1. Tardigrade
  2. Question
  3. Chemistry
  4. A hydrogenation reaction is carried out at 500 K. CH2 = CH2 + H2 xrightarrow[ textno catalyst]500 K CH3 - CH3 Activation energy - Ea kJ mol -1 CH2 = CH2 + H2 xrightarrowPd. 400 K CH3 - CH3 Activation energy = (Ea - 20) kJ mol -1 If rate remains constant, then Ea is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A hydrogenation reaction is carried out at $500 K$.
$CH_{2} = CH_{2} + H_{2} \xrightarrow[{\text{no catalyst}}]{500 \,K} CH_{3} - CH_{3}$
Activation energy $- E_{a} kJ$ mol $^{-1}$
$CH_{2} = CH_{2} + H_{2} \xrightarrow{Pd. 400\, K} CH_{3} - CH_{3}$
Activation energy $ = \left(E_{a} - 20\right) kJ$ mol $^{-1}$
If rate remains constant, then $E_{a}$ is

Chemical Kinetics

Solution:

$k = Ae^{-E_{a}/RT}$
$k =Ae^{-E_{a}/R \times500}$
$k' = Ae^{-\left(E_{a}=20\right)/R\times400}$
Rates are equal, hence
$\therefore Ae^{-E_{a}/R \times500} =Ae^{-\left(E_{a}-20\right)R \times400}$
$\therefore \frac{E_{a}}{500} = \frac{E_{a } -20}{400}$
$4E_{a} = 5E_{a} -100$
$\therefore E_{a} = 100 kJ$ mol $^{-1}$