A gas sample is heated from 300 K to 1200 K (temperature quadrupled). By what factor does the most probable molecular speed increase?
AIt doubles, because most probable speed scales as the square root of temperature
BIt quadruples, because most probable speed scales linearly with temperature
CIt increases by a factor of 16, because kinetic energy scales as temperature squared
DIt stays the same — only the distribution broadens, the peak doesn't move
The most probable speed v_p = √(2kT/m) scales as √T. Quadrupling T multiplies v_p by √4 = 2. The common mistake (option B) is to confuse the linear T-dependence of average kinetic energy with the speed — since KE ∝ T and KE ∝ v², we get v ∝ √T, not v ∝ T. Option D is wrong because both the peak position and the width shift with temperature.
Question 2 Multiple Choice
Which ordering correctly ranks the three characteristic speeds of the Maxwell-Boltzmann distribution?
Av_p < v_avg < v_rms
Bv_avg < v_p < v_rms
Cv_rms < v_avg < v_p
DAll three are equal for an ideal gas at equilibrium
v_p (most probable) = √(2kT/m) < v_avg (mean) = √(8kT/πm) < v_rms = √(3kT/m). The rms speed is highest because squaring before averaging gives extra weight to the high-speed tail. The ratios v_p : v_avg : v_rms = 1 : 1.128 : 1.225. They would only be equal for a delta-function distribution, not the asymmetric Maxwell-Boltzmann.
Question 3 True / False
The Maxwell-Boltzmann speed distribution is symmetric around the most probable speed — equal fractions of molecules have speeds above and below v_p.
TTrue
FFalse
Answer: False
The distribution is distinctly asymmetric. It starts at zero (no molecules at zero speed), rises to a peak at v_p, then falls off with a longer tail at high speeds than at low speeds. This asymmetry arises because the geometric phase-space factor (∝ v²) pulls the distribution above zero at the peak, and the exponential Boltzmann factor decays but doesn't cut off as sharply as the low-speed side. More molecules have speeds above v_p than below it.
Question 4 True / False
Even a small temperature increase can dramatically accelerate chemical reaction rates because the fraction of molecules above the activation energy threshold grows exponentially with temperature.
TTrue
FFalse
Answer: True
The fraction of molecules with energy above E_a is proportional to e^(−E_a/kT) from the Boltzmann factor in the Maxwell-Boltzmann distribution. Because E_a appears in the exponent, a small ΔT produces a large multiplicative change in this fraction. For a typical activation energy of ~50 kJ/mol, heating from 300 K to 310 K roughly doubles the reactive fraction — explaining why reaction rates approximately double per 10°C. This exponential sensitivity underlies the Arrhenius equation.
Question 5 Short Answer
Why is the high-speed tail of the Maxwell-Boltzmann distribution disproportionately important for chemical reaction rates, even though it represents a tiny fraction of all molecules?
Think about your answer, then reveal below.
Model answer: Chemical reactions require collisions with energy above the activation energy E_a. Only molecules in the high-speed tail have enough kinetic energy to react. Because the fraction of molecules above E_a depends on e^(−E_a/kT), even a small increase in the tail population produces a large increase in reaction rate — the tail fraction changes exponentially with temperature while average speed changes only as √T.
This is the physical basis of the Arrhenius equation: k = A·e^(−E_a/kT). The pre-exponential factor A reflects collision frequency (which scales slowly with √T), but the exponential term reflects how the tail population changes. A reaction rate that doubles for every 10°C isn't about average speed increasing — average speed only increases by ~1.6% per 10°C at 300 K. It's entirely about the exponential sensitivity of the high-energy tail.