Q: Driving unfavorable reactions in cells: Reactions with positive ΔG° can still proceed in vivo by which strategy?

Introduction / Context: Cells routinely drive thermodynamically unfavorable steps by harnessing energy from favorable processes. This is central to metabolism, transport, and biosynthesis, where pathway design and enzyme mechanisms exploit coupling and mass-action effects. Given Data / Assumptions: Standard free energy change ΔG° > 0 for the isolated reaction. Cellular systems are open and maintain non-equilibrium concentrations. ATP hydrolysis is a common exergonic currency (ΔG°′ ≈ −30.5 kJ/mol). Concept / Approach: Actual free energy change is ΔG = ΔG° + RT ln Q. By altering Q (ratio of products to reactants), ΔG can become negative. Enzymes also couple unfavorable reactions to favorable ones (including ATP hydrolysis), often via phosphorylated or otherwise high-energy intermediates, making the overall sum exergonic. Step-by-Step Solution: Use mass action: keep products low and reactants high → RT ln Q sufficiently negative. Couple reactions: A ⇌ B (unfavorable) + B ⇌ C (highly favorable) → A ⇌ C overall favorable. Exploit ATP: enzyme forms phosphorylated intermediate, transfers group to drive the endergonic step. Therefore, all listed strategies enable progress of an otherwise unfavorable step. Verification / Alternative check: Glycolysis and amino acid activation (aminoacyl-tRNA synthetases) exemplify ATP-coupled steps; metabolic channeling maintains favorable Q values. Why Other Options Are Wrong: Choosing only one strategy ignores the complementary biochemical tactics cells actually use. Common Pitfalls: Equating ΔG° with inevitability in vivo; forgetting that ΔG depends on concentrations and coupling. Final Answer: All of the above.

Q: Standard Gibbs energy when enthalpy change is zero: If ΔH° = 0 for a reaction, what is the correct expression for ΔG°?

Introduction / Context: Gibbs free energy (ΔG°) predicts the spontaneity of reactions at constant temperature and pressure. Understanding how ΔH° (enthalpy) and ΔS° (entropy) contribute helps interpret biochemical processes such as protein folding and ligand binding. Given Data / Assumptions: ΔH° = 0 (no net enthalpy change under standard conditions). Temperature T is constant and in kelvin. Standard thermodynamic identity applies. Concept / Approach: The fundamental relation is ΔG° = ΔH° − TΔS°. If ΔH° = 0, then ΔG° simplifies to −TΔS°. A positive ΔS° gives a negative ΔG° (favorable), while a negative ΔS° gives positive ΔG° (unfavorable) at constant T. Step-by-Step Solution: Write identity: ΔG° = ΔH° − TΔS°.Substitute ΔH° = 0 → ΔG° = −TΔS°.Interpret sign: if ΔS° > 0, reaction tends to be spontaneous. Verification / Alternative check: Statistical thermodynamics similarly predicts that increased microstates (higher entropy) lowers free energy when enthalpy is unchanged. Why Other Options Are Wrong: TΔS°: Wrong sign. −ΔH°: Ignores entropy term. RT ln Keq: Correct general relation (ΔG° = −RT ln Keq) but the prompt asks the expression given ΔH° = 0; the simplified form is −TΔS°. ΔH° − T: Dimensionally incorrect. Common Pitfalls: Dropping the negative sign or mixing ΔG with ΔG°; standard vs nonstandard conditions differ. Final Answer: -TΔS°.

Question 1

Thermodynamics sign conventions: If the standard Gibbs free energy change (ΔG°) for a reaction is positive, what is favored at equilibrium under standard conditions?

Accepted Answer

Introduction / Context: Gibbs free energy links thermodynamics to chemical equilibrium. The sign and magnitude of ΔG° determine the direction of spontaneity under standard conditions and relate quantitatively to the equilibrium constant Keq via ΔG° = −RT ln Keq. Given Data / Assumptions: ΔG° > 0 (positive standard free energy change). Standard temperature and pressure; concentrations at 1 M for solutes (activities idealized). Concept / Approach: If ΔG° = −RT ln Keq is positive, then ln Keq is negative, meaning Keq < 1. This implies that, at equilibrium under standard conditions, the mixture contains more reactants than products—reactants are favored. Step-by-Step Solution: Start: ΔG° = −RT ln Keq. Given ΔG° > 0 ⇒ −RT ln Keq > 0 ⇒ ln Keq < 0. Therefore Keq < 1 (reactants favored at equilibrium). Select the statement consistent with Keq ≪ 1: reactants favored. Verification / Alternative check: Graphing ΔG° vs ln Keq shows the monotonic inverse relationship: positive ΔG° corresponds to small Keq. Why Other Options Are Wrong: Products favored requires Keq > 1 (ΔG° < 0); Keq = 1 implies ΔG° = 0; complete conversion ignores equilibrium; rate is kinetic, not governed by ΔG° alone. Common Pitfalls: Confusing thermodynamic favorability with speed; overlooking the log relationship between ΔG° and Keq. Final Answer: Reactants will be favored (Keq << 1).

