1. Tardigrade
  2. Question
  3. Chemistry
  4. For a given reaction, energy of activation for forward reaction ((Ea)f) is 80KJmol- 1 and ΔH=-40KJmol- 1 for the reaction. A catalyst lowers (Ea)f by 20KJmol- 1. The ratio of energy of activation for reverse reaction before and after addition of catalyst is:-

Q. For a given reaction, energy of activation for forward reaction $\left(\left(Ea\right)_{f}\right)$ is $80KJmol^{- 1}$ and $ΔH=-40KJmol^{- 1}$ for the reaction. A catalyst lowers $\left(Ea\right)_{f}$ by $20KJmol^{- 1}.$ The ratio of energy of activation for reverse reaction before and after addition of catalyst is:-

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Solution:

$ΔH=\left(\right.Ea\left(\left.\right)_{f}-\left(\right.Ea\left(\left.\right)_{b}$
$ΔH$ remains same after addition of catalyst
$-40=80-\left(\right.Ea\left(\left.\right)_{b}$
$\left(\right.Ea\left(\left.\right)_{b}$ before catalyst addition $=120$
$ΔH=\left(\right.Ea\left(\left.\right)_{f}-\left(\right.Ea\left(\left.\right)_{b}$
$-40=60-\left(\right.Ea\left(\left.\right)_{b}\Rightarrow \left(\right.Ea\left(\left.\right)_{b}=100$
$\left(\right.Ea\left(\left.\right)_{b}$ after addition of catalyst $=100$
$\therefore $ Ratio $=\frac{120}{100}=1.2$

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