JEE PYQ: Chemical Kinetics Question 38
Question 38 - 2019 (11 Jan Shift 1)
If a reaction follows the Arrhenius equation, the plot of $\ln k$ vs $\frac{1}{(\text{RT})}$ gives a straight line with a gradient $(-y)$ unit. The energy required to activate the reactant is:
(1) $y/R$ unit
(2) $y$ unit
(3) $yR$ unit
(4) $-y$ unit
Show Answer
Answer: (2)
Solution
From Arrhenius equation: $k = Ae^{-E_a/RT}$. $\ln k = \ln A - \frac{E_a}{RT}$. slope $= -y$ (given). $-y = -E_a$. $E_a = y$.
BETA
