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The standard enthalpy of combustion at \(25^{\circ} \mathrm{C}\) of hydrogen, cyclohexene \(\left(\mathrm{C}_{6} \mathrm{H}_{10}\right)\) and cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) are \(-241,-3800\) and \(-\) \(3920 \mathrm{~kJ} /\) mole respectively. Calculate the heat of hydrogenation of cyclohexene.

Short Answer

Expert verified
The heat of hydrogenation of cyclohexene is 121 kJ/mol.

Step by step solution

01

Write the Balanced Chemical Equations for Combustion

The combustion reactions for each compound are as follows:- Hydrogen: \[2\text{H}_2 (g) + \text{O}_2 (g) \rightarrow 2\text{H}_2\text{O} (g)\]- Cyclohexene: \[\text{C}_6\text{H}_{10} (g) + \frac{17}{2}\text{O}_2 (g) \rightarrow 6\text{CO}_2 (g) + 5\text{H}_2\text{O} (g)\]- Cyclohexane: \[\text{C}_6\text{H}_{12} (g) + 9\text{O}_2 (g) \rightarrow 6\text{CO}_2 (g) + 6\text{H}_2\text{O} (g)\]
02

Write the Reaction for Hydrogenation

Cyclohexene is hydrogenated to form cyclohexane, as represented by the equation:\[\text{C}_6\text{H}_{10} (g) + \text{H}_2 (g) \rightarrow \text{C}_6\text{H}_{12} (g)\]
03

Apply Hess's Law

According to Hess's Law, the total enthalpy change for the reaction can be calculated by:\[\Delta H_{\text{reaction}} = \Delta H_{\text{comb, cyclohexane}} - \Delta H_{\text{comb, cyclohexene}} - \Delta H_{\text{comb, H}_2}\]Substitute the given values:\[\Delta H_{\text{hydrogenation}} = (-3920) - (-3800) - (-241)\]
04

Calculate the Enthalpy Change

Calculate the enthalpy change using the equation from Step 3:\[\Delta H_{\text{hydrogenation}} = -3920 + 3800 + 241 = 121 \text{ kJ/mol}\]
05

Conclusion

The heat of hydrogenation of cyclohexene is 121 kJ/mol. This value represents the energy released or absorbed during the hydrogenation process under standard conditions.

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Key Concepts

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

Hess's Law Explained
Hess's Law is a principle in chemistry that helps us understand how energy is conserved in chemical reactions. It states that the total enthalpy change for a chemical reaction is the same, no matter how many steps are taken to complete the reaction. In simple words, it doesn't matter if a reaction occurs in one big step or several smaller ones, the total energy change remains constant. This is because enthalpy is a state function, which means its value is determined only by the initial and final states of the system, not the path taken between them.
In the exercise given, Hess's Law was used to calculate the enthalpy change, or the heat of reaction, for the hydrogenation of cyclohexene by combining the enthalpy changes of known combustion reactions. This showcases the power of Hess's Law in solving thermochemical problems, especially when direct measurement is difficult or impractical.
Understanding Heat of Hydrogenation
Heat of hydrogenation is the energy change that occurs when a molecule, typically an alkene or alkyne, reacts with hydrogen to become a more saturated compound. It's an exothermic process, meaning it releases heat. This is because the products formed, like cyclohexane from cyclohexene, are usually more stable and lower in energy than the reactants.
The original exercise focuses on determining the heat of hydrogenation of cyclohexene. Using Hess's Law, we were able to calculate this value indirectly based on known data of combustion reactions. The heat of hydrogenation indicates how much energy is released when cyclohexene turns into cyclohexane, reflecting the stability added by the saturation process.
The Basics of Combustion Reactions
Combustion reactions are a crucial category of chemical reactions, involving a substance combining with oxygen and releasing energy in the form of heat and light. These reactions are generally exothermic, which means they release energy.
The combustion reactions for hydrogen, cyclohexene, and cyclohexane were used in the exercise to ultimately determine the heat of hydrogenation. Each of these reactions involves the reactant burning in oxygen to produce carbon dioxide and water as products. Understanding these reactions helps in recognizing their role in energy transfer, essential for calculations involving Hess's Law and enthalpy.
Enthalpy of Combustion - A Key Concept
Enthalpy of combustion is the heat energy released when one mole of a substance completely reacts with oxygen to form products under standard conditions. It's a specific type of exothermic reaction important for energy calculations in chemistry.
In the context of the exercise, the enthalpy of combustion was provided for hydrogen, cyclohexene, and cyclohexane. These values were crucial for applying Hess's Law to find the desired heat of hydrogenation. By knowing the enthalpy of combustion, we can assess the energy content of the substances involved, thus understanding their potential energy changes during reactions. This concept is vital for both practical applications like energy production and theoretical studies of reaction mechanisms.

