For a first order reaction,the rate constant | vedclass

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(A) The Arrhenius equation is given by: $\ln \frac{K_{2}}{K_{1}} = \frac{E_{a}}{R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right)$.Given: $K_{1} = 0

Vedclass · Accepted Answer

$13.02$

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(A) The Arrhenius equation is given by: $\ln \frac{K_{2}}{K_{1}} = \frac{E_{a}}{R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} ight)$.Given: $K_{1} = 0.04 \, 	ext{min}^{-1}$,$K_{2} = 0.08 \, 	ext{min}^{-1}$,$T_{1} = 27 + 273 = 300 \, 	ext{K}$,$T_{2} = 37 + 273 = 310 \, 	ext{K}$,and $R = 2 \, 	ext{cal} \, 	ext{K}^{-1} \, 	ext{mol}^{-1}$.Substituting the values: $\ln \left( \frac{0.08}{0.04} ight) = \frac{E_{a}}{2} \left( \frac{310 - 300}{300 	imes 310} ight)$.$\ln(2) = \frac{E_{a}}{2} \left( \frac{10}{93000} ight)$.$0.7 = \frac{E_{a}}{2} 	imes \frac{1}{9300}$.$E_{a} = 0.7 	imes 2 	imes 9300 = 13020 \, 	ext{cal} / 	ext{mol}$.Converting to $	ext{kcal} / 	ext{mol}$: $E_{a} = 13.02 \, 	ext{kcal} / 	ext{mol}$.

For a first order reaction,the rate constant is $0.04 \, \text{min}^{-1}$ at $27 \, ^\circ\text{C}$ and $0.08 \, \text{min}^{-1}$ at $37 \, ^\circ\text{C}.$ The activation energy of reaction is .......... $\text{kcal} / \text{mol}$ $(\ln \, 2 = 0.7)$

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