1. Tardigrade
  2. Question
  3. Chemistry
  4. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 20KJmol- 1 . If K1 and K2 are rate constants for reactions R1 and R2 respectively at 300K , then ln((K2/ K1)) is equal to (.R=8.314J(mol)- 1K- 1.) :

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Q. Two reactions $R_{1}$ and $R_{2}$ have identical pre-exponential factors. Activation energy of $R_{1}$ exceeds that of $R_{2}$ by $20KJmol^{- 1}$ . If $$ $K_{1}$ and $K_{2}$ are rate constants for reactions $R_{1}$ and $R_{2}$ respectively at $300K$ , then $ln\left(\frac{K_{2}}{ K_{1}}\right)$ is equal to $\left(\right.R=8.314J\left(mol\right)^{- 1}K^{- 1}\left.\right)$ :

NTA AbhyasNTA Abhyas 2020

Solution:

$Ea_{1}=Ea_{2}+20$
$\ell nK_{1}=cnA-\frac{E_{1}}{R \times 300}$
$\ell nK_{2}=lnA-\frac{E_{2}}{R \times 300}$
eq.(2) $-$ eq.(1)
$\ell n\left(\frac{K_{2}}{ K_{1}}\right)=\frac{E_{1} - E_{2}}{R \times 300}$
$\ell n\left(\frac{K_{2}}{ K_{1}}\right)=\frac{20 \times \left(10\right)^{3}}{8 . 314 \times 300}\approx8$