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The enthalpy of hydrogenation of cyclohexene is \(-119.5\) \(\mathrm{kJ} \mathrm{mol}^{-1}\). If resonance energy of benzene is \(-150.4 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\), its enthalpy of hydrogenation would be (a) \(-269.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-358.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-508.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-208.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Short Answer

Expert verified
The enthalpy of hydrogenation of benzene is \(-208.1\ \mathrm{kJ/mol}\) matching option (d).

Step by step solution

01

Understanding the Problem

We are provided with the enthalpy of hydrogenation of cyclohexene and the resonance energy of benzene. We need to calculate the enthalpy of hydrogenation of benzene based on these values. Enthalpy of hydrogenation typically measures the heat released when hydrogen is added to unsaturated molecules, like benzene. Resonance energy reflects the stability gained from the resonance.
02

Calculating Expected Enthalpy Without Resonance

First, calculate the hypothetical enthalpy of hydrogenation for benzene if it did not have resonance stabilization. Since benzene has 3 double bonds, we multiply the enthalpy of hydrogenation of cyclohexene by 3. Thus, the calculation is: \(-119.5 \times 3 = -358.5\ \mathrm{kJ/mol}\).
03

Incorporating Resonance Energy

The real enthalpy of hydrogenation of benzene needs to consider the resonance stabilization. Therefore, we add the resonance energy to adjust for this stabilization: \(-358.5 + 150.4 = -208.1\ \mathrm{kJ/mol}\).
04

Identifying the Correct Option

After calculation, the enthalpy of hydrogenation of benzene is \(-208.1\ \mathrm{kJ/mol}\). Thus, the correct option is the one that matches this value.

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Key Concepts

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

Resonance Energy
Resonance energy is a concept in chemistry that helps explain the stability of certain molecules beyond what might be expected from their structural arrangement alone. It's the additional stability a molecule gains from resonance, which occurs when there are multiple ways to draw the structure of a molecule without changing its actual chemical framework. This is common in aromatic compounds, like benzene, where electrons are delocalized over several atoms.
When benzene is involved, the electrons in the carbon atoms' π bonds are not confined to a single pair of carbon atoms. Instead, they are spread over the entire ring. This results in a lower overall energy state than would be the case if the electrons were localized. The difference in this energy state is what is known as the resonance energy.
For benzene, this energy is quite significant, around -150.4 ext{ kJ/mol}. This large energy value highlights just how much additional stability the resonance imparts to benzene's structure.
Benzene Stability
The stability of benzene is one of its most fascinating and defining characteristics. Benzene rings are a classic example of an aromatic system, where all atoms participate in a delocalized bonding system. This delocalization forms a cloud of electrons above and below the plane of the hexagon formed by the carbon atoms.
The stability of benzene is largely attributed to this delocalization of its electrons, which allows benzene to maintain an even distribution of electrons across the entire molecule. This results in a lower reactivity compared to alkenes and other unsaturated hydrocarbons.
  • The presence of resonance means benzene does not participate in addition reactions as readily as other unsaturated compounds might.
  • Instead, substitution reactions are more common, preserving the aromatic ring structure.
Benzene's stability is a direct consequence of both its geometric ring structure and resonance, making it a cornerstone in the study of organic chemistry.
Chemical Thermodynamics
Understanding chemical thermodynamics is key to grasping the concepts related to energy changes in chemical reactions. Thermodynamics considers the energy transitions in reactions, which include enthalpy changes, such as those in hydrogenation.
When discussing the enthalpy of hydrogenation, we're looking at the change in heat content when hydrogen is added to unsaturated molecules. This is an exothermic process, meaning it releases energy.
For benzene, calculating the real enthalpy of hydrogenation requires considering the resonance energy. Without this, benzene would release -358.5 ext{ kJ/mol} in energy upon hydrogenation. However, the actual enthalpy is less exothermic due to the stabilizing effect of resonance.
  • This example highlights the integral role energy changes (enthalpy) play in chemical stability and reaction spontaneity.
  • Thermodynamics helps predict and explain why certain reactions occur and their potential energy flow.
Unsaturated Molecules
Unsaturated molecules are those that contain double or triple bonds, which allow them to participate in addition reactions, where elements are added across their unsaturated bonds. The unsaturation in these molecules provides locations for potential chemical reactivity, especially in reactions with hydrogen or halogens.
In the context of benzene, each carbon-carbon bond is somewhere between a single and a double bond due to resonance. This uncommon scenario means benzene behaves differently compared to other typical unsaturated hydrocarbons.
While typical unsaturated molecules like alkenes readily participate in addition reactions, the aromatic nature of benzene leads to different behaviors, favoring substitution reactions instead.
  • Its rings are stabilized by a delocalized electronic structure, resulting in unique chemical properties.
  • This makes benzene less reactive towards addition than expected for typical unsaturated compounds but quite stable overall.
Understanding how unsaturated molecules behave differently depending on their structure is critical in predicting their chemical reactivity and uses.

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Most popular questions from this chapter

For the reaction, \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}, \Delta \mathrm{S}\) and \(\mathrm{T}\) are \(40.657 \mathrm{~kJ} \mathrm{~mol}^{-1}, 109 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) and \(373 \mathrm{~K}\) respectively. Find the free energy change \((\Delta \mathrm{G})\) of the reaction.

For an endothermic reaction, where \(\Delta \mathrm{H}\) represents the enthalpy of the reaction in \(\mathrm{kJ} / \mathrm{mol}\), the minimum value for the energy of activation will be (a) less than \(\Delta \mathrm{H}\) (b) zero (c) more than \(\Delta \mathrm{H}\) (d) equal to \(\Delta \mathrm{H}\).

The entropy values in \(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) of \(\mathrm{H}_{2}(\mathrm{~g})=130.6\), \(\mathrm{Cl}_{2}(\mathrm{~g})=223\) and \(\mathrm{HC} 1(\mathrm{~g})=186.7\) at \(298 \mathrm{~K}\) and 1 atm pressure. Then entropy change for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HC} 1(\mathrm{~g})\) is (a) \(+540.3\) (b) \(+727.3\) (c) \(-166.9\) (d) \(+19.8\)

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of \(10 \mathrm{dm}^{3}\) to a volume of \(100 \mathrm{dm}^{3}\) at \(27^{\circ} \mathrm{C}\) is: (a) \(35.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) (b) \(38.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) (c) \(45.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) (d) \(23.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant \(K_{c}\) is (a) \(\Delta \mathrm{G}=\mathrm{RT}\) In \(\mathrm{K}\) (b) \(-\Delta \mathrm{G}=\mathrm{RT}\) In \(\mathrm{K}\) (c) \(\Delta \mathrm{G}^{\circ}=\mathrm{RT} \operatorname{In} \mathrm{K}_{\mathrm{c}}\) (d) \(-\Delta \mathrm{G}^{\circ}=\mathrm{RT} \operatorname{In} \mathrm{K}_{\mathrm{c}}\)

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