If a reaction follows the Arrhenius equation, the plot ln k vs ( RT ) 1 gives straight line with a gradient ( − y ) unit. The energy required to activate the reactant is:

From Arrhenius equation, k = A e − E n / RT ln k = ln A − RT E a slope = − y (given) − y = − E s ⇒ E a = y

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ChemistryChemical KineticsZero Order ReactionsMedium2 minPYQ_2019

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If a reaction follows the Arrhenius equation, the plot $ln k$ vs $\frac{1}{( RT )}$ gives straight line with a gradient $(- y)$ unit. The energy required to activate the reactant is:

Options:

Answer:

Solution:

From Arrhenius equation, $k = A e^{- E_{n} / RT}$ $ln k = ln A - \frac{E _{a}}{RT}$ slope $= - y$ (given) $- y = - E_{s}$ $\Rightarrow E_{a} = y$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Zero Order Reactions

⏱ 2mℹ️ Source: PYQ_2019

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