When temperature is increased from 27,{}^{cir | vedclass

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Question

(A) The Arrhenius equation is given by: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
Given: $T_1 = 27 + 273

Vedclass · Accepted Answer

$1.66$

Vedclass · Answer

(A) The Arrhenius equation is given by: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$
Given: $T_1 = 27 + 273 = 300 \, K$,$T_2 = 127 + 273 = 400 \, K$,$k_2 = 2k_1$,and $R = 2 	imes 10^{-3} \, kCal \, K^{-1} \, mol^{-1}$.
Substituting the values: $\log 2 = \frac{E_a}{2.303 	imes 2 	imes 10^{-3}} 	imes \left[ \frac{400 - 300}{300 	imes 400} ight]$
$0.3010 = \frac{E_a}{4.606 	imes 10^{-3}} 	imes \frac{100}{120000}$
$0.3010 = \frac{E_a}{4.606 	imes 10^{-3}} 	imes \frac{1}{1200}$
$E_a = 0.3010 	imes 4.606 	imes 10^{-3} 	imes 1200 \approx 1.66 \, kCal$.

When temperature is increased from $27\,{}^{\circ}C$ to $127\,{}^{\circ}C$,the rate of reaction becomes doubled. The activation energy $E_a$ will be ........ $kCal$.

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