Question 2

Driving unfavorable reactions in cells: Reactions with positive ΔG° can still proceed in vivo by which strategy?

Accepted Answer

Introduction / Context: Cells routinely drive thermodynamically unfavorable steps by harnessing energy from favorable processes. This is central to metabolism, transport, and biosynthesis, where pathway design and enzyme mechanisms exploit coupling and mass-action effects. Given Data / Assumptions: Standard free energy change ΔG° > 0 for the isolated reaction. Cellular systems are open and maintain non-equilibrium concentrations. ATP hydrolysis is a common exergonic currency (ΔG°′ ≈ −30.5 kJ/mol). Concept / Approach: Actual free energy change is ΔG = ΔG° + RT ln Q. By altering Q (ratio of products to reactants), ΔG can become negative. Enzymes also couple unfavorable reactions to favorable ones (including ATP hydrolysis), often via phosphorylated or otherwise high-energy intermediates, making the overall sum exergonic. Step-by-Step Solution: Use mass action: keep products low and reactants high → RT ln Q sufficiently negative. Couple reactions: A ⇌ B (unfavorable) + B ⇌ C (highly favorable) → A ⇌ C overall favorable. Exploit ATP: enzyme forms phosphorylated intermediate, transfers group to drive the endergonic step. Therefore, all listed strategies enable progress of an otherwise unfavorable step. Verification / Alternative check: Glycolysis and amino acid activation (aminoacyl-tRNA synthetases) exemplify ATP-coupled steps; metabolic channeling maintains favorable Q values. Why Other Options Are Wrong: Choosing only one strategy ignores the complementary biochemical tactics cells actually use. Common Pitfalls: Equating ΔG° with inevitability in vivo; forgetting that ΔG depends on concentrations and coupling. Final Answer: All of the above.

Question 3

Protein disorder upon unfolding: What is the effect on entropy when regular secondary structure (α-helix/β-sheet) unfolds into a random coil?

Accepted Answer

Introduction / Context: Entropy (disorder) plays a central role in protein stability. While folding can be enthalpically favorable due to noncovalent interactions, the unfolded chain has many more accessible conformations, dramatically affecting the entropy term in the free energy balance. Given Data / Assumptions: Regular secondary structures constrain backbone dihedral angles (ϕ, ψ) to narrow regions. Unfolding releases these constraints, increasing conformational freedom. We consider the protein’s conformational entropy specifically. Concept / Approach: Entropy S reflects the number of accessible microstates. An ordered helix/sheet permits far fewer microstates than a flexible random coil. Thus, unfolding increases protein conformational entropy substantially, although solvent entropy can counterbalance during hydrophobic collapse in folding. Step-by-Step Solution: Helix/sheet → restricted ϕ/ψ angles, low conformational variance. Unfolding → many backbone and side-chain rotamers become accessible. Microstate count rises sharply → entropy increases. Therefore, unfolding produces a large increase in protein entropy. Verification / Alternative check: Statistical thermodynamics and calorimetry support that conformational entropy is a major opposing term to folding; heat-denaturation increases configurational freedom. Why Other Options Are Wrong: Decrease/no change contradicts the definition of conformational entropy; “little increase” underestimates the substantial rise. Common Pitfalls: Ignoring solvent entropy (hydrophobic effect) and overgeneralizing—here the question focuses on the protein’s entropy. Final Answer: Large increase in the entropy of the protein.

Question 4

Meaning of standard Gibbs free energy: Which description best captures what ΔG° represents for a chemical reaction?

Accepted Answer

Introduction / Context: ΔG° is a thermodynamic state function that predicts spontaneity under standard conditions and links directly to the equilibrium constant Keq by ΔG° = −RT ln Keq. It is not a kinetic parameter and not a vague “residual energy.” Given Data / Assumptions: Standard state: 1 bar pressure, 1 M activities for solutes (idealized). Temperature typically 298 K unless otherwise specified. Equilibrium composition depends on Keq, which depends on ΔG°. Concept / Approach: Gibbs free energy is a state function; ΔG° compares the standard molar Gibbs energies of products and reactants, weighted by stoichiometry. Negative ΔG° favors products (Keq > 1); positive ΔG° favors reactants (Keq < 1). It is distinct from activation energy (Ea), which governs rate. Step-by-Step Solution: Recall definition: ΔG° = Σν G°(products) − Σν G°(reactants). Relate to equilibrium: ΔG° = −RT ln Keq. Interpretation: sign/magnitude indicate position of equilibrium under standard conditions. Thus, “difference in standard Gibbs energies” best captures the meaning. Verification / Alternative check: Thermodynamic tables list standard Gibbs energies of formation; summation gives ΔG° and predicts Keq consistent with experiments. Why Other Options Are Wrong: “Residual energy” is undefined; “energy required per mole regardless of conditions” ignores dependence on standard state; activation energy is kinetic, not ΔG°. Common Pitfalls: Confusing ΔG° with ΔG (which depends on Q) or with the activation barrier; forgetting that ΔG° is condition-specific (standard state). Final Answer: Difference in standard Gibbs energies of products and reactants (related to Keq).

