Questions: The Arrhenius Equation and Activation Energy
3 questions to test your understanding
Score: 0 / 3
Question 1 Multiple Choice
Two reactions have the same frequency factor A. Reaction X has Ea = 40 kJ/mol; Reaction Y has Ea = 80 kJ/mol. At the same temperature, which reaction is faster and why?
AReaction X, because lower activation energy means a greater fraction of collisions have sufficient energy to react
BReaction Y, because higher activation energy means the transition state releases more energy when products form
CReaction X, because the frequency factor dominates and A is the same for both
DReaction Y, because more energy input means more molecules are activated per unit time
In the Arrhenius equation k = Ae^(-Ea/RT), a larger Ea makes the exponent more negative, which makes e^(-Ea/RT) smaller and therefore k smaller. Physically, the Boltzmann factor e^(-Ea/RT) represents the fraction of molecular collisions with enough energy to overcome the activation barrier — lower Ea means a larger fraction clears the bar. With equal A, Reaction X has a larger k and is faster.
Question 2 True / False
Adding a catalyst to a reaction increases the rate by raising the temperature of the reaction mixture, which increases the average kinetic energy of the molecules.
TTrue
FFalse
Answer: False
Catalysts do not raise the temperature of a reaction. They provide an alternative reaction pathway with a lower activation energy Ea. At the same temperature, a smaller Ea makes the Boltzmann factor e^(-Ea/RT) larger, increasing k. Temperature and activation energy are two independent ways to change k in the Arrhenius equation. Confusing these two mechanisms is a common error — a catalyst affects Ea (and often A), not T.
Question 3 Short Answer
A student plots ln k versus 1/T for a reaction and measures a slope of −9600 K. What is the activation energy in kJ/mol? (R = 8.314 J/mol·K)
Think about your answer, then reveal below.
Model answer: Ea ≈ 79.8 kJ/mol. From the linearized Arrhenius equation ln k = ln A − (Ea/R)(1/T), the slope equals −Ea/R. Therefore Ea = −slope × R = 9600 K × 8.314 J/mol·K = 79,814 J/mol ≈ 79.8 kJ/mol.
The linearized form is the practical tool for extracting Ea from experimental data. The slope of a ln k vs 1/T plot is always −Ea/R (negative because higher temperature means lower 1/T and higher k). Multiplying the magnitude of the slope by R converts from units of K to J/mol. Dividing by 1000 converts to kJ/mol, the conventional unit for activation energies.