
(a)
Interpretation:
The stability of
Concept introduction:
Gibbs free energy is the useful work obtained from the system. It is the difference between the enthalpy and the product of entropy and absolute temperature. It is always negative. Therefore, more negative the gibbs free energy more stable the compound.
(b)
Interpretation:
The hydration reaction between
Concept introduction:
In hydration reaction, water molecules attack on a compound and it is an exothermic reaction. The molecule has standard activation energy to get hydrated. Lower the activation energy faster will be the hydration of that molecule.
(c)
Interpretation:
The free-energy-diagrams representing the hydration reaction of
Concept introduction:
Gibbs free energy is the useful work obtained from the system. The free energy diagram represents the gibbs free energy of reactants and products. The lower the gibbs free energy more stable the compound.
(d)
Interpretation:
The difference in the standard free energies of the transition states for the hydration reactions of
Concept introduction:
Gibbs free energy is the useful work obtained from the system. In hydration, the compound reacts with water to release energy. Hydration reaction has activation energy. Higher the activation energy, lower will be the

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Chapter 4 Solutions
Organic Chemistry Study Guide and Solutions
- In a benzene derivative that has -CH2CH3, indicate how it can be substituted by -COOH.arrow_forwardIn a sulfonated derivative of benzene, indicate how -SO3H can be eliminated.arrow_forwardWhat is the equilibrium expression (law of mass action) for the following reaction:CO2 (g) + H2O (l) ⇋ H+ (aq) + HCO3- (aq)arrow_forward
- Indicate the compound resulting from adding NaOH cyclopentane-CH2-CHO.arrow_forwardUse the provided information to calculate Kc for the following reaction at 550 °C: H2(g) + CO2(g) ⇌ CO(g) + H2O(g) Kc = ?CoO(s) + CO(g) ⇌ Co(s) + CO2(g) Kc1 = 490CoO(s) + H2(g) ⇌ Co(s) + H2O(g) Kc2 = 67arrow_forwardCalculate Kc for the reaction: I2 (g) ⇋ 2 I (g) Kp = 6.26 x 10-22 at 298Karrow_forward
- For each scenario below, select the color of the solution using the indicator thymol blue during the titration. When you first add indicator to your Na2CO3solution, the solution is basic (pH ~10), and the color is ["", "", "", "", ""] . At the equivalence point for the titration, the moles of added HCl are equal to the moles of Na2CO3. One drop (or less!) past this is called the endpoint. The added HCl begins to titrate the thymol blue indicator itself. At the endpoint, the indicator color is ["", "", "", "", ""] . When you weren't paying attention and added too much HCl (~12 mL extra), the color is ["", "", "", "", ""] . When you really weren't paying attention and reached the second equivalence point of Na2CO3, the color isarrow_forwardTo convert cyclopentane-CH2-CHO to cyclopentane-CH2-CH3, compound A is added, followed by (CH3)3CO-K+, DMS at 100oC. Indicate which compound A is.arrow_forwardIndicate how to obtain the compound 2-Hydroxy-2-phenylacetonitrile from phenylmethanol.arrow_forward
- Indicate the reagent needed to go from cyclopentane-CH2-CHO to cyclopentane-CH2-CH=CH-C6H5.arrow_forwardesc Write the systematic name of each organic molecule: structure CH3 CH3-C=CH2 CH3-CH2-C-CH2-CH3 CH-CH3 CH3 ☐ ☐ ☐ CI-CH-CH=CH2 Explanation Check F1 F2 name 80 F3 F4 F5 F6 A 7 ! 2 # 3 4 % 5 6 & 7 Q W E R Y FT 2025 Mcarrow_forwardTwo reactants X and Z are required to convert the compound CH3-CH2-CH2Br to the compound CH3-CH2-CH=P(C6H5)3. State reactants X and Z.arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
