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If at \(298 \mathrm{~K}\) the bond energies of \(\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}\) and \(\mathrm{H}-\mathrm{H}\) bonds are respectively \(414,347,615\) and \(435 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\), the value of enthalpy change for the reaction \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}(\mathrm{~g})\) at \(298 \mathrm{~K}\) will be: (a) \(+250 \mathrm{~kJ}\) (b) \(-250 \mathrm{~kJ}\) (c) \(+125 \mathrm{~kJ}\) (d) \(-125 \mathrm{~kJ}\)

Short Answer

Expert verified
The enthalpy change is \(-125\, \text{kJ/mol}\), option (d).

Step by step solution

01

Identify the Bonds Involved

In the reaction, we're breaking 1 C=C bond and 1 H-H bond to form 2 C-H bonds and 1 C-C bond. The reaction converts ethene and hydrogen gas into ethane.
02

Calculate Energy Required to Break Bonds

The energy required to break the bonds is the sum of the bond energies of the C=C and H-H bonds. Calculate this as follows: \[\text{Energy to Break} = 615\, \text{kJ/mol} (\text{C=C}) + 435\, \text{kJ/mol} (\text{H-H}) = 1050\, \text{kJ/mol}.\]
03

Calculate Energy Released by Forming Bonds

The energy released during bond formation is the sum of the bond energies of the newly formed C-H and C-C bonds. Calculate this as follows: \[\text{Energy to Form} = 2 \times 414\, \text{kJ/mol} (\text{C-H}) + 347\, \text{kJ/mol} (\text{C-C}) = 1175\, \text{kJ/mol}.\]
04

Determine the Enthalpy Change

The enthalpy change, \(\Delta H\), is calculated by subtracting the energy to break bonds from the energy to form bonds: \[\Delta H = 1175\, \text{kJ/mol} - 1050\, \text{kJ/mol} = -125\, \text{kJ/mol}.\]
05

Evaluate the Sign of \(\Delta H\)

Since the enthalpy change \(\Delta H\) is negative, the reaction releases heat and is exothermic.

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Key Concepts

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

Bond Energy
Bond energy is a vital concept in chemical reactions. It refers to the amount of energy needed to break one mole of bonds in a molecule. The bond energy values give us insights into how strong a bond is. For instance, in the original exercise, the bond energies provided for the C-H, C-C, C=C, and H-H bonds help determine the enthalpy change for the reaction. The higher the bond energy, the more energy is required to break it. When you break a C=C bond, like in ethene, and an H-H bond, like in hydrogen gas, you need to use a certain amount of energy.
  • C=C bond has a bond energy of 615 kJ/mol.
  • H-H bond has a bond energy of 435 kJ/mol.
When bonds such as C-H and C-C are formed, they release energy. It's this balance between breaking and forming bonds that determines the overall energy change in a reaction. Understanding bond energy allows us to quantitatively analyze and predict whether a reaction will absorb or release energy.
Exothermic Reaction
An exothermic reaction is a chemical reaction that releases energy into the surroundings, usually in the form of heat. In other words, the energy released from forming new bonds is greater than the energy needed to break the existing bonds. In our exercise, the calculated enthalpy change (95 H) was -125 kJ/mol, indicating that it is an exothermic reaction.
  • Signs of exothermic reactions include a negative enthalpy change.
  • During such reactions, the temperature of the surroundings typically increases.
For students, it is important to remember that exothermic reactions are often more spontaneous due to the release of energy, creating a more stable product. The initial ethene and hydrogen gas react to form ethane, and energy is released to make up for the newly formed C-H and C-C bonds.
Chemical Thermodynamics
Chemical thermodynamics involves the study of energy changes occurring in chemical reactions. It extends beyond just observing changes; it quantitatively analyzes them. Key elements of chemical thermodynamics include enthalpy, entropy, and free energy, with enthalpy being the focus here. Enthalpy (95 H) is the heat content of a system at constant pressure. Calculating enthalpy changes helps in understanding the energy dynamics of reactions. From bond energy calculations, you can deduce whether a reaction absorbs heat (endothermic) or releases heat (exothermic).
  • Enthalpy change is crucial for predicting reaction feasibility and spontaneity.
  • Knowing 95 H allows chemists to optimize conditions for industrial and laboratory processes.
By understanding where energy is going or being released, chemical thermodynamics allows chemists to manipulate reactions for desired outcomes, ensuring reactions progress in a controlled and efficient manner.

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Most popular questions from this chapter

The entropy change when \(36 \mathrm{~g}\) of water evaporates at \(373 \mathrm{~K}\) is \(\left(\Delta \mathrm{H}=40.63 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) : (a) \(218 \mathrm{~J} \mathrm{~K}^{-1}\) (b) \(150 \mathrm{~J} \mathrm{~K}^{-1}\) (c) \(118 \mathrm{~J} \mathrm{~K}^{-1}\) (d) \(200 \mathrm{~J} \mathrm{~K}^{-1}\)

The amount of heat required to raise the temperature of one mole of the substance through \(1 \mathrm{~K}\) is called, its: (a) Molar heat (b) Entropy (c) Thermal capacity (d) Specific heat

Which of the following is not an endothermic reaction? (a) Combustion of methane (b) Decomposition of water (c) Dehydrogenation of ethane or ethylene (d) Conversion of graphite to diamond

Anhydrous \(\mathrm{AlCl}_{3}\) is covalent. From the data given below, predict whether it would remain covalent or become ionic in aqueous solution (ionization energy of \(\mathrm{Al}=5137 \mathrm{~kJ} \mathrm{~mol}^{-1} \Delta \mathrm{H}_{\text {hydratian }}\) for \(\mathrm{Al}^{+3}=-4665 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}, \Delta \mathrm{H}_{\text {hydration }}\) for \(\left.\mathrm{Cl}^{-}=-381 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) (a) Ionic (b) Covalent (c) Both (a) and (b) (d) None of these

The standard enthalpy of decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) is \(58.04 \mathrm{~kJ}\) and standard entropy of this reaction is \(176.7 \mathrm{~J} \mathrm{~K}^{-1}\). The standard free energy change for this reaction at \(25^{\circ} \mathrm{C}\), is: (a) \(5.39 \mathrm{~kJ}\) (b) \(-5.39 \mathrm{~kJ}\) (c) \(539 \mathrm{~kJ}\) (d) \(53.9 \mathrm{~kJ}\)

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