The following reaction has a value of ΔG= -2.1 kJ/mol (-0.50 kcal/mol). CH3Br + H2S ⇌ CH3SH + HBr a. Calculate Keq at room temperature (25 °C) for this reaction as written.

Which of the following parameters would be different for a reaction carried out in the presence of a catalyst compared with the same reaction carried out in the absence of a catalyst?

∆G°, ∆H^‡, E_a, ∆S^‡, ∆H°, K_eq, ∆G^‡, ∆S°, k

Under base-catalyzed conditions, two molecules of acetone can condense to form diacetone alcohol. At room temperature (25 °C), about 5% of the acetone is converted to diacetone alcohol. Determine the value of ΔG° for this reaction.

In light of your answers to parts (b) and (c), where both were shown to be quite favorable, imagine a scenario where either reaction is possible. Of the two, which would you expect to be faster? Which would you expect to be more favored? Explain each in the context of the important thermodynamic and/or kinetic parameters.

Compounds containing deuterium (D = ²H) are useful for kinetic studies and metabolic studies with new pharmaceuticals. One way to introduce deuterium is by using the reagent LiAlD₄, equivalent in reactivity to LiAlH₄. Show how to make these deuterium-labeled compounds, using LiAlD₄ and D₂O as your sources of deuterium, and any non-deuterated starting materials you wish.

a. CH₃CHDOH

Looking ahead In Chapter 5, we explain that the equilibrium constant (K_eq) for a reaction can be calculated based on the difference in energy between reactants and products, according to the following equation:

$K_{eq}=e^{-\frac{\Delta E}{RT}}$

Using this equation, calculate the equilibrium constant for the 'reaction' shown. [For the rest of the book, if not otherwise specified, assume room temperature (298K).]

The following reaction has a value of ΔG° = –2.1 kJ/mol (–0.50 kcal/mol).

CH₃Br + H₂S ⇌ CH₃SH + HBr

a. Calculate K_eq at room temperature (25 °C) for this reaction as written.