Question 5

Protein folding and hydrogen bonds: During the unfolding of an α-helix, breaking each intramolecular hydrogen bond requires about 2 kJ/mol. What does this indicate about the relative strength of hydrogen bonds in proteins compared with hydrogen bond…

Accepted Answer

Introduction / Context: Hydrogen bonding stabilizes secondary structure (α-helices and β-sheets) in proteins. However, in aqueous solution a hydrogen bond broken inside a protein can often be replaced by new hydrogen bonds to water. The small energetic cost reported for bond breakage during helix unfolding helps interpret the balance between intraprotein and water-mediated hydrogen bonding. Given Data / Assumptions: Energetic cost to break each helix hydrogen bond is ~2 kJ/mol. Process occurs in water where donor/acceptor groups can hydrogen-bond to solvent. We compare relative strengths qualitatively (protein interior vs bulk water). Concept / Approach: In thermodynamics of folding, intramolecular hydrogen bonds compete with hydrogen bonds to water. If breaking a helix hydrogen bond required a large energy, that would imply markedly stronger bonds in proteins. A modest ~2 kJ/mol cost indicates that when one H-bond is disrupted, new H-bonds with water almost compensate, making the net cost small. Thus, protein H-bonds are not drastically stronger; if anything, they are slightly weaker or only marginally favorable relative to solvent H-bonding. Step-by-Step Solution: Identify the reported bond breakage energy: ~2 kJ/mol (small).Infer that water quickly re-forms H-bonds with exposed donors/acceptors.Conclude that intraprotein H-bonds are slightly less favorable than water H-bonds. Verification / Alternative check: Experimental unfolding energetics and molecular dynamics show that hydrogen bonds contribute modestly to stability in water; hydrophobic packing and side-chain interactions often dominate. Why Other Options Are Wrong: Much stronger / slightly stronger: Would imply a large energetic penalty, not observed. Not reformed with water: In water, exposed groups readily form new hydrogen bonds. Identical in all environments: Local environment modulates H-bond strength. Common Pitfalls: Assuming that any hydrogen bond automatically confers large stability; context (solvent competition) matters. Final Answer: Slightly weaker in proteins than in water.

Question 6

Standard Gibbs energy when enthalpy change is zero: If ΔH° = 0 for a reaction, what is the correct expression for ΔG°?

Accepted Answer

Introduction / Context: Gibbs free energy (ΔG°) predicts the spontaneity of reactions at constant temperature and pressure. Understanding how ΔH° (enthalpy) and ΔS° (entropy) contribute helps interpret biochemical processes such as protein folding and ligand binding. Given Data / Assumptions: ΔH° = 0 (no net enthalpy change under standard conditions). Temperature T is constant and in kelvin. Standard thermodynamic identity applies. Concept / Approach: The fundamental relation is ΔG° = ΔH° − TΔS°. If ΔH° = 0, then ΔG° simplifies to −TΔS°. A positive ΔS° gives a negative ΔG° (favorable), while a negative ΔS° gives positive ΔG° (unfavorable) at constant T. Step-by-Step Solution: Write identity: ΔG° = ΔH° − TΔS°.Substitute ΔH° = 0 → ΔG° = −TΔS°.Interpret sign: if ΔS° > 0, reaction tends to be spontaneous. Verification / Alternative check: Statistical thermodynamics similarly predicts that increased microstates (higher entropy) lowers free energy when enthalpy is unchanged. Why Other Options Are Wrong: TΔS°: Wrong sign. −ΔH°: Ignores entropy term. RT ln Keq: Correct general relation (ΔG° = −RT ln Keq) but the prompt asks the expression given ΔH° = 0; the simplified form is −TΔS°. ΔH° − T: Dimensionally incorrect. Common Pitfalls: Dropping the negative sign or mixing ΔG with ΔG°; standard vs nonstandard conditions differ. Final Answer: -TΔS°.

Thermodynamics and Free Energy Questions