JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Chemical Kinetics

From the Arrhenius equation $k=A{e}^{-E_a/RT}$, a larger activation energy makes the exponential term smaller and so lowers the rate constant, which makes the first statement correct. The temperature coefficient $k_{T+10}/k_T$ grows with activation energy, because a larger $E_a$ makes $k$ more sensitive to temperature, so the second statement is correct. Taking logarithms gives $\ln k=\ln A-\frac{E_a}{R}\cdot \frac{1}{T}$, a straight line of slope $-E_a/R$ against $\frac{1}{T}$, so the fourth statement is correct. The actual rise in $k$ for a ten-degree increase is larger at higher temperatures, where $k$ itself is large, so the claim that the change is greater at lower temperatures does not hold. Three of the statements are correct.