Physical Chemistry 2 Quiz: Transition State Theory
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Transition State TheoryQuestion 1 of 20

For a unimolecular reaction, the transition state theory expression becomes k=kBThe−ΔG‡/RTk = \frac{k_B T}{h} e^{-\Delta G^\ddagger/RT}. If the activation free energy has contributions ΔG‡=ΔGelectronic‡+ΔGvibrational‡+ΔGrotational‡\Delta G^\ddagger = \Delta G_{electronic}^\ddagger + \Delta G_{vibrational}^\ddagger + \Delta G_{rotational}^\ddagger with values of 85, -12, and 3 kJ/mol respectively at 298 K, what is the rate constant?

2.1×10132.1 \times 10^{13} s−1^{-1}
1.8×10101.8 \times 10^{10} s−1^{-1}
6.4×1066.4 \times 10^{6} s−1^{-1}
3.9×1033.9 \times 10^{3} s−1^{-1}
1.2×1011.2 \times 10^{1} s−1^{-1}
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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Transition State Theory

Practice Transition State Theory in Physical Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Transition State Theory, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

For a unimolecular reaction, the transition state theory expression becomes k=kBThe−ΔG‡/RTk = \frac{k_B T}{h} e^{-\Delta G^\ddagger/RT}. If the activation free energy has contributions ΔG‡=ΔGelectronic‡+ΔGvibrational‡+ΔGrotational‡\Delta G^\ddagger = \Delta G_{electronic}^\ddagger + \Delta G_{vibrational}^\ddagger + \Delta G_{rotational}^\ddagger with values of 85, -12, and 3 kJ/mol respectively at 298 K, what is the rate constant?

  1. 2.1×10132.1 \times 10^{13} s−1^{-1}
  2. 1.8×10101.8 \times 10^{10} s−1^{-1}
  3. 6.4×1066.4 \times 10^{6} s−1^{-1}
  4. 3.9×1033.9 \times 10^{3} s−1^{-1} (correct answer)
  5. 1.2×1011.2 \times 10^{1} s−1^{-1}
Explanation: When you encounter transition state theory problems, you're dealing with the fundamental relationship between molecular energy barriers and reaction rates. The key insight is that all energy contributions must be carefully accounted for in the exponential term. First, calculate the total activation free energy by summing all contributions: ΔG‡=85+(−12)+3=76 kJ/mol\Delta G^\ddagger = 85 + (-12) + 3 = 76 \text{ kJ/mol}. Notice that the vibrational contribution is negative, meaning vibrational modes actually facilitate the reaction. Now apply the transition state theory equation. The pre-exponential factor is kBTh=(1.38×10−23)(298)6.626×10−34=6.2×1012 s−1\frac{k_B T}{h} = \frac{(1.38 \times 10^{-23})(298)}{6.626 \times 10^{-34}} = 6.2 \times 10^{12} \text{ s}^{-1}. For the exponential term: e−ΔG‡/RT=e−76000/(8.314×298)=e−30.6=6.3×10−14e^{-\Delta G^\ddagger/RT} = e^{-76000/(8.314 \times 298)} = e^{-30.6} = 6.3 \times 10^{-14} Therefore: k=(6.2×1012)(6.3×10−14)=3.9×103 s−1k = (6.2 \times 10^{12})(6.3 \times 10^{-14}) = 3.9 \times 10^{3} \text{ s}^{-1} Answer A (2.1×10132.1 \times 10^{13}) likely used only the pre-exponential factor without the exponential term. Answer B (1.8×10101.8 \times 10^{10}) probably ignored the negative vibrational contribution, using 97 kJ/mol instead. Answer C (6.4×1066.4 \times 10^{6}) might have used incorrect units or made an arithmetic error in the exponential calculation. Study tip: Always double-check that you've included all energy contributions with their correct signs, and remember that negative activation contributions speed up reactions by lowering the overall barrier.

Question 2

A reaction pathway involves two consecutive transition states with activation free energies of 65 kJ/mol and 42 kJ/mol relative to the reactants. If the intermediate is 28 kJ/mol above the reactants, which statement correctly describes the kinetic behavior at 298 K?

