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

Question

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

Aicharya · Accepted Answer

Arrhenius suggested an equation which describes rate constant, $k$ as a function of temperature, T.
$k = {Ae}^{- Ea / RT}$
A=frequency factor
$E_{a} = Activation energy$
In the Arrhenius equation the factor $e^{- Ea / RT}$ corresponds to the fraction of molecules that have kinetic energy greater than E_a
$lnk = \frac{-}{Ea} + lnA$
$Y = C + M X$
On plotting $lnk against 1 / T$ we get a straight line.The slope of the graph gives activation energy and intercept gives frequency factor

slope $= - \frac{Ea}{R} = \frac{-}{Δy}$
Slope $= \frac{Ea}{R} = \frac{20}{5}$
$E_{a} = 4 R = 8 Cal / mol$

For a reaction, given below is the graph of lnk vs 1 T . The activation energy for the reaction is equal…

Solution