4. The percentage (%) increase in the rate of a chemical reaction will be maximum when
the temperature is increased from:
a. 270 K to 280 K
b. 280 K to 290 K
c. 290 K to 300 K
d. In all these cases the increase will be the same
Question Analysis
Reaction rates follow the Arrhenius equation k = A e^(-Ea/RT), where higher temperatures exponentially increase the rate constant k by boosting molecular collisions with sufficient activation energy Ea.
(k₂ - k₁)/k₁ × 100% = (e^(Ea/R × (1/T₁ - 1/T₂)) - 1) × 100%The percentage increase depends on Δ(1/T), which decreases as absolute temperature rises despite fixed 10 K intervals.
Option Breakdown
| Option | Temperature Range | Δ(1/T) | % Increase (Ea ≈ 50 kJ/mol) | Explanation |
|---|---|---|---|---|
| a | 270 K to 280 K | 0.000132 | ~122% | Largest Δ(1/T) yields maximum exponential increase |
| b | 280 K to 290 K | 0.000123 | ~110% | Less than option a |
| c | 290 K to 300 K | 0.000115 | ~100% (doubling) | Smallest Δ(1/T) |
| d | Same increase | – | – | Incorrect: Δ(1/T) varies |
Correct Answer: a. 270 K to 280 K
Why Lower Temperatures Maximize Increase
At lower Kelvin ranges like 270K-280K, the relative change in 1/T is largest for equal ΔT=10K intervals, amplifying k’s exponential growth. Calculations confirm ~122% rise here versus ~100% at 290K-300K for Ea = 50 kJ/mol. Rates roughly double every 10°C rise near room temperature, but percentage gains diminish higher up.
CSIR NET Exam Insights
This MCQ tests Arrhenius understanding: option a wins as 1/270 – 1/280 > 1/290 – 1/300. Practice with ln(k₂/k₁) = Ea/R × (1/T₁ - 1/T₂) to solve similar problems.
