Use bond-dissociation enthalpies [TABLE 4-2], p. 167) to calculate values of ΔH° for the following reactions.a. CH3—CH3 + I2 —> CH3CH2I + HI

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Identify the bonds broken and formed in the reaction: CH3—CH3 + I2 —> CH3CH2I + HI.

Calculate the total energy required to break the bonds in the reactants: C—C bond in ethane and I—I bond in iodine.

Calculate the total energy released when new bonds are formed in the products: C—I bond in ethyl iodide and H—I bond in hydrogen iodide.

Use the bond-dissociation enthalpies from the table to find the energy values for each bond broken and formed.

Apply the formula for the change in enthalpy: \( \Delta H^\circ = \text{(sum of bond energies of bonds broken)} - \text{(sum of bond energies of bonds formed)} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Bond-Dissociation Enthalpy

Bond-dissociation enthalpy (BDE) is the energy required to break a specific bond in a molecule, resulting in the formation of two radicals. It is a crucial concept in thermochemistry, as it helps predict the stability of molecules and the energy changes during chemical reactions. BDE values are typically provided in kilojoules per mole (kJ/mol) and vary depending on the type of bond and the molecular environment.

Enthalpy Change (ΔH°)

The enthalpy change (ΔH°) of a reaction is the difference in enthalpy between the products and reactants under standard conditions. It indicates whether a reaction is exothermic (releases heat, ΔH° < 0) or endothermic (absorbs heat, ΔH° > 0). Calculating ΔH° using bond-dissociation enthalpies involves summing the BDEs of bonds broken in the reactants and subtracting the BDEs of bonds formed in the products.

Reaction Mechanism

A reaction mechanism describes the step-by-step sequence of elementary reactions by which a chemical change occurs. Understanding the mechanism is essential for predicting the products and energy changes in a reaction. In the given reaction, the mechanism involves the homolytic cleavage of the I—I bond and the formation of new C—I and H—I bonds, which can be analyzed using bond-dissociation enthalpies.

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Each of the following proposed mechanisms for the free-radical chlorination of methane is wrong. Explain how the experimental evidence disproves each mechanism.

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Deuterium (D) is the hydrogen isotope of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C―D bond is slightly stronger than the C―H bond by 5.0 kJ/mol (1.2 kcal/mol). Reaction rates tend to be slower when a C―D bond (as opposed to a C―H bond) is broken in a rate-limiting step.This effect, called a kinetic isotope effect, is clearly seen in the chlorination of methane. Methane undergoes free-radical chlorination 12 times as fast as tetradeuteriomethane (CD4)Faster: CH4 + Cl⋅ —> CH3Cl + HCl relative rate= 12Slower: CD4 + Cl⋅ —> CD3Cl + DClrelative rate= 1 c. Consider the thermodynamics of the chlorination of methane and the chlorination of ethane, and use the Hammond postulate to explain why one of these reactions has a much larger isotope effect than the other.

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The reaction of tert-butyl chloride with methanol(CH3)3C—Cl Tert-butylchloride + CH3—OH methanol —> (CH3)C—OCH3 methyltert-butylether + HCl is found to follow the rate equation rate= Kt[(CH3)3C—Cl] a. What is the kinetic order with respect to tert-butyl chloride?

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The bromination of methane proceeds through the following steps:1. Br2 + 2 Br• ΔH° (per mole)/+190 kJ (45 kcal)Ea (per mole)/ 190 kJ (45 kcal)2. CH4 + Br• —> CH3+ HBr +73 kJ (17 kcal) 79 kJ (19 kcal) 3. • CH3 + Br2 —> CH3Br + Br -112 kJ (-27 kcal) 4 kJ (1 kcal) d. Compute the overall value of ΔH° for the bromination