Questions: Transition State Theory and Reaction Rate Constants

Question 1

Two reactions have identical activation enthalpies ΔH‡ = 50 kJ/mol, but reaction A involves two freely diffusing molecules combining into a tight cyclic transition state, while reaction B is a unimolecular rearrangement. At the same temperature, which reaction is faster?

• A. Reaction A — bimolecular reactions always have more collision opportunities
• B. Reaction B — the unimolecular rearrangement has a less negative activation entropy ΔS‡
• C. They are identical — only ΔH‡ determines the rate constant in TST
• D. Reaction A — higher molecularity means higher pre-exponential factor

Answer: (B)

TST gives k = (κ k_BT/h) · exp(−ΔG‡/RT) where ΔG‡ = ΔH‡ − TΔS‡. Even with equal ΔH‡, reaction A has a large negative ΔS‡ because two freely translating, rotating molecules must come together in a highly ordered, constrained transition state — losing translational and rotational degrees of freedom. This raises ΔG‡ and slows the reaction. The unimolecular rearrangement has a much smaller entropic penalty (ΔS‡ ≈ 0 or mildly negative). Ignoring entropy in TST is a common error that leads to wrong rate predictions.

Question 2

An enzyme reduces ΔG‡ from 80 kJ/mol to 60 kJ/mol at 310 K (body temperature). By what factor does the rate increase? (Use RT ≈ 2.57 kJ/mol)

• A. About 2-fold — a 20 kJ/mol reduction is modest
• B. About 40-fold — reflecting the exponential sensitivity
• C. About 2,800-fold — the exponential of 20/2.57 ≈ e^7.8
• D. About 10^7-fold — catalysis always gives astronomical rate enhancements

Answer: (C)

The rate ratio is exp(ΔΔG‡/RT) = exp(20/2.57) = exp(7.78) ≈ 2,400. Option C is closest (e^7.8 ≈ 2,440; using 2.57 gives ~2,400). This illustrates exponential sensitivity to ΔG‡: a seemingly modest 20 kJ/mol reduction translates to roughly a 2,000-fold rate enhancement. The statement in the explainer — 5.7 kJ/mol reduction gives 10-fold increase — follows from the same logic: exp(5.7/2.57·ln10) ≈ 10. Enzymes achieving reductions of 20–50 kJ/mol can accelerate reactions by factors of 10^3 to 10^8.

Question 3

According to transition state theory, a catalyst increases the reaction rate by raising the energy of the reactants relative to the transition state.

• T. True
• F. False

Answer: False

A catalyst lowers the activation free energy ΔG‡ by providing an alternative reaction pathway with a lower-energy transition state — it does not change the energy of the reactants or products. The free energy difference between reactants and products (reaction thermodynamics, ΔG) is unaffected by catalysis. Raising the reactant energy would also increase the reverse reaction rate identically, and it is not how catalysts function. Enzymes work by stabilizing the transition state through electrostatic interactions, precise substrate positioning, and covalent intermediates.

Question 4

A reaction can be slow even when the activation enthalpy ΔH‡ is small, if the activation entropy ΔS‡ is large and negative.

• T. True
• F. False

Answer: True

ΔG‡ = ΔH‡ − TΔS‡. A large negative ΔS‡ (highly organized transition state) adds a positive contribution to ΔG‡, raising the barrier and slowing the reaction regardless of how small ΔH‡ is. This is particularly important for reactions that bring together two large molecules in a precisely ordered geometry. The entropic cost of organizing a bimolecular transition state can easily be 50–100 J/mol·K, which at 300 K contributes 15–30 kJ/mol to ΔG‡ — enough to slow a reaction by 3–5 orders of magnitude relative to what ΔH‡ alone would predict.

Question 5

Why can a bimolecular reaction be slow even when its activation enthalpy ΔH‡ is relatively small?

Model answer: The activation free energy ΔG‡ = ΔH‡ − TΔS‡ depends on both enthalpic and entropic terms. A bimolecular reaction that brings two freely moving molecules into a single, tightly ordered transition state pays a large entropic penalty (ΔS‡ << 0), because translational and rotational degrees of freedom are lost. Even if ΔH‡ is small, the TΔS‡ term adds a substantial positive contribution to ΔG‡, slowing the reaction by orders of magnitude. A slow reaction with small ΔH‡ is often entropically controlled.

This insight is why many enzyme mechanisms are designed to reduce the entropic cost of forming the transition state — by binding both substrates in the correct orientation, the enzyme effectively pre-organizes the transition state geometry, reducing the ΔS‡ penalty. Without the enzyme, both the enthalpic and entropic contributions to ΔG‡ must be overcome; with the enzyme, the entropic portion is largely paid by tight binding rather than by thermal fluctuation.