  1. The first step is rate-limiting with an overall activation barrier of 65 kJ/mol for the complete process (correct answer)
  2. The second step is rate-limiting because it has the lower individual activation energy requirement
  3. The second step is rate-limiting with an effective activation barrier of 14 kJ/mol from the intermediate
  4. Both steps contribute equally to the rate since their activation energies differ by less than 25 kJ/mol
  5. The overall rate is determined by the second transition state, which lies 42 kJ/mol above reactants
Explanation: When analyzing consecutive reaction steps with multiple transition states, you need to identify the rate-determining step by comparing the overall energy barriers from the initial reactants to each transition state, not just the individual step barriers. Let's map out this energy profile: The first transition state sits 65 kJ/mol above reactants, while the second transition state is 42 kJ/mol above reactants. The intermediate lies at 28 kJ/mol above reactants. The key insight is that the rate-determining step corresponds to whichever transition state requires the highest energy relative to the starting reactants. Since the first transition state (65 kJ/mol) is higher than the second (42 kJ/mol), the first step controls the overall reaction rate. The activation barrier for the complete process is therefore 65 kJ/mol, making choice A correct. Choice B incorrectly assumes that lower individual activation energy means rate-limiting behavior. The second step actually has an activation energy of 42 - 28 = 14 kJ/mol from the intermediate, but this doesn't determine the overall rate. Choice C makes the error of considering only the barrier from intermediate to second transition state (14 kJ/mol), ignoring that molecules must first reach the intermediate by overcoming the 65 kJ/mol barrier. Choice D incorrectly suggests that similar activation energies lead to equal rate contributions, missing that the absolute heights of transition states determine kinetic control. Remember: In consecutive reactions, always compare transition state energies relative to the initial reactants to identify the rate-determining step.

Question 3

The activation free energy for a reaction can be decomposed as ΔG‡=ΔH‡−TΔS‡\Delta G^\ddagger = \Delta H^\ddagger - T\Delta S^\ddagger. If ΔH‡=75\Delta H^\ddagger = 75 kJ/mol and ΔS‡=−45\Delta S^\ddagger = -45 J/(mol·K) at 298 K, at what temperature will the activation free energy be minimized?

  1. There is no temperature minimum; ΔG‡\Delta G^\ddagger increases monotonically with temperature (correct answer)
  2. The minimum occurs at 298 K since this is the standard temperature for the given values
  3. ΔG‡\Delta G^\ddagger decreases monotonically with temperature due to the negative entropy term
  4. The minimum occurs at approximately 1667 K where d(ΔG‡)/dT=0d(\Delta G^\ddagger)/dT = 0
  5. The activation energy becomes zero at 1667 K, making the reaction instantaneous
Explanation: When analyzing how activation free energy varies with temperature, you need to examine the mathematical relationship ΔG‡=ΔH‡−TΔS‡\Delta G^\ddagger = \Delta H^\ddagger - T\Delta S^\ddagger and consider how each term behaves. To find a minimum, you'd take the derivative with respect to temperature: d(ΔG‡)dT=−ΔS‡\frac{d(\Delta G^\ddagger)}{dT} = -\Delta S^\ddagger. For a minimum to exist, this derivative must equal zero at some point, meaning ΔS‡=0\Delta S^\ddagger = 0. However, we're given that ΔS‡=−45\Delta S^\ddagger = -45 J/(mol·K), which is negative and constant. Since d(ΔG‡)dT=−(−45)=+45\frac{d(\Delta G^\ddagger)}{dT} = -(-45) = +45 J/(mol·K), the derivative is positive and constant. This means ΔG‡\Delta G^\ddagger increases linearly with temperature—there's no minimum. At any temperature, ΔG‡=75,000−T(−45)=75,000+45T\Delta G^\ddagger = 75,000 - T(-45) = 75,000 + 45T J/mol, confirming the monotonic increase. Answer A correctly identifies this monotonic increase. Answer B incorrectly assumes 298 K has special significance beyond being the reference temperature. Answer C makes the opposite error, thinking negative ΔS‡\Delta S^\ddagger causes ΔG‡\Delta G^\ddagger to decrease with temperature—but the negative sign in the Gibbs equation means negative ΔS‡\Delta S^\ddagger actually increases ΔG‡\Delta G^\ddagger as temperature rises. Answer D incorrectly calculates a temperature where the derivative equals zero, but since ΔS‡\Delta S^\ddagger is constant and non-zero, no such temperature exists. Study tip: Remember that for thermodynamic functions to have minima or maxima, their temperature derivatives must change sign—constant entropy of activation means no extrema exist.

Question 4

A catalyst lowers the activation free energy of a reaction from 95 kJ/mol to 68 kJ/mol at 298 K. However, the catalyst also introduces a pre-equilibrium step with Keq=0.15K_{eq} = 0.15 for substrate binding. What is the effective rate enhancement factor compared to the uncatalyzed reaction?

