For a reaction, given below is the graph of $lnkvs\dfrac{1}{T}$ . The activation energy for the reaction is equal to____ $cal{mol}^{- 1}$ .(Given : $R = 2{calK}^{- 1}{mol}^{- 1}$ )

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For a reaction, given below is the graph of lnkvs(1)/(T) . The activation energy for the reaction is equal to____ cal mol ^ - 1 .(Given : R = 2 calK ^ - 1 mol ^ - 1 )

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Arrhenius suggested an equation which describes rate constant, $k$ as a function of temperature, T.
$k = {Ae}^{- Ea / RT}$
A=frequency factor
${E}_{a} = Activationenergy$
In the Arrhenius equation the factor $e{}^{- Ea / RT}$ corresponds to the fraction of molecules that have kinetic energy greater than E_a
$lnk = \dfrac{-}{Ea} + lnA$
$Y = C + MX$
On plotting $lnkagainst1 / T$ we get a straight line.The slope of the graph gives activation energy and intercept gives frequency factor
slope $= - \dfrac{Ea}{R} = \dfrac{-}{Δy}$
Slope $= \dfrac{Ea}{R} = \dfrac{20}{5}$
${E}_{a} = 4R = 8Cal / mol$

For a reaction, given below is the graph of lnkvs(1)/(T) . The...

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