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

Chemical Kinetics: JEE Main 2019 Chemistry question

JEE Main 2019 · 11 Jan, Morning shift · Q446

JEE Main 201911 Jan, Morning shiftChemistryMediumChemical Kinetics62× since 2002

If a reaction follows the Arrhenius equation, the plot ln k vs

{1 \over {\left( {RT} \right)}}

gives straight line with a gradient (

-

y) unit. The energy required to active the reactant is :

JEE Main 2019 · 11 Jan, Morning shift · Q446

JEE Main 201911 Jan, Morning shiftChemistryMediumChemical Kinetics62× since 2002

If a reaction follows the Arrhenius equation, the plot ln k vs

{1 \over {\left( {RT} \right)}}

gives straight line with a gradient (

-

y) unit. The energy required to active the reactant is :