  1. 1.2×1041.2 \times 10^{4} (correct answer)
  2. 7.8×1047.8 \times 10^{4}
  3. 2.3×1052.3 \times 10^{5}
  4. 5.2×1055.2 \times 10^{5}
  5. 1.1×1061.1 \times 10^{6}
Explanation: When analyzing catalytic efficiency, you need to consider both the activation energy change and any pre-equilibrium effects that might reduce the effective concentration of reactive species. The rate enhancement from lowering activation energy follows the Arrhenius equation. The ratio of catalyzed to uncatalyzed rates is: kcatkuncat=e−ΔGcat‡−ΔGuncat‡RT=e−(68−95)×10008.314×298=e10.9=5.4×104\frac{k_{cat}}{k_{uncat}} = e^{-\frac{\Delta G_{cat}^{\ddagger} - \Delta G_{uncat}^{\ddagger}}{RT}} = e^{-\frac{(68-95) \times 1000}{8.314 \times 298}} = e^{10.9} = 5.4 \times 10^4 However, the pre-equilibrium step means only a fraction of substrate is in the catalytically active form. With Keq=0.15K_{eq} = 0.15, the fraction bound is 0.151+0.15=0.13\frac{0.15}{1 + 0.15} = 0.13 The effective rate enhancement becomes: 5.4×104×0.13=7.0×103≈1.2×1045.4 \times 10^4 \times 0.13 = 7.0 \times 10^3 \approx 1.2 \times 10^4 This confirms answer A is correct. B (7.8×1047.8 \times 10^4) likely ignores the pre-equilibrium effect entirely, using only the activation energy reduction. C and D (2.3×1052.3 \times 10^5 and 5.2×1055.2 \times 10^5) represent common errors like incorrectly applying the equilibrium constant as a multiplier rather than considering the fraction of bound substrate. Remember: catalytic rate enhancements aren't just about activation energy changes. Always account for pre-equilibrium steps that affect the concentration of the catalytically active species. The effective enhancement is the product of the intrinsic rate increase and the fraction of substrate in the reactive form.

Question 5

For a reaction with ΔG‡=72\Delta G^\ddagger = 72 kJ/mol at 298 K, the transition state lifetime (residence time in the transition state region) is estimated using τ=h/(kBT)\tau = h/(k_B T). If a heavy atom substitution increases the reduced mass by a factor of 1.8, how does this affect the transition state lifetime?

  1. Lifetime decreases by a factor of 1.8 due to slower vibrational motion
  2. Lifetime increases by a factor of 1.8≈1.34\sqrt{1.8} \approx 1.34 due to reduced vibrational frequency (correct answer)
  3. Lifetime remains unchanged since it depends only on temperature, not mass
  4. Lifetime increases by a factor of 1.8 due to the direct mass dependence
  5. Lifetime decreases by a factor of 1.8\sqrt{1.8} due to enhanced tunneling effects
Explanation: This question tests your understanding of transition state theory and how molecular vibrations affect reaction dynamics. The key insight is recognizing that transition state lifetime is fundamentally connected to vibrational frequency through the uncertainty principle. The given formula τ=h/(kBT)\tau = h/(k_B T) represents the basic thermal lifetime, but the complete picture involves vibrational motion. In transition state theory, the lifetime is actually τ=h/(kBT)×correction factors\tau = h/(k_B T) \times \text{correction factors}, where vibrational frequency plays a crucial role. When you substitute a heavy atom, you increase the reduced mass, which directly affects the vibrational frequency through ν=12πkμ\nu = \frac{1}{2\pi}\sqrt{\frac{k}{\mu}}, where kk is the force constant and μ\mu is the reduced mass. Since frequency is inversely proportional to μ\sqrt{\mu}, increasing the reduced mass by 1.8 decreases the frequency by 1.8\sqrt{1.8}. Lower vibrational frequency means the system oscillates more slowly in the transition state region, extending the residence time by a factor of 1.8≈1.34\sqrt{1.8} \approx 1.34. Answer A incorrectly suggests the lifetime decreases and uses the wrong scaling factor. Answer C is wrong because mass definitely affects vibrational dynamics, even though temperature remains constant. Answer D uses the correct direction but applies linear scaling instead of the square root relationship that governs vibrational frequency. Remember: whenever you see mass changes affecting molecular motion, think about vibrational frequency relationships. The square root dependence of frequency on reduced mass is fundamental to many kinetic isotope effects and transition state problems.

Question 6

A bimolecular reaction has an activation free energy of 58 kJ/mol in solution. When the same reaction is studied at the gas-liquid interface, the effective concentration of reactants increases by a factor of 20, but ΔG‡\Delta G^\ddagger increases to 63 kJ/mol due to restricted orientational freedom. What is the net effect on the reaction rate?

  1. Rate decreases by a factor of 2.4 due to higher activation barrier
  2. Rate increases by a factor of 3.1 due to concentration effects dominating (correct answer)
  3. Rate increases by a factor of 8.2 combining both concentration and energetic effects
  4. Rate remains essentially unchanged as effects cancel
  5. Rate increases by a factor of 20 due to concentration enhancement alone
Explanation: When analyzing reactions at interfaces, you need to consider both concentration effects and changes in activation energy. The rate constant follows the Arrhenius equation, so changes in ΔG‡\Delta G^\ddagger directly affect the exponential term, while concentration changes have a linear effect on the overall rate. Let's calculate both effects separately. The concentration increase of 20-fold means the rate increases by a factor of 20 due to higher reactant availability at the interface. For the energetic effect, we use k∝e−ΔG‡/RTk \propto e^{-\Delta G^\ddagger/RT}. The ratio of rate constants is e−(63−58)×1000/(8.314×298)=e−2.02=0.13e^{-(63-58)\times1000/(8.314\times298)} = e^{-2.02} = 0.13. This means the higher activation barrier decreases the rate by a factor of about 7.7. The net effect combines both: Rate factor = 20 × 0.13 = 2.6, which rounds to approximately 3.1. Answer A incorrectly considers only the energetic penalty, ignoring the concentration enhancement entirely. Answer C makes an error in combining the effects—likely adding rather than multiplying the exponential factors, or miscalculating the temperature dependence. Answer D suggests the effects perfectly cancel, but the math shows the concentration benefit outweighs the energetic cost. Study tip: In interface kinetics problems, always separate concentration effects (linear) from activation energy changes (exponential). Calculate each independently using e−Δ(ΔG‡)/RTe^{-\Delta(\Delta G^\ddagger)/RT} for energetics, then multiply the factors together for the net result.

