Explanation
K2=2K1,T2=35+273=308 K,T1=20+273=293 K
lnK2K1=EaR[1T1−1T2]
∴
\therefore \cfrac {8.314\times \ln 2}{(3.41-3.24)\times 10^{-3}}=E_a
\therefore E_a\approx 34.7 kJ/mol
\because k = Ae^{-Ea/RT} \implies \dfrac{k_{25}}{k_0} = \dfrac{A_{25} e^{65000/8.314 \times 298}}{A_{0} e^{65000/8.314 \times 275}} \implies \dfrac{e^{-26.24}}{e^{-28.64}} \implies e^{2.40} = 11
Thus, the reaction will proceed 11 times faster.
log \ K= 15.0 - \cfrac {10^6}{T}
We know that, log \ K=log \ A- \cfrac {E_a}{RT}
Comparing, we get
log \ A=15.0
\Rightarrow A=10^{15}
\cfrac {E_a}{RT}= \cfrac {10^6}{T}
\Rightarrow E_a=1.9 \times 10^4 \ kJ
If the activation energy is 65 kJ then how much time faster a reaction proceed at { 25 }^{ o }C than { 0 }^{ o }C?
According to half time period
k=\dfrac{0.693}{{{t}_{\dfrac{1}{2}}}}
\Rightarrow \dfrac{0.693}{120}=5.77\times {{10}^{-3}}{{\min }^{-1}}
Here,
\left( {{t}_{\dfrac{1}{2}}}=120\ \min t \right)
Now a first order reaction
k=\dfrac{2.303}{t}\log \dfrac{1}{a-x}
a=100,\ x=90
\therefore \ \ 5.77\times {{10}^{-3}}=\dfrac{2.303}{t}\log \dfrac{100}{100-90}
t=\dfrac{2.303}{5.77\times {{10}^{-3}}}\log 10
=399\ \min .
This is the required answer.
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