Chemistry - Chemical Kinetics Question with Solution | TestHub

$\log k$ versus T will give a straight line

$\log k$ versus $1/\mathrm{T}$ gives straight line whose slope is $-2500/2.303\mathrm{R}$

Half time of reaction will be more at higher temperatures

$\log k$ versus $1/\mathrm{T}$ gives a straight line with slope $2500/\mathrm{R}$