) : (a) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\) See Appendix 2 for thermodynamic data.

Consider the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Fe}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Fe}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) $$ Calculate \(\Delta G^{\circ}\) for the reaction. \(K_{\mathrm{sp}}\) for \(\mathrm{Fe}(\mathrm{OH})_{2}\) is \(1.6 \times 10^{-14}\).

The reaction \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) proceeds spontaneously at \(25^{\circ} \mathrm{C}\) even though there is a decrease in the number of microstates of the system (gases are converted to a solid). Explain.
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