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The standard enthalpy of formation \(\left(\Delta_{t} \mathrm{H}^{\circ}\right)\) at \(298 \mathrm{~K}\) for methane, \(\mathrm{CH}_{4}(\mathrm{~g})\) is \(-74.8 \mathrm{~kJ} \mathrm{~mol}-1\), the additional information required to determine the average energy for \(\mathrm{C}-\mathrm{H}\) bond formation would be: (a) The dissociation energy of \(\mathrm{H}_{2}\) and enthalpy of sublimation of carbon (b) Latent heat of vaporization of methane (c) The first four ionization energies of carbon and electron gain enthalpy of hydrogen (d) The dissociation energy of hydrogen molecule, \(\mathrm{H}_{2}\)

Short Answer

Expert verified
Option (a): Dissociation energy of H₂ and enthalpy of sublimation of carbon.

Step by step solution

01

Understand the Question

We need to determine what additional information is necessary to calculate the average energy of the carbon-hydrogen (C-H) bond formation from CH₄.
02

Recall the Bond Formation Process

The standard enthalpy of formation of methane (CH₄) involves forming C-H bonds from separate carbon and hydrogen atoms. Breakdown the process into necessary steps: sublimating carbon solid to gaseous atoms, dissociating hydrogen molecules (00H₂) into atoms, and forming C-H bonds.
03

Identify Information for Calculating C-H Bond Energy

To calculate the C-H bond energy, we need the enthalpy changes for sublimating carbon (to provide gaseous C atoms) and dissociating hydrogen (to provide H atoms), because both pieces are then used in forming C-H bonds. Thus, we need the dissociation energy of H₂ and the enthalpy of sublimation of carbon.
04

Reviewing Options

Among the options, only (a) provides both the dissociation energy of H₂ and the enthalpy of sublimation of carbon, which are key in calculating the C-H bond energy.

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Key Concepts

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

CH4 Bond Formation
Understanding how methane (CH\(_4\)) forms is crucial to grasping the fundamentals of chemical bonding. Methane is a simple molecule composed of one carbon atom bonded to four hydrogen atoms. The formation of these C-H bonds involves several steps.
  • First, carbon in its solid form needs to be converted to gaseous form. This is achieved through sublimation, where solid carbon is transformed into gaseous carbon atoms.
  • Next, diatomic hydrogen molecules (H\(_2\)) must be split into individual hydrogen atoms. This process is called dissociation and requires a specific amount of energy.
  • Finally, the gaseous carbon reacts with the hydrogen atoms to form C-H bonds, resulting in methane.
Each of these steps involves energy changes, and understanding them helps us calculate the energy needed to form the C-H bonds in methane.
Enthalpy Changes
Enthalpy changes refer to the energy absorbed or released during a chemical reaction at constant pressure. When discussing methane formation, several enthalpy changes are involved:
  • Enthalpy of Sublimation: This is the energy required to convert solid carbon into gaseous carbon atoms.
  • Bond Dissociation Energy: The energy needed to break the hydrogen molecule into individual atoms.
  • Finally, there's the enthalpy change associated with forming the C-H bonds, leading to the creation of methane.
All these enthalpy changes are interconnected. By understanding them, you can grasp how much energy is involved in creating the stable structure of methane. Each change gives insight into the energetics of forming or breaking chemical bonds, which is vital in various chemical processes.
Bond Dissociation Energy
Bond dissociation energy is the amount of energy required to break a specific bond in a molecule. For methane, determining the bond dissociation energy is essential for understanding how C-H bonds are formed. Dissociating hydrogen involves breaking the H-H bond in hydrogen molecules. This energy is critical because, without supplying enough energy to break this bond, hydrogen atoms cannot react with the gaseous carbon atoms. The energy released or needed during this bond breaking is called dissociation energy. Knowing the dissociation energy helps predict how easily certain bonds can be formed or broken during reactions. It's a key factor in calculating the energy dynamics of chemical reactions and is pivotal in understanding chemical stability and reactivity.
Sublimation Enthalpy
Sublimation enthalpy is the energy required to change a substance from its solid state to a gas without passing through the liquid phase. This concept is crucial when forming methane, as solid carbon must transition into gaseous carbon before bonding can occur. In practical terms, you need to know how much energy is needed to break the lattice energy of solid carbon to obtain isolated atoms. This enthalpy change is one of the first steps in the process of methane formation. The value of sublimation enthalpy influences the calculation of overall bond formation energy. Knowing and understanding this energy helps to estimate the energy balance involved in the transformation of elements into compounds like methane. Sublimation is a physical process but plays a pivotal role in chemical reactions where gaseous atoms are necessary.

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Most popular questions from this chapter

In thermodynamics, a process is called reversible when: (a) The surroundings are always in equilibrium with the system (b) There is no boundary between system and sur roundings (c) The surroundings and system change into each other (d) The system changes into the surroundings sponta neously

\((\Delta \mathrm{H}-\Delta \mathrm{U})\) for the formation of carbon monoxide (CO) from its elements at \(298 \mathrm{~K}\) is: (a) \(-1238.78 \mathrm{~J} \mathrm{~mol}^{-1}\) (b) \(1238.78 \mathrm{~J} \mathrm{~mol}^{-1}\) (c) \(-2477.57 \mathrm{~J} \mathrm{~mol}^{-1}\) (d) \(2477.57 \mathrm{~J} \mathrm{~mol}^{-1}\)

For the reaction, \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}(\mathrm{g})+3 \mathrm{D}(\mathrm{g})\) The value of \(\Delta \mathrm{H}\) at \(27^{\circ} \mathrm{C}\) is \(19.0 \mathrm{kcal}\). The value of \(\Delta \mathrm{E}\) for the reaction would be (given \(\mathrm{R}=2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ) (a) \(20.8 \mathrm{kcal}\) (b) \(19.8 \mathrm{kcal}\) (c) \(18.8 \mathrm{kcal}\) (d) \(17.8 \mathrm{kcal}\)

If the standard entropies of \(\mathrm{CH}_{4}, \mathrm{O}_{2}, \mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are \(186.2,205.3,213.6\) and \(69.96 \mathrm{JK}^{-1} \mathrm{~mol}^{-\mathrm{t}}\) respectively, then standard entropy change for the reaction: \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is: (a) \(-215.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (b) \(-243.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (c) \(-130.5 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (d) \(-85.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

Calculate \(Q\) and \(W\) for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of \(1.0\) bar to a final pressure of \(0.1\) bar at a constant temperature of \(273 \mathrm{~K}\). (a) \(5.22 \mathrm{~kJ},-5.22 \mathrm{~kJ}\) (b) \(-27.3 \mathrm{~kJ}, 27.3 \mathrm{~kJ}\) (c) \(27.3 \mathrm{~kJ},-27.3 \mathrm{~kJ}\) (d) \(-5.22 \mathrm{~kJ}, 5.22 \mathrm{~kJ}\)

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