Explanation
$$K_2=2K_1, T_2=35+273=308$$ $$K, T_1=20+273=293$$ $$K$$
$$\ln \cfrac {K_2}{K_1}=\cfrac {E_a}{R}\left [\cfrac {1}{T_1}-\cfrac {1}{T_2}\right]$$
$$\therefore \ln 2=\cfrac {E_a}{8.314}\left[\cfrac {1}{293}-\cfrac {1}{308}\right] $$
$$\therefore \cfrac {8.314\times \ln 2}{(3.41-3.24)\times 10^{-3}}=E_a$$
$$\therefore E_a\approx 34.7$$ $$kJ/mol$$
$$ \because k = Ae^{-Ea/RT}$$ $$ \implies \dfrac{k_{25}}{k_0} = \dfrac{A_{25} e^{65000/8.314 \times 298}}{A_{0} e^{65000/8.314 \times 275}}$$ $$ \implies \dfrac{e^{-26.24}}{e^{-28.64}}$$ $$ \implies e^{2.40} = 11 $$
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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