JEE Main Solved Paper 2017 Question 13
Question: Two reactions $ R _1 $ and $ R _2 $ have identical preexponential factors. Activation energy of $ R _1 $ exceeds that of $ R _2 $ by $ 10,kJ,mo{l^{-1}}. $ If $ k _1 $ and $ k _2 $ are rate constants for reactions $ R _1 $ and $ R _2 $ respectively at 300 K, then ln $ (k _2/k _1) $ is equal to:- $ (R=8.314,J,mo{l^{-1}}{K^{-1}})\ [JEE Main Solved Paper-2017]
Options:
A) 8
B) 12
C) 6
D) 4
Show Answer
Answer:
Correct Answer: D
Solution:
From arrhenius equation, $ K=A.{e^{\frac{-Ea}{RT}}} $
So, $ K _1=A.{e^{-{E _{a _1}}/RT}} $ ?.. $ K _2=A.{e^{-{E _{a _2}}/RT}} $ ?..
So, equation $ (2)(1)\Rightarrow \frac{K _2}{K _1}={e^{\frac{({E _{a _1}}-{E _{a _2}})}{RT}}} $ (As pre-exponential factors of both reactions is same) $ \ln ( \frac{K _2}{K _1} )=\frac{{E _{a _1}}-{E _{a _2}}}{RT}=\frac{10,000}{8.314\times 3000}=4 $
