Q. The rate of a reaction quadruples when temperature changes from $27^\circ C$ to $57^\circ C$. Calculate the energy of activation. Given R = 8.314 J K$^{-1}$ mol$^{-1}$, log 4 = 0.6021
A) 3.80 kJ/mol
B) 3804 kJ/mol
C) 38.04 kJ/mol
D) 380.4 kJ/mol
Explanation
[Solution by Claude - not from source] $\log4=\frac{E_a}{2.303\times8.314}\cdot\frac{30}{300\times330}\Rightarrow E_a\approx38{,}046$ J/mol $=38.04$ kJ/mol. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.