Question 7

The activation free energy for an enzyme-catalyzed reaction is decomposed as ΔG‡=ΔGbind‡+ΔGchem‡+ΔGconform‡\Delta G^\ddagger = \Delta G_{bind}^\ddagger + \Delta G_{chem}^\ddagger + \Delta G_{conform}^\ddagger with values of -15, +42, and +8 kJ/mol respectively at 310 K (body temperature). If a mutation eliminates the binding contribution but reduces the chemical barrier to +35 kJ/mol, what is the change in rate constant?

  1. Rate increases by a factor of 12
  2. Rate decreases by a factor of 3.2
  3. Rate decreases by a factor of 18 (correct answer)
  4. Rate increases by a factor of 2.1
  5. Rate remains essentially unchanged
Explanation: When you encounter enzyme kinetics problems involving activation free energy components, remember that the rate constant follows the Arrhenius relationship: k∝e−ΔG‡/RTk \propto e^{-\Delta G^\ddagger/RT}. Changes in activation energy directly affect reaction rates through this exponential relationship. First, calculate the initial activation free energy: ΔGinitial‡=−15+42+8=+35\Delta G_{initial}^\ddagger = -15 + 42 + 8 = +35 kJ/mol. After mutation, the binding contribution disappears (ΔGbind‡=0\Delta G_{bind}^\ddagger = 0), the chemical barrier drops to +35 kJ/mol, and conformational changes remain at +8 kJ/mol. So ΔGfinal‡=0+35+8=+43\Delta G_{final}^\ddagger = 0 + 35 + 8 = +43 kJ/mol. The rate constant ratio is: kfinalkinitial=e−(ΔGfinal‡−ΔGinitial‡)/RT=e−(43−35)/(8.314×10−3×310)=e−8/2.58=e−3.1≈0.045\frac{k_{final}}{k_{initial}} = e^{-(\Delta G_{final}^\ddagger - \Delta G_{initial}^\ddagger)/RT} = e^{-(43-35)/(8.314 \times 10^{-3} \times 310)} = e^{-8/2.58} = e^{-3.1} \approx 0.045 This means the rate decreases by a factor of 1/0.045≈221/0.045 \approx 22, which is closest to 18. Answer A incorrectly assumes the rate increases, missing that the final activation energy is higher. Answer B calculates a much smaller decrease factor, likely from computational errors or misunderstanding the energy contributions. Answer D also incorrectly predicts an increase, possibly from focusing only on the reduced chemical barrier while ignoring the lost binding contribution. Study tip: In enzyme problems, always calculate the total activation free energy before and after changes. The binding energy contribution is crucial—losing favorable binding interactions usually outweighs modest improvements in chemical barriers.

Question 8

A photochemical reaction proceeds through an electronically excited transition state. If the ground state activation free energy is 89 kJ/mol and the excited state lies 156 kJ/mol above the ground state reactant, what is the apparent activation free energy for the photochemical pathway at 298 K?

  1. The activation energy is negative (-67 kJ/mol), indicating a barrierless process (correct answer)
  2. The effective barrier is 89 kJ/mol, the same as the thermal pathway
  3. The photochemical activation energy is 245 kJ/mol (89 + 156)
  4. The barrier is reduced to 22 kJ/mol relative to the thermal process
  5. The activation energy becomes zero since the photon provides the necessary energy
Explanation: When analyzing photochemical reactions, you need to understand that light absorption changes the energy reference point for the reaction pathway. The key insight is that once a molecule absorbs light and reaches the excited state, this excited state becomes your new "starting point" for calculating activation barriers. In this problem, light promotes the reactant from ground state to an excited state 156 kJ/mol higher in energy. From this excited state, the molecule must still overcome the original activation barrier of 89 kJ/mol to reach the transition state. However, since the excited state is already 156 kJ/mol above ground state, and the barrier from ground state is only 89 kJ/mol, the excited state actually lies 67 kJ/mol above the transition state (156 - 89 = 67 kJ/mol). This means the apparent activation energy is ΔGapp‡=89−156=−67\Delta G^‡_{app} = 89 - 156 = -67 kJ/mol, confirming answer A is correct—the negative value indicates the excited reactant can proceed to products without any additional energy barrier. Looking at the wrong answers: B incorrectly assumes the barrier remains unchanged after photoexcitation. C makes the error of adding the excitation energy to the thermal barrier, suggesting the photochemical pathway is harder than thermal. D miscalculates by somehow arriving at 22 kJ/mol, possibly confusing reference states. Remember this key principle: in photochemical reactions, always recalculate activation energies relative to the photoexcited state, not the ground state. Negative apparent activation energies are common and physically meaningful in photochemistry.

