Maxwell Boltzmann Distribution Pogil Answer Key Extension Questions May 2026

Students must perform a qualitative calculation to see the exponential effect.

Step-by-step calculation of the fraction ratio:

The Ratio of Rates: [ \frac\textRate at 400K\textRate at 300K = \frace^-15.03e^-20.05 = e^5.02 \approx 152 ]

Conclusion: Even though the temperature increased by only 100K, the reaction rate is 150 times faster. The M-B extension question forces students to realize that kinetic energy distributions are mercilessly exponential.

POGIL Acceptable Answer: "The fraction of molecules with sufficient energy is exquisitely sensitive to temperature because (E_a / RT) appears in the exponent. A 100K increase reduces the exponent magnitude, yielding a 150-fold increase in reactive collisions."

Question: Given that the fraction of molecules with kinetic energy greater than (E_a) is roughly ( e^-E_a / RT ), explain why a reaction with (E_a = 50 \text kJ/mol) proceeds very slowly at 300K but rapidly at 400K. (Use (R = 8.314 \text J/mol·K)). Students must perform a qualitative calculation to see

Answer: At very low speeds, very few molecules have exactly zero velocity because kinetic energy is quantized in terms of molecular motion; also, the probability density function ( f(v) \propto v^2 e^-mv^2/(2kT) ) gives ( f(v) \to 0 ) as ( v \to 0 ).

This is a trick question to test if students confuse distribution with total number.

The Answer: No, the shape does not change.

Reasoning:

POGIL Acceptable Answer: "The M-B distribution depends only on temperature and mass (and the fundamental constants). Vacuum reduces the number of molecules but does not change the fraction of molecules at a given speed. The curve's shape is invariant under changes in pressure or volume." The Ratio of Rates: [ \frac\textRate at 400K\textRate

The statement is approximately true but not strictly true for a real gas.

An advanced extension question modified from standard POGILs:

Question: A soccer ball (mass 0.43 kg) is treated as a "molecule" at 300 K. Calculate its most probable speed. Why does it not appear to move even though the M-B distribution applies?

Answer: Using ( v_p = \sqrt\frac2RTM ) — but here we use ( R = 8.314 , J/(mol·K) ) and mass in kg/mol. Molar mass of soccer ball = ( 0.43 , kg \times 6.022 \times 10^23 = 2.59 \times 10^23 , kg/mol ).

[ v_p = \sqrt\frac2(8.314)(300)2.59 \times 10^23 \approx \sqrt1.93 \times 10^-20 \approx 1.39 \times 10^-10 , m/s ] Question: Given that the fraction of molecules with

This is slower than a nanometer per second. The reason we don't see the ball move is that the velocity is infinitesimally small due to the enormous "molar mass" of a macroscopic object, and the ball is constantly bombarded asymmetrically by air molecules (Brownian motion), but the net thermal velocity is dwarfed by friction and gravity.

When students are stuck on the Extension Questions, use these guided inquiry prompts:

Prompt: As temperature increases, what happens to the peak of the curve? Why does this violate a simple "shift to the right" explanation?

Answer: The peak (most probable speed) increases and shifts to the right, but the height of the peak decreases.

Reasoning: Students often mistakenly think the peak simply moves right and up. In reality, because the total area (number of molecules) is constant, the curve must "spread out." To maintain the same area, the curve must flatten. Mathematically, the most probable speed ( v_p = \sqrt\frac2RTM ) increases with T. However, the peak height is proportional to ( \frac1\sqrtT ), meaning it drops as temperature rises.