Thermodynamic and Kinetic Control - College Chemistry

1. A reversible reaction system shows the following behavior: at 25°C, 85% Product A and 15% Product B; at 100°C, 35% Product A and 65% Product B. After heating to 100°C and then cooling back to 25°C, the final composition is 40% Product A and 60% Product B. What does this indicate?
   Answer: Product A is kinetically favored at 25°C, but Product B is thermodynamically favored When you encounter problems about product distributions at different temperatures, you're dealing with the fundamental distinction between kinetic control (which product forms faster) and thermodynamic control (which product is more stable at equilibrium). The key insight comes from analyzing what happens after the heating-cooling cycle. Initially at 25°C, you have 85% A and 15% B. After heating to 100°C, the composition shifts to 35% A and 65% B. Most importantly, when cooled back to 25°C, the final composition is 40% A and 60% B—not the original 85% A and 15% B. This tells you that the original 25°C composition wasn't at true equilibrium. If it were, you'd return to 85% A after cooling. Instead, the heating provided enough energy to overcome kinetic barriers and reach closer to the true thermodynamic equilibrium, where B is favored. At 25°C, A forms faster (kinetic control), but B is actually more thermodynamically stable. Option A is wrong because A isn't thermodynamically favored—the equilibrium actually favors B. Option C incorrectly states B is kinetically favored; the initial high concentration of A shows A forms faster at 25°C. Option D is incorrect because the original 25°C state represents kinetic, not thermodynamic control. The correct answer is B: Product A is kinetically favored at 25°C (forms faster initially), but Product B is thermodynamically favored (more stable at equilibrium). Study tip: When product ratios change irreversibly after heating-cooling cycles, suspect that kinetic control was masking the true thermodynamic preference.
2. Two parallel reactions compete for the same starting material. At 300 K, the rate constants are $k_1 = 2.3 \times 10^{-3} \ \text{s}^{-1}$ ($E_a = 85 \ \text{kJ/mol}$) and $k_2 = 1.8 \times 10^{-4} \ \text{s}^{-1}$ ($E_a = 95 \ \text{kJ/mol}$). The equilibrium constants are $K_1 = 150$ and $K_2 = 850$. Which product dominates under kinetic control?
   Answer: Product 1, because it has the larger rate constant and lower activation energy When you encounter competing parallel reactions, you need to distinguish between kinetic control and thermodynamic control. Under kinetic control, the product that forms faster dominates, regardless of which is more thermodynamically stable. Under thermodynamic control, the more stable (lower energy) product dominates. Since this question asks about kinetic control, you should focus on reaction rates, not equilibrium constants. The reaction with the larger rate constant will produce its product faster. Here, $k_1 = 2.3 \times 10^{-3} \ \text{s}^{-1}$ is about 13 times larger than $k_2 = 1.8 \times 10^{-4} \ \text{s}^{-1}$. Additionally, the lower activation energy (85 kJ/mol vs 95 kJ/mol) confirms that reaction 1 proceeds faster, making Product 1 kinetically favored. A is correct because it properly identifies that under kinetic control, the larger rate constant and lower activation energy determine which product dominates. B incorrectly focuses on equilibrium constants, which determine thermodynamic stability, not kinetic favorability. C makes a conceptual error by claiming lower activation energy affects thermodynamic favorability—activation energy only affects reaction rate, not product stability. D is wrong because higher activation energy indicates a slower reaction and less stable transition state, not more stable. Study tip: Remember the key distinction—kinetic control = faster reaction wins (look at rate constants), thermodynamic control = more stable product wins (look at equilibrium constants or ΔG values). The activation energy tells you about reaction speed, not product stability.

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