Questions: Arrhenius Equation and Temperature Dependence of Rate Constants

Question 1

An enzyme lowers the activation energy of a reaction from 80 kJ/mol to 50 kJ/mol at 310 K. The rate constant increases dramatically because:

• A. Lowering Eₐ increases the frequency of molecular collisions, so more reactions occur per second
• B. Eₐ appears in the exponent of the Arrhenius equation, so even a moderate reduction in Eₐ produces an exponential increase in k
• C. The enzyme raises the temperature of the local reaction environment, effectively increasing RT
• D. Lowering Eₐ changes the thermodynamics of the reaction, making the products more stable

Answer: (B)

The Arrhenius equation k = A·exp(−Eₐ/RT) places Eₐ in the exponent. At 310 K (body temperature), RT ≈ 2.58 kJ/mol. A reduction of 30 kJ/mol in Eₐ changes the exponent by 30/2.58 ≈ 11.6 units. Since exp(11.6) ≈ 110,000, the rate constant increases by roughly five orders of magnitude — not because collisions are more frequent, but because a vastly larger fraction of collisions now have sufficient energy to overcome the barrier. Critically, the enzyme does not change ΔG (the thermodynamics are unchanged); it only lowers the kinetic barrier. This exponential sensitivity to Eₐ is why enzymes are so spectacularly effective.

Question 2

On an Arrhenius plot (ln k vs 1/T), Reaction A has a steeper negative slope than Reaction B. This means:

• A. Reaction A is faster than Reaction B at all temperatures
• B. Reaction A has a higher activation energy than Reaction B
• C. Reaction A is more temperature-sensitive at low temperatures, but less so at high temperatures
• D. Reaction A has a smaller pre-exponential factor A than Reaction B

Answer: (B)

The slope of the Arrhenius plot is −Eₐ/R. A steeper (more negative) slope means a larger Eₐ/R ratio, hence higher activation energy. A higher Eₐ means the rate constant changes more dramatically per degree of temperature change — the reaction is more temperature-sensitive. Note that slope tells you nothing about which reaction is faster at any given temperature (that also depends on the pre-exponential factor A); two reactions can have different slopes but the same rate at some temperature. The Arrhenius plot cleanly separates the two contributions: slope gives Eₐ, y-intercept gives ln(A).

Question 3

A catalyst increases the rate of a reaction without being consumed, and it does so by lowering the activation energy without changing the overall thermodynamics (ΔG) of the reaction.

• T. True
• F. False

Answer: True

This is the precise mechanistic definition of catalysis. A catalyst provides an alternative reaction pathway with a lower Eₐ, so more collisions carry enough energy to proceed — increasing k exponentially. However, ΔG = ΔH − TΔS for the overall reaction is set by the reactant and product identities, which the catalyst does not change. The equilibrium constant K is related to ΔG (not to the kinetics), so the catalyst does not shift the equilibrium position — it only reaches it faster. A catalyst speeds the forward and reverse reactions equally, leaving Keq unchanged. This is why a catalyst cannot cause a thermodynamically unfavorable reaction to proceed — it can only accelerate thermodynamically allowed reactions.

Question 4

Doubling the absolute temperature of a reaction typically approximately doubles the rate constant.

• T. True
• F. False

Answer: False

The relationship is exponential, not linear. Doubling T from 300 K to 600 K changes exp(−Eₐ/RT) from exp(−Eₐ/300R) to exp(−Eₐ/600R). For a typical Eₐ of 60 kJ/mol, this changes the exponent from −24 to −12 — increasing k by exp(12) ≈ 160,000-fold, far more than doubling. For a smaller Eₐ or a smaller temperature jump, the factor is smaller. The common empirical rule of thumb that a 10°C rise roughly doubles the rate works for reactions with Eₐ ≈ 50–80 kJ/mol near room temperature, but it is an approximation that breaks down at other temperatures and activation energies.

Question 5

Why does the Arrhenius plot (ln k versus 1/T) yield a straight line, and what information can be extracted from its slope and intercept?

Model answer: Taking the natural log of k = A·exp(−Eₐ/RT) gives ln(k) = ln(A) − Eₐ/(RT). Written as ln(k) = ln(A) − (Eₐ/R)·(1/T), this is in the form y = b + mx, a linear equation where y = ln(k), x = 1/T, slope m = −Eₐ/R, and intercept b = ln(A). A straight line results because the log transformation converts the exponential relationship into a linear one. The slope gives the activation energy (Eₐ = −slope × R), and the y-intercept gives ln(A), from which the pre-exponential factor A can be extracted.

The Arrhenius plot is the standard experimental method for determining activation energies. Rate constants are measured at several temperatures, plotted as ln(k) vs 1/T, and the best-fit line is drawn. A curved Arrhenius plot signals that the mechanism changes with temperature (e.g., a different rate-limiting step dominates at high vs. low T), which is diagnostically important. The linearity assumption underlies most kinetic analyses, so curvature is always worth investigating rather than dismissing.