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  • Two reactions R1 and R2 have identical pre-exponential factors.  Activation energy of R1 exceeds that of R2 by 10 kJ mol−1.  If k1 and k2 are rate constants fo

Two reactions R1 and R2 have identical pre-exponential factors.  Activation energy of R1 exceeds that of R2 by 10 kJ mol−1.  If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to : (R=8.314 J mol−1K−1)  
Option: 1 6
Option: 2 4
Option: 3 8
Option: 4 12
 

Answers (1)

best_answer

K=Ae^{\frac{-Ea}{RT}}

\frac{K_{1}}{K_{2}}=e\left ( \frac{Ea_{1}-Ea_{2}}{RT} \right )

Taking log both the sides

ln\left ( \frac{K_{2}}{K_{1}} \right )=\frac{E_{a1}-E_{a2}}{RT}= \frac{10000}{8.314\times 300}=4

Posted by

vishal kumar

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