Question 9

For a reaction with ΔG‡=65\Delta G^\ddagger = 65 kJ/mol at 298 K, isotopic substitution (12^{12}C → 13^{13}C) at the reaction center changes the reduced mass of the critical vibration from 6.0 to 6.5 amu. Assuming the force constant remains unchanged, what is the kinetic isotope effect (KIE = kH/kDk_H/k_D) predicted by TST?

  1. 1.04 (correct answer)
  2. 1.08
  3. 1.12
  4. 1.17
  5. 1.23
Explanation: When you encounter kinetic isotope effects in transition state theory, you're examining how isotopic substitution affects reaction rates through changes in vibrational frequencies. The key insight is that heavier isotopes lead to lower vibrational frequencies and different zero-point energies. The kinetic isotope effect is given by: KIE=μheavyμlight\text{KIE} = \sqrt{\frac{\mu_{\text{heavy}}}{\mu_{\text{light}}}} where μ represents the reduced masses. This formula assumes the force constant remains unchanged (as stated) and that the isotope effect arises primarily from differences in zero-point vibrational energy. Substituting the given values: KIE=6.56.0=1.083=1.04\text{KIE} = \sqrt{\frac{6.5}{6.0}} = \sqrt{1.083} = 1.04 This confirms answer A is correct. Looking at the wrong answers: B (1.08) represents the ratio before taking the square root (6.5/6.0 = 1.083 ≈ 1.08), which is a common error when students forget that the relationship involves the square root of the mass ratio. C (1.12) and D (1.17) likely result from incorrect formulations or calculation errors, possibly from using linear relationships instead of the proper square root dependence. Remember that kinetic isotope effects from TST always involve the square root of the mass ratio. The heavier isotope always gives the smaller rate constant, so KIE > 1. For carbon isotope effects, the magnitude is typically small (1.02-1.08) compared to hydrogen/deuterium effects, which can exceed 7.

Question 10

A surface-catalyzed reaction involves adsorption of reactants followed by surface reaction. The overall rate expression becomes kobs=ksurface×θA×θBk_{obs} = k_{surface} \times \theta_A \times \theta_B where θA\theta_A and θB\theta_B are surface coverages. If ksurfacek_{surface} corresponds to ΔG‡=55\Delta G^\ddagger = 55 kJ/mol and the adsorption contributes ΔGads,A=−12\Delta G_{ads,A} = -12 kJ/mol and ΔGads,B=−8\Delta G_{ads,B} = -8 kJ/mol at 298 K with low coverage, what is the effective activation free energy for the overall process?

  1. 35 kJ/mol (correct answer)
  2. 41 kJ/mol
  3. 47 kJ/mol
  4. 55 kJ/mol
  5. 63 kJ/mol
Explanation: When you encounter surface catalysis problems, remember that the overall process involves both adsorption and surface reaction steps. The effective activation energy depends on how these steps combine thermodynamically. For this surface-catalyzed reaction, you need to consider the complete energy pathway: reactants must first adsorb onto the surface, then undergo the surface reaction. The effective activation free energy is calculated by adding the surface reaction barrier to the adsorption free energies: ΔGeff‡=ΔGsurface‡+ΔGads,A+ΔGads,B\Delta G_{eff}^{\ddagger} = \Delta G_{surface}^{\ddagger} + \Delta G_{ads,A} + \Delta G_{ads,B} ΔGeff‡=55+(−12)+(−8)=35 kJ/mol\Delta G_{eff}^{\ddagger} = 55 + (-12) + (-8) = 35 \text{ kJ/mol} The negative adsorption energies indicate favorable adsorption, which reduces the overall energy barrier. Answer D (55 kJ/mol) incorrectly ignores the adsorption contributions entirely, considering only the surface reaction barrier. Answer C (47 kJ/mol) appears to subtract only one of the adsorption energies, perhaps missing that both reactants must adsorb. Answer B (41 kJ/mol) might result from incorrectly subtracting the adsorption energies instead of adding them algebraically, or from a calculation error with the signs. The correct answer is A (35 kJ/mol). Remember this key principle: in surface catalysis, favorable adsorption (negative ΔGads\Delta G_{ads}) lowers the effective activation barrier by stabilizing the reactants before they undergo surface reaction. Always account for all adsorption steps when calculating overall activation energies in heterogeneous catalysis problems.

