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Chapter 5: Problem 50

What is the enthalpy change for the isomerization reaction: \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}(\mathrm{~A})\) \(\underset{\Delta}{\stackrel{\mathrm{NaNH}_{2}}{\longrightarrow}} \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}\) \(=\mathrm{CH}-\mathrm{CH}_{3}(\mathrm{~B})\) Magnitude of resonance energies of \(\mathrm{A}\) and \(\mathrm{B}\) are 50 and \(70 \mathrm{~kJ} / \mathrm{mol}\), respectively. Enthalpies of formation of \(\mathrm{A}\) and \(\mathrm{B}\) are \(-2275.2\) and \(-2839.2 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) \(-584 \mathrm{~kJ}\) (b) \(-564 \mathrm{~kJ}\) (c) \(-544 \mathrm{~kJ}\) (d) \(-20 \mathrm{~kJ}\)

Short Answer

Expert verified
-544 kJ/mol

Step by step solution

01

Understand the isomerization reaction

The given reaction is an isomerization in which the molecular formula remains the same, but the structure rearranges from A to B. The reactant is A and the product is B.
02

Define the enthalpy change of reaction

The enthalpy change for the isomerization reaction, \(\Delta H_{reaction}\), is defined as the difference between the enthalpy of formation of the products and the reactants.
03

Calculate the enthalpy change

Substitute the given enthalpies of formation for A and B, and the resonance energies for A and B into the enthalpy change formula:\[\Delta H_{reaction} = (\Delta H_{f, \text{B}} + E_{\text{resonance, B}}) - (\Delta H_{f, \text{A}} + E_{\text{resonance, A}})\]where \(\Delta H_{f, \text{A}}\) and \(\Delta H_{f, \text{B}}\) are enthalpies of formation for A and B, and \(E_{\text{resonance, A}}\) and \(E_{\text{resonance, B}}\) are the resonance energies for A and B, respectively.
04

Substitute in the given values

Using the provided values:\[\Delta H_{reaction} = (-2839.2 \text{ kJ/mol} + 70 \text{ kJ/mol}) - (-2275.2 \text{ kJ/mol} + 50 \text{ kJ/mol})\]
05

Perform the calculation

Now, calculate the enthalpy change:\[\Delta H_{reaction} = (-2839.2 + 70) - (-2275.2 + 50)\]\[\Delta H_{reaction} = -2769.2 \text{ kJ/mol} - (-2225.2 \text{ kJ/mol})\]\[\Delta H_{reaction} = -2769.2 \text{ kJ/mol} + 2225.2 \text{ kJ/mol}\]\[\Delta H_{reaction} = -544 \text{ kJ/mol}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isomerization Reaction
An isomerization reaction involves the rearrangement of atoms within a molecule to form a different molecule with the same molecular formula but a different structure. Despite having the same types and numbers of atoms, isomers can exhibit vastly different chemical and physical properties. In the context of organic chemistry, such as the provided exercise, isomerization can significantly impact the stability and reactivity of the compounds involved.

An easy way to picture isomerization is to think of it like rearranging the rooms in a house. Even though the same rooms are present, changing their connections and layout can result in a very different home structure. Similarly, in isomerization reactions, the 'rooms' (atoms or groups of atoms) are rearranged to form a 'new house' (isomer).

The reason behind studying isomerization reactions in chemistry is paramount since it helps us understand the conversion processes that occur in various chemical and biological systems, such as the energy-releasing pathways in our bodies and the manufacturing of different synthetic materials.
Resonance Energy
Resonance energy is an important concept in chemistry that describes the stability of a molecule compared to its 'canonical' forms鈥攖he different ways we can draw the molecule's structure on paper using Lewis structures. The actual structure of a molecule is a hybrid of these canonical forms and does not change or resonate between them; instead, the electrons are delocalized over the entire structure, giving it a stability that is not represented in any one canonical form.

Think of resonance energy like the extra stability you'd have if you could spread your weight evenly on a hammock, rather than sitting on a single, wobbly stool. In molecular terms, the more ways we can draw a stable structure for a molecule (without changing the positions of the atoms), the more stable the molecule is due to this delocalization of electrons.

