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Chapter 19: Problem 6

Isomersare moleculesthat havethesamechemical 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) The difference in the enthalpy of combustion between the two isomers would not be significant because they have similar bond energies in their structures. (b) The linear isomer (n-pentane) is expected to have a higher standard molar entropy than the branched isomer (2,2-dimethylpropane) due to its flexible structure, which allows for more conformational possibilities and contributes to higher disorder in the system.

Step by step solution

01

Understanding Enthalpy of Combustion #

Enthalpy of combustion is the heat released when one mole of a substance undergoes complete combustion with oxygen at constant pressure. In this case, the combustion reaction for pentane isomers will be: \(C_5H_{12} + \frac{15}{2} O_2 \rightarrow 5 CO_2 + 6 H_2O\) Both isomers have the same formula (C5H12), so they will have the same number of moles of reactants and products. Therefore, we will analyze the bond energy to find the difference between their enthalpy of combustion.
02

Analyzing Bond Energy #

The enthalpy change in a chemical reaction depends mainly on the difference in bond energies between the reactants and the products. The combustion of both isomers will lead to the formation of CO2 and H2O, which have the same bond energies. Since both isomers have the same chemical formula, they will have similar bond energies in terms of carbon-carbon and carbon-hydrogen bonds.
03

Answer for Part (a) #

Based on the analysis of bond energies, we can expect that the difference in the enthalpy of combustion between the two isomers would not be significant because they have similar bond energies in their structures.
04

Understanding Entropy #

Entropy is a thermodynamic property that represents the randomness or disorder of a system. In the context of isomers, entropy is influenced by the number and nature of atoms, molecular geometry, and molecular flexibility. The higher the entropy, the higher the disorder in the system.
05

Analyzing Isomers' Structure #

We are given two isomers of pentane that have different arrangements of atoms: 1. n-pentane (linear chain): CH3-CH2-CH2-CH2-CH3 2. 2,2-dimethylpropane (branched chain): (CH3)3C-CH3 Isomer 1 has a linear structure, which can rotate around carbon-carbon single bonds, resulting in more conformational possibilities. On the other hand, isomer 2 has a more static structure due to the central carbon being bonded to three identical methyl groups.
06

Answer for Part (b) #

The linear isomer (n-pentane) is expected to have a higher standard molar entropy than the branched isomer (2,2-dimethylpropane) because of its flexible structure, which allows for more conformational possibilities and therefore contributes to higher disorder in the system.

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Most popular questions from this chapter

Consider a system that consists of two standard playing dice, with the state of the system defined by the sum of the values shown on the top faces. (a) The two arrangements of top faces shown here can be viewed as two possible microstates of the system. Explain. (b) To which state does each microstate correspond? (c) How many possible states are there for the system? (d) Which state or states have the highest entropy? Explain. (e) Which state or states have the lowest entropy? Explain. (f) Calculate the absolute entropy of the two-dice system.

Using data from Appendix \(\mathrm{C}\), write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free- energy change for these reactions at \(298 \mathrm{~K}:\) (a) $\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)$ (b) $2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)$ (c) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$

Use data from Appendix \(C\) to calculate the equilibrium constant, \(K,\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) (b) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ (c) $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)$

Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta H\) zero for the process?

Using the data in Appendix \(C\) and given the pressures listed, calculate \(K_{\mathrm{p}}\) and \(\Delta G\) for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \\ \quad R_{\mathrm{N}_{2}}=263.4 \mathrm{kPa}, P_{\mathrm{H}_{2}}=597.8 \mathrm{kPa}, P_{\mathrm{NH}_{3}}=101.3 \mathrm{kPa} \\ \text { (b) } 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.07 \mathrm{kPa}\) $$ \begin{array}{l} \quad R_{\mathrm{N}_{2}}=50.7 \mathrm{kPa}, P_{\mathrm{H}_{2} \mathrm{O}}=30.4 \mathrm{kPa} \\ \text { (c) } \mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \\ P_{\mathrm{N}_{2} \mathrm{H}_{4}}=101.3 \mathrm{kPa}, P_{\mathrm{N}_{2}}=152.0 \mathrm{kPa}, P_{\mathrm{H}_{2}}=253.3 \mathrm{kPa} \end{array} $$
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