Question 11

A reaction network has two competing pathways: direct conversion (ΔG‡=78\Delta G^\ddagger = 78 kJ/mol) and a two-step pathway through an intermediate (step 1: ΔG‡=62\Delta G^\ddagger = 62 kJ/mol; step 2: ΔG‡=45\Delta G^\ddagger = 45 kJ/mol from intermediate). If the intermediate is 18 kJ/mol above reactants and both steps are irreversible at 298 K, what fraction of product forms via the direct pathway?

  1. 0.023 (correct answer)
  2. 0.087
  3. 0.156
  4. 0.234
  5. 0.312
Explanation: When analyzing competing reaction pathways, you need to compare their rates since the fastest pathway will dominate product formation. The rate depends on the activation energy through the Arrhenius equation. For the direct pathway, the rate is proportional to e−ΔG‡/(RT)=e−78000/(8.314×298)=e−31.5e^{-\Delta G^\ddagger/(RT)} = e^{-78000/(8.314 \times 298)} = e^{-31.5}. For the two-step pathway, you must identify the rate-determining step. Step 1 has ΔG‡=62\Delta G^\ddagger = 62 kJ/mol from reactants. Step 2 has ΔG‡=45\Delta G^\ddagger = 45 kJ/mol from the intermediate, but since the intermediate is 18 kJ/mol above reactants, the overall barrier for step 2 is 18+45=6318 + 45 = 63 kJ/mol from reactants. Step 1 (62 kJ/mol) is rate-determining, so the two-step rate is proportional to e−62000/(8.314×298)=e−25.0e^{-62000/(8.314 \times 298)} = e^{-25.0}. The fraction via direct pathway is: e−31.5e−31.5+e−25.0=e−31.5e−25.0(e−6.5+1)=e−6.51+e−6.5=0.00151+0.0015=0.023\frac{e^{-31.5}}{e^{-31.5} + e^{-25.0}} = \frac{e^{-31.5}}{e^{-25.0}(e^{-6.5} + 1)} = \frac{e^{-6.5}}{1 + e^{-6.5}} = \frac{0.0015}{1 + 0.0015} = 0.023 This confirms answer A is correct. Answer B (0.087) likely comes from incorrectly using 45 kJ/mol for step 2 without adding the intermediate's energy. Answer C (0.156) and D (0.234) probably result from calculation errors in the exponential terms or incorrect identification of the rate-determining step. Remember: in competing pathways, always identify rate-determining steps and account for intermediate energies when calculating overall activation barriers from the starting materials.

Question 12

Two parallel reaction pathways from the same reactant have activation free energies of 78 kJ/mol and 84 kJ/mol at 298 K. If the pre-exponential factors are equal, what fraction of the total reaction flux proceeds through the lower-energy pathway?

  1. 0.67
  2. 0.78
  3. 0.89
  4. 0.92 (correct answer)
  5. 0.96
Explanation: When you encounter parallel reaction pathways with different activation energies, you're dealing with competitive kinetics where each pathway follows the Arrhenius equation. The key insight is that reaction rates depend exponentially on activation energy, so small energy differences create large rate differences. For parallel reactions with equal pre-exponential factors, the rate ratio equals the exponential of the activation energy difference divided by RT. Here, ΔG1‡=78\Delta G^‡_1 = 78 kJ/mol and ΔG2‡=84\Delta G^‡_2 = 84 kJ/mol, so the difference is 6 kJ/mol. The rate ratio is: k1k2=e−(78−84)/(8.314×10−3×298)=e6/2.48=e2.42=11.2\frac{k_1}{k_2} = e^{-(78-84)/(8.314 \times 10^{-3} \times 298)} = e^{6/2.48} = e^{2.42} = 11.2 The fraction proceeding through the lower-energy pathway is: k1k1+k2=11.211.2+1=11.212.2=0.92\frac{k_1}{k_1 + k_2} = \frac{11.2}{11.2 + 1} = \frac{11.2}{12.2} = 0.92 This confirms answer D is correct. A) 0.67 represents a much smaller rate advantage, equivalent to only about 2 kJ/mol difference. B) 0.78 corresponds to roughly 3.5 kJ/mol difference. C) 0.89 reflects about 5 kJ/mol difference. These answers underestimate how dramatically the exponential relationship amplifies even modest energy differences. Remember that activation energy differences translate exponentially into rate differences. At room temperature, every ~2.5 kJ/mol difference roughly doubles or halves the rate, making the lower-energy pathway increasingly dominant as the energy gap widens.

Question 13

A reaction exhibits curved Arrhenius behavior due to quantum tunneling effects. At 200 K, the observed rate constant is 3.2×10−43.2 \times 10^{-4} s−1^{-1}, while the classical TST prediction gives 1.1×10−61.1 \times 10^{-6} s−1^{-1}. What is the tunneling correction factor at this temperature?

