Thermodynamics check: If the current reactant concentration exceeds the equilibrium condition, what happens to Gibbs free energy and reaction direction?

Difficulty: Easy

Correct Answer: Both (b) and (c)

Explanation:


Introduction / Context:
Chemical and biochemical reactions are driven by the Gibbs free energy change ΔG, which depends on both intrinsic standard free energy (ΔG°) and the instantaneous mass-action ratio (reaction quotient, Q). Understanding how concentrations relative to equilibrium determine spontaneity is central to metabolism.



Given Data / Assumptions:

  • Reactant concentration is higher than the equilibrium composition (i.e., there is an excess of reactants relative to products).
  • Temperature is constant; ΔG = ΔG° + R*T*ln(Q).
  • Keq = exp(−ΔG°/(R*T)).



Concept / Approach:
When there are more reactants than at equilibrium, Q = [products]/[reactants] is smaller than Keq. Because ΔG can also be written as R*T*ln(Q/Keq), if Q < Keq then ln(Q/Keq) is negative, making ΔG negative. A negative ΔG indicates the forward reaction is spontaneous and will convert reactants into products until Q approaches Keq.



Step-by-Step Solution:
Define Q = [products]/[reactants]. Excess reactants → Q decreases.Compute ΔG = R*T*ln(Q/Keq). With Q < Keq → ln term is negative.Therefore ΔG < 0 (thermodynamically favorable forward direction).As the system evolves, product concentration rises and ΔG approaches 0 at equilibrium.



Verification / Alternative check:
Le Chatelier’s principle predicts that adding reactants drives the reaction toward products to re-establish equilibrium, consistent with ΔG becoming negative.



Why Other Options Are Wrong:

  • Positive ΔG: Would require Q > Keq (excess products), not the case here.
  • Reaction must stop: Q = Keq only at equilibrium; the prompt states a deviation from equilibrium.



Common Pitfalls:
Confusing ΔG° (a constant at given conditions) with ΔG (state-dependent). In cells, ΔG often differs markedly from ΔG° due to metabolite concentrations.



Final Answer:
Both (b) and (c).


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