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(a)
Interpretation:
The reason for disruption of lipid bilayer of cell membrane by ethanol and
Concept Introduction:
The lipid bilayer of the cell membrane is made of both polar and non-polar part. The polar head contains ionic phosphate groups. The non-polar tail contains long chain hydrocarbons. The lipid bilayer is both hydrophobic as well as hydrophilic in nature. The presence of long hydrocarbon chain does not allow any ionic group to cross the lipid bilayer.
(b)
Interpretation:
Sugars like glucose do not diffuse through cell membrane is to be explained.
Concept Introduction:
The lipid bilayer of the cell membrane is made of both polar and non-polar part. The polar head contains ionic phosphate groups. The non-polar tail contains long chain hydrocarbons. The lipid bilayer is both hydrophobic as well as hydrophilic in nature. The presence of long hydrocarbon chain does not allow any ionic group to cross the lipid bilayer.
(c)
Interpretation:
The value
Concept Introduction:
The favorable interaction of dissolved molecule with solvent is known as solvation. The favorable interaction with solvent and solute results in the solvating the ion. Thus the interactions are hydrogen bonding, dipole-dipole and electrostatic interaction.
(d)
Interpretation:
The higher solubility of diethyl ether in water than
Concept Introduction:
The solubility of different compounds in water depends on the interaction between the compound and water molecule. Water forms hydrogen bonding with many compounds, which increases their solubility in water. The molecules which form hydrogen bonding among itself are less soluble in water. They require high salvation energy to break the hydrogen bonding and then solvates in water.
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Chapter 8 Solutions
EBK ORGANIC CHEMISTRY
- Another step in the metabolism of glucose, which occurs after the formation of glucose6-phosphate, is the conversion of fructose6-phosphate to fructose1,6-bisphosphate(bis meanstwo): Fructose6-phosphate(aq) + H2PO4(aq) fructose l,6-bisphosphate(aq) + H2O() + H+(aq) (a) This reaction has a Gibbs free energy change of +16.7 kJ/mol of fructose6-phosphate. Is it endergonic or exergonic? (b) Write the equation for the formation of 1 mol ADP fromATR for which rG = 30.5 kJ/mol. (c) Couple these two reactions to get an exergonic process;write its overall chemical equation, and calculate theGibbs free energy change.arrow_forwardActually, the carbon in CO2(g) is thermodynamically unstable with respect to the carbon in calcium carbonate(limestone). Verify this by determining the standardGibbs free energy change for the reaction of lime,CaO(s), with CO2(g) to make CaCO3(s).arrow_forwardWhy is ΔTmetal < 0? Why is ΔTwaterarrow_forward
- Consider the following isomerization reactions of some simple sugars and values for their standard Gibbs free energy AG°: reaction A: glucose-1-phosphate → glucose-6-phosphate, AG° –7.28 kJ/mol reaction B: fructose-6-phosphate → glucose-6-phosphate, AG° = –1.67 kJ/molarrow_forward(a) Write the chemical equation for the equilibriumthat corresponds to Kb. (b) By using the value ofKb, calculate ΔG° for the equilibrium in part (a). (c) What isthe value of ΔG at equilibrium? (d) What is the value of ΔGwhen [H+] = 6.7x 10-9 M, [CH3NH3 +] = 2.4 x 10-3 M,and [CH3NH2] = 0.098 M?arrow_forwardOne of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):G6P ⇌ F6P ΔG°298 = 1.7 kJ(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.arrow_forward
- For the reaction CO(NH2)2 +H2O(l) -> CO2(g) + 2NH3(g) The ∆S^2 and ∆H^2 is 354.8KJ/mol and 119.2KJ/mol respectively. Calculate (a) the standard entropy of urea(b) the standard enthalpy of formation of urea(c) the standard free energy change of the reaction and (d) the standard free energy of formation of urea.arrow_forward(a) Calculate the value of Kc for the reaction: PCl5 (g) PCl3 (g) + Cl2 (g) ΔH = Positive Given that when 8.4 mol of PCl5 (g) is mixed with 1.8 mol of PCl3 (g) and allowed to come to equilibrium in a 10 dm3container the amount of PCl5 (g) at equilibrium is 7.2 mol. Kc = (b) Explain the effect of the following changes below on the value of Kc: Increasing temperature Lowering the concentration of chlorine (Cl2) Addition of a catalystarrow_forwardWithout doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.(a) N2(g) + 3H2(g) ⟶ 2NH3(g)(b) HCl(g) + NH3(g) ⟶ NH4 Cl(s)(c) (NH4)2 Cr2 O7(s) ⟶ Cr2 O3(s) + 4H2 O(g) + N2(g)(d) 2Fe(s) + 3O2(g) ⟶ Fe2 O3(s)arrow_forward
- Hi, Can someone help me with this question? I cannot seem to figure out how to solve this. Thanks! Calculate ΔG° for the equation below using the thermodynamic data given: N2(g) + 3 H2(g) --> 2 NH3(g) ΔH° (NH3) = -46 kJ; ΔS° values (J/molK) are NH3 = 192.5, N2 = 191.5, H2 = 130.6.arrow_forwardto calculate the ΔH° of the unknown reaction:arrow_forwardWhen heated, the DNA double helix separates into two random-coil single strands. When cooled, the random coils re-form the double helix: double helix ⇌2 random coils.(a) What is the sign of ΔS for the forward process? Why?(b) Energy must be added to overcome ΔH bonds and dispersion forces between the strands. What is the sign of ΔG for the for-ward process when TΔS is smaller than ΔH?(c) Write an expression that shows T in terms of ΔH and ΔS when the reaction is at equilibrium. (This temperature is calledthe melting temperatureof the nucleic acid.)arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage Learning
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