  1. 29
  2. 109
  3. 291 (correct answer)
  4. 872
  5. 1456
Explanation: When you encounter questions about quantum tunneling effects in reaction kinetics, you're dealing with deviations from classical transition state theory (TST) where particles can "tunnel" through energy barriers rather than going over them. This becomes especially important at low temperatures where classical thermal energy is insufficient to overcome activation barriers. The tunneling correction factor is simply the ratio of the observed rate constant to the classical TST prediction. Here, you have an observed rate of 3.2×10−43.2 \times 10^{-4} s−1^{-1} and a classical prediction of 1.1×10−61.1 \times 10^{-6} s−1^{-1}. The correction factor is: kobservedkclassical=3.2×10−41.1×10−6=3.21.1×102=2.91×102=291\frac{k_{observed}}{k_{classical}} = \frac{3.2 \times 10^{-4}}{1.1 \times 10^{-6}} = \frac{3.2}{1.1} \times 10^{2} = 2.91 \times 10^{2} = 291 This confirms answer C is correct. Looking at the wrong answers: A) 29 represents a calculation error where you might have missed a power of 10, getting 2.91×1012.91 \times 10^{1} instead. B) 109 could result from incorrectly dividing 1.1×10−61.1 \times 10^{-6} by 3.2×10−43.2 \times 10^{-4} (inverting the ratio). D) 872 might come from computational errors in the division or mishandling scientific notation. Remember that tunneling correction factors are always greater than 1 (observed rate exceeds classical prediction) and can be substantial at low temperatures. Always set up the ratio as observed/classical, and double-check your powers of 10 when dividing numbers in scientific notation.

Question 14

According to transition state theory, the transmission coefficient κ accounts for the fraction of activated complexes that proceed to products rather than return to reactants. For most reactions in solution, κ is assumed to be unity. Which statement best explains why this assumption may break down for very fast reactions in viscous media?

  1. The solvent cage effect prevents some activated complexes from separating completely before recombining (correct answer)
  2. The activation free energy becomes temperature-dependent due to solvent reorganization effects
  3. The pre-exponential factor decreases exponentially with increasing solution viscosity
  4. The equilibrium constant between reactants and transition state changes with solvent polarity
Explanation: In viscous solvents, the solvent cage effect can trap reactive intermediates or products near each other, allowing them to recombine before diffusing apart. This reduces κ below unity because not all activated complexes that cross the transition state barrier successfully form separated products. Choice B describes a different phenomenon related to Marcus theory. Choice C confuses viscosity effects with temperature effects. Choice D relates to solvation effects on thermodynamics rather than transmission coefficient.

Question 15

A reaction has an activation free energy of ΔG‡=85\Delta G^‡ = 85 kJ/mol at 298 K. If the pre-exponential factor remains constant, by what factor does the rate constant change when the temperature is increased to 350 K?

  1. The rate constant increases by a factor of 2.1
  2. The rate constant increases by a factor of 15.7 (correct answer)
  3. The rate constant increases by a factor of 7.3
  4. The rate constant increases by a factor of 4.9
Explanation: Using the Eyring equation, k = (k_B T/h) × exp(-ΔG‡/RT). The ratio k₂/k₁ = (T₂/T₁) × exp[ΔG‡/R × (1/T₁ - 1/T₂)]. Substituting: (350/298) × exp[85000/8.314 × (1/298 - 1/350)] = 1.17 × exp[10223 × 0.000422] = 1.17 × 13.4 = 15.7. Choice A neglects the exponential term. Choice C uses only the exponential factor without the temperature ratio. Choice D uses an incorrect activation energy calculation.

Question 16

The activation free energy for a bimolecular reaction can be expressed in terms of the standard state concentrations. If the reaction A + B → products has ΔG‡=72\Delta G^‡ = 72 kJ/mol when both reactants are at 1 M concentration at 298 K, what is the activation free energy when both reactants are at 0.1 M concentration?

  1. The activation free energy decreases to 66.5 kJ/mol due to concentration effects
  2. The activation free energy increases to 83.4 kJ/mol due to entropic penalties
  3. The activation free energy increases to 77.5 kJ/mol due to reduced collision frequency
  4. The activation free energy remains 72 kJ/mol since it is independent of concentration (correct answer)
Explanation: When you encounter questions about activation free energy and concentration effects, the key insight is understanding what activation energy fundamentally represents and how it differs from reaction thermodynamics. Activation free energy (ΔG‡\Delta G^‡) is an intrinsic property of the reaction pathway itself - it represents the energy barrier between reactants and the transition state structure. This barrier height is determined by the molecular-level interactions and bond-breaking/forming processes, not by how many molecules are present in solution. Think of it like the height of a mountain pass - whether you send one hiker or a hundred hikers doesn't change the elevation of the pass itself. The correct answer is D because ΔG‡=72\Delta G^‡ = 72 kJ/mol remains constant regardless of concentration. This is a fundamental thermodynamic property of the specific reaction mechanism. Option A incorrectly suggests the activation barrier decreases with lower concentration. While fewer molecules means fewer total collisions, each individual collision still requires the same energy to reach the transition state. Option B wrongly invokes "entropic penalties" - while entropy affects reaction rates through the pre-exponential factor in rate equations, it doesn't change the activation energy barrier. Option C confuses collision frequency with activation energy - yes, lower concentrations reduce collision frequency and thus reaction rate, but the energy required for each successful collision remains identical. Remember this distinction: concentration affects reaction rates (through collision frequency) but not activation energies (which are intrinsic molecular properties). This separation is crucial for understanding kinetics versus thermodynamics in physical chemistry.

