JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Chemical Kinetics

From the Arrhenius equation $\log k=\log A-\frac{E_a}{2.303RT}$, a plot of $\log k$ against $1/T$ is a straight line of slope $-\frac{E_a}{2.303R}$, so a steeper downward slope corresponds to a larger activation energy. In the graph the line labelled $2$ descends most steeply, the line labelled $1$ is of intermediate steepness, and the line labelled $3$ is almost flat. The activation energies therefore decrease in the order $E_{a2}>E_{a1}>E_{a3}$.