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Chapter 18: Problem 51

Under what conditions does a substance have a standard entropy of zero? Can a substance ever have a negative standard entropy?

Short Answer

Expert verified
A substance has a standard entropy of zero at absolute zero temperature (0 Kelvin). As for the possibility of a negative standard entropy, it is not possible because entropy measures the disorder within a system and cannot drop below the most ordered state, which is represented by zero.

Step by step solution

01

Zero Entropy Condition

Standard entropy of a substance is zero at absolute zero temperature (0 Kelvin). This is a consequence of the Third Law of Thermodynamics, which states that the entropy of a perfectly organized crystalline substance is zero at absolute zero temperature. This is because, at 0 Kelvin, a perfect crystal has only one possible microstate and its energy is well defined, so it is completely ordered.
02

Possibility of Negative Standard Entropy

A substance cannot have a negative standard entropy. The second law of thermodynamics states that in an isolated system, entropy can only increase over time. In addition, entropy measures the number of specific ways in which a system may be arranged, often taken to be a measure of 'disorder'. The most orderly situation can be represented with a 0 value (all particles in one state), so you cannot have 'less than the most orderly state', hence negative entropy doesn't make physical sense.

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Most popular questions from this chapter

Consider two carboxylic acids (acids that contain the \(-\mathrm{COOH}\) group \(): \mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid, \(K_{\mathrm{a}}=1.8 \times 10^{-5}\) ) and \(\mathrm{CH}_{2} \mathrm{ClCOOH}\) (chloroacetic acid, \(K_{\mathrm{a}}=1.4 \times 10^{-3}\) ). (a) Calculate \(\Delta G^{\circ}\) for the ionization of these acids at \(25^{\circ} \mathrm{C}\) (b) From the equation \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ},\) we see that the contributions to the \(\Delta G^{\circ}\) term are an enthalpy term \(\left(\Delta H^{\circ}\right)\) and a temperature times entropy term \(\left(T \Delta S^{\circ}\right)\). These contributions are listed below for the two acids: Which is the dominant term in determining the value of \(\Delta G^{\circ}\) (and hence \(K_{\mathrm{a}}\) of the acid)? (c) What processes contribute to \(\Delta H^{\circ} ?\) (Consider the ionization of the acids as a Bronsted acid-base reaction.) (d) Explain why the \(T \Delta S^{\circ}\) term is more negative for \(\mathrm{CH}_{3} \mathrm{COOH}\).

Define entropy. What are the units of entropy?

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce \(\mathrm{Ni}(\mathrm{CO})_{4},\) which is a gas and can therefore be separated from solid impurities: $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ Given that the standard free energies of formation of \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) are \(-137.3 \mathrm{~kJ} / \mathrm{mol}\) and \(-587.4 \mathrm{~kJ} / \mathrm{mol}\), respectively, calculate the equilibrium constant of the reaction at \(80^{\circ} \mathrm{C}\). Assume that \(\Delta G_{f}^{\circ}\) is temperature independent.

The standard enthalpy of formation and the standard entropy of gaseous benzene are \(82.93 \mathrm{~kJ} / \mathrm{mol}\) and \(269.2 \mathrm{~J} / \mathrm{K} \cdot\) mol, respectively. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}\) and \(\Delta G^{\circ}\) for the process at \(25^{\circ} \mathrm{C}\). Comment on your answers. $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$

From the values of \(\Delta H\) and \(\Delta S,\) predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}\) : Reaction \(\mathrm{A}: \Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) reaction \(\mathrm{B}: \Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C}\), at what temperature might it become spontaneous?
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