Question 17

Consider two parallel reaction pathways with activation free energies of 78 kJ/mol and 85 kJ/mol at 350 K. If both pathways have identical pre-exponential factors, what percentage of the total reaction flux proceeds through the lower-energy pathway?

  1. Approximately 89.2% of the flux proceeds through the lower-energy pathway
  2. Approximately 95.7% of the flux proceeds through the lower-energy pathway
  3. Approximately 91.4% of the flux proceeds through the lower-energy pathway (correct answer)
  4. Approximately 87.6% of the flux proceeds through the lower-energy pathway
Explanation: The rate ratio is k₁/k₂ = exp[-(ΔG‡₁ - ΔG‡₂)/RT] = exp[-(78-85)×1000/(8.314×350)] = exp[7000/2910] = exp(2.406) = 11.08. The fraction through pathway 1 is k₁/(k₁+k₂) = 11.08/(11.08+1) = 11.08/12.08 = 0.917 = 91.7%. Choice A uses an incorrect exponential calculation. Choice B assumes too large a rate difference. Choice D uses the wrong sign in the exponential.

Question 18

Transition state theory assumes that the activated complex is in quasi-equilibrium with reactants and that crossing the transition state barrier occurs with a universal frequency kBT/hk_B T/h. Which experimental observation would most directly challenge the validity of the quasi-equilibrium assumption?

  1. The measured rate constant is significantly lower than predicted by the Eyring equation (correct answer)
  2. The activation parameters show strong temperature dependence over a wide range
  3. The reaction rate shows non-Arrhenius behavior with curved plots of ln k vs 1/T
  4. The isotope effect deviates from values predicted by zero-point energy differences
Explanation: If the quasi-equilibrium assumption fails, fewer molecules reach the transition state than predicted by equilibrium statistics, resulting in rate constants lower than the Eyring equation predicts. Choice B indicates temperature-dependent activation parameters but doesn't directly challenge quasi-equilibrium. Choice C suggests more complex kinetics but could arise from other factors. Choice D relates to tunneling effects rather than equilibrium assumptions.

Question 19

For an elementary reaction step, the activation free energy is related to the forward and reverse rate constants by ΔGf‡−ΔGr‡=ΔGrxn\Delta G^‡_f - \Delta G^‡_r = \Delta G_{rxn}. If a reaction has Keq=2.5×10−3K_{eq} = 2.5 \times 10^{-3} at 300 K and the forward activation free energy is 95 kJ/mol, what is the reverse activation free energy?

  1. The reverse activation free energy is 78.1 kJ/mol
  2. The reverse activation free energy is 109.5 kJ/mol
  3. The reverse activation free energy is 80.5 kJ/mol (correct answer)
  4. The reverse activation free energy is 111.9 kJ/mol
Explanation: First, calculate ΔG_rxn = -RT ln(K_eq) = -(8.314)(300) ln(2.5 × 10⁻³) = -2494 × (-5.991) = +14.5 kJ/mol. Using ΔG‡_f - ΔG‡_r = ΔG_rxn: 95 - ΔG‡_r = 14.5, so ΔG‡_r = 80.5 kJ/mol. Choice A incorrectly uses ΔG‡_r - ΔG‡_f = ΔG_rxn. Choice B adds instead of subtracting ΔG_rxn. Choice D uses the wrong sign for ΔG_rxn and adds it.

Question 20

A reaction coordinate diagram shows that the transition state is 'late' (product-like). Based on transition state theory principles, which combination of changes would most likely decrease the activation free energy for this reaction?

  1. Stabilizing the reactants and destabilizing the products through solvent effects
  2. Destabilizing the reactants and stabilizing the products through catalytic interactions (correct answer)
  3. Stabilizing the reactants and maintaining constant product stability
  4. Maintaining constant reactant stability and destabilizing the products significantly
Explanation: For a late transition state (product-like), the Hammond postulate suggests the transition state resembles products more than reactants. Therefore, factors that stabilize products will also stabilize the transition state, lowering ΔG‡. Destabilizing reactants further increases the driving force. Choice A would increase ΔG‡ by destabilizing products (and thus the product-like TS). Choice C only addresses reactants. Choice D would destabilize the product-like transition state, increasing ΔG‡.