Question

$A \rightarrow B$ The rate constants of the above reaction at $200 K$ and $300 K$ are $0.03 \min ^{-1}$ and $0.05 \min ^{-1}$ respectively The activation energy for the reaction is __$J$ (Nearest integer) (Given $: \ln 10=2.3$

$R =8.3 \,JK ^{-1} \,mol ^{-1}$

$\log 5=0.70$

$\log 3=0.48$

$\log 2=0.30)$

Answer 1

The correct answer is 2520.
logK200​K300​​=2.3×8.314Ea​​(T1​1​−T2​1​)
log0.030.05​=2.305×8.314Ea​×[2001​−3001​]
Ea​=2519.88J⇒Ea​=2520J