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Chapter 19: Problem 6

Isomers are molecules that have the same chemical formula but different arrangements of atoms, as shown here for two isomers of pentane, \(\mathrm{C}_{5} \mathrm{H}_{12} .\) (a) Do you expect a significant difference in the enthalpy of combustion of the two isomers? Explain. (b) Which isomer do you expect to have the higher standard molar entropy? Explain. [Section 19.4\(]\)

Short Answer

Expert verified
(a) We expect a slight difference in the enthalpy of combustion of the two isomers due to the different arrangements of atoms affecting the strength of their chemical bonds. However, this difference is generally small since their overall chemical compositions remain the same. (b) Without the specific structures of the two isomers, it is impossible to definitively determine which one has a higher standard molar entropy. However, the isomer with a more complex structure, more possible conformations, or more rotational freedom will have a higher standard molar entropy.

Step by step solution

01

Understanding Enthalpy of Combustion

Enthalpy of combustion is the amount of heat released when a substance is completely burned in excess oxygen. It depends on the strength and number of chemical bonds in the molecules, as well as their arrangement.
02

Understanding Standard Molar Entropy

Standard molar entropy is a measure of the randomness or disorder of a substance. In general, larger and more complex molecules have higher standard molar entropy. Molecules with more possible conformations or arrangements also tend to have higher standard molar entropy.
03

Addressing Enthalpy of Combustion (Part a)

Since isomers have the same chemical formula, the overall number and types of atoms are the same. However, their arrangement of atoms may affect the strength of their chemical bonds to some extent. If the difference in the arrangement of atoms is significant, it can result in a small difference in the enthalpy of combustion. However, it is generally expected that the enthalpy of combustion of isomers remains relatively similar because the overall chemical composition remains the same.
04

Addressing Standard Molar Entropy (Part b)

To determine which isomer has a higher standard molar entropy, we need to analyze their molecular structures. The isomer with a more complex structure, more possible conformations, or more rotational freedom will have a higher standard molar entropy. Without knowing the specific structures of the two isomers, it is impossible to definitively determine which one has a higher standard molar entropy. However, we can make general predictions based on our understanding of entropy and its dependence on molecular structure.

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Most popular questions from this chapter

(a) Express the second law of thermodynamics in words. (b) If the entropy of the system increases during a reversible process, what can you say about the entropy change of the surroundings? (c) In a certain spontaneous process the system undergoes an entropy change, \(\Delta S=42 \mathrm{~J} / \mathrm{K} .\) What can you conclude about ?

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of \(\mathrm{I}_{2} .\) (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?

Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(\mathrm{C}\) to predict how \(\Delta \mathrm{G}^{\circ}\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G^{\circ}\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at \(800 \mathrm{~K} ?\) (c) Calculate \(\Delta G^{\circ}\) at \(1000 \mathrm{~K}\). Is the reaction spontaneous under standard conditions at this temperature?

A particular constant-pressure reaction is spontaneous at \(390 \mathrm{~K}\). The enthalpy change for the reaction is \(+23.7 \mathrm{~kJ}\). What can you conclude about the sign and magnitude of \(\Delta S\) for the reaction?

Using data in Appendix C, calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} .\) (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) \(\mathrm{C}(s,\) graphite \()+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)\) (c) \(2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{POCl}_{3}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)
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