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Most popular questions from this chapter

An ideal gas is expanded from \(\left(P_{1}, V_{1}, T_{1}\right)\) to \(\left(P_{2}, V_{2}, T_{2}\right)\) under different conditions. The correct statement(s) among the following is (are) (a) The work done on the gas is maximum when it is compressed irreversibly from \(\left(P_{2}, V_{2}\right)\) to \(\left(P_{1}, V_{1}\right)\) against constant pressure \(P_{1}\) (b) If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic (c) The work done by the gas is less when it is expanded reversibly from \(V_{1}\) to \(V_{2}\) under adiabatic conditions as compared to that when expanded reversibly from \(V_{1}\) to \(V_{2}\) under isothermal conditions (d) The change in internal energy of the gas is (i) zero, if it is expanded reversibly with \(T_{1}=T_{2}\), and (ii) positive, if it is expanded reversibly under adiabatic conditions with \(T_{1} \neq T_{2}\)

First law of thermodynamics is not adequate in predicting the direction of a process.

Choose the reaction(s) from the following options, for which the standard enthalpy of reaction is equal to the standard enthalpy of formation [Adv. 2019] (a) \(\frac{1}{8} \mathrm{~S}_{8}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{2}(\mathrm{~g})\) (b) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(1)\) (c) \(\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{O}_{3}(\mathrm{~g})\) (d) \(2 \mathrm{C}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\)

The standard state Gibbs free energies of formation of C(graphite) and \(\mathrm{C}\) (diamond) at \(T=298 \mathrm{~K}\) are \(\Delta_{f} G^{0}[\mathrm{C}\) (graphite) \(]=0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta_{f} G^{0}[\mathrm{C}\) (diamond) \(]=2.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite \([\mathrm{C}\) (graphite) \(]\) to diamond \(\left[C\right.\) (diamond)] reduces its volume by \(2 \times 10^{-}\) \({ }^{6} \mathrm{~m}^{3} \mathrm{~mol}^{-1} .\) If \(C\) (graphite) is converted to \(\mathrm{C}\) (diamond) isothermally at \(T\) \(=298 \mathrm{~K}\), the pressure at which \(\mathrm{C}\) (graphite) is in equilibrium with \(\mathrm{C}\) (diamond), is [Useful information : \(\left.1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2} ; 1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} ; 1 \mathrm{bar}=10^{5} \mathrm{~Pa}\right]\) [Adv. 2017] (a) 14501 bar (b) 58001 bar (c) \(1450 \mathrm{bar}\) (d) 29001 bar

The difference between \(\Delta \mathrm{H}\) and \(\Delta \mathrm{U}(\Delta \mathrm{H}-\Delta \mathrm{U})\), when the combustion of one mole of heptane (I) is carried out at a temperature \(\mathrm{T}\), is equal to [Main April \(10, \mathbf{2 0 1 9}\) (II)] (a) \(-4 \mathrm{RT}\) (b) \(-3 \mathrm{RT}\) (c) \(4 \mathrm{RT}\) (d) \(3 \mathrm{RT}\)

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