In the given exercise, the resonance energies of isomers A and B significantly influence the overall stability of these compounds. Isomer B, with a higher resonance energy, is inherently more stable than isomer A, resulting in a lower potential energy state.
Enthalpy of Formation
The enthalpy of formation, represented by ewline ewline

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Most popular questions from this chapter

Enthalpy of neutralization of \(\mathrm{H}_{3} \mathrm{PO}_{3}\) by \(\mathrm{NaOH}\) is \(-106.68 \mathrm{~kJ} / \mathrm{mol}\). If the enthalpy of neutralization of \(\mathrm{HCl}\) by \(\mathrm{NaOH}\) is \(-55.84 \mathrm{~kJ} / \mathrm{mol}\). The \(\Delta H_{\text {ionization }}\) of \(\mathrm{H}_{3} \mathrm{PO}_{3}\) into its ions is (a) \(50.84 \mathrm{~kJ} / \mathrm{mol}\) (b) \(5 \mathrm{~kJ} / \mathrm{mol}\) (c) \(10 \mathrm{~kJ} / \mathrm{mol}\) (d) \(2.5 \mathrm{~kJ} / \mathrm{mol}\)

The value of \(\Delta H_{\text {sol }}\) of anhydrous \(\begin{array}{lllll}\text { copper (II) sulphate } & \text { is } & -66.11 & \mathrm{~kJ}\end{array}\) Dissolution of 1 mole of blue vitriol, [Copper (II) sulphate pentahydrate] is followed by absorption of \(11.5 \mathrm{~kJ}\) of heat. The enthalpy of dehydration of blue vitriol is (a) \(-77.61 \mathrm{~kJ}\) (b) \(+77.61 \mathrm{~kJ}\) (c) \(-54.61 \mathrm{~kJ}\) (d) \(+54.61 \mathrm{~kJ}\)

The intermediate \(\mathrm{SiH}_{2}\) is formed in the thermal decomposition of silicon hydrides. Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) of \(\mathrm{SiH}_{2}\) from the following reactions: \(\mathrm{Si}_{2} \mathrm{H}_{6}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SiH}_{4}(\mathrm{~g})\) \(\Delta H^{\circ}=-11.7 \mathrm{~kJ} / \mathrm{mol}\) \(\mathrm{SiH}_{4}(\mathrm{~g}) \rightarrow \mathrm{SiH}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) \(\Delta H^{\circ}=+239.7 \mathrm{~kJ} / \mathrm{mol}\) \(\Delta H_{\mathrm{f}}^{\circ}, \mathrm{Si}_{2} \mathrm{H}_{6}(\mathrm{~g})=+80.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (a) \(353 \mathrm{~kJ} / \mathrm{mol}\) (b) \(321 \mathrm{~kJ} / \mathrm{mol}\) (c) \(198 \mathrm{~kJ} / \mathrm{mol}\) (d) \(274 \mathrm{~kJ} / \mathrm{mol}\)

A geyser, operating on LPG (liquefied petroleum gas) heats water flowing at the rate of \(3.0\) litres per minutes, from \(27^{\circ} \mathrm{C}\) to \(77^{\circ} \mathrm{C}\). If the heat of combustion of LPG is \(40,000 \mathrm{~J} / \mathrm{g}\), how much fuel, in \(\mathrm{g}\), is consumed per minute? (Specific heat capacity of water is \(4200 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) ) (a) \(15.25\) (b) \(15.50\) (c) \(15.75\) (d) \(16.00\)

The enthalpy of neutralization of a strong monobasic acid by a strong monoacidic base is \(-13,700\) cal. A certain monobasic weak acid is \(10 \%\) ionized in a molar solution. If the enthalpy of ionization of the weak acid is \(+400 \mathrm{cal} / \mathrm{mole}\), what is the enthalpy of neutralization of one molar solution of the weak acid? (a) \(-13,700 \mathrm{cal}\) (b) \(-13,340 \mathrm{cal}\) (c) \(-13,660\) cal (d) \(-13,300\) cal
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