A reaction has an activation energy of 209 , | vedclass

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(A) Given,activation energy $E_a = 209 \, kJ \, mol^{-1} = 209000 \, J \, mol^{-1}$.Rate increases $10$-fold,so $K_2 / K_1 = 10$.Initial temperature $

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(A) Given,activation energy $E_a = 209 \, kJ \, mol^{-1} = 209000 \, J \, mol^{-1}$.Rate increases $10$-fold,so $K_2 / K_1 = 10$.Initial temperature $T_1 = 27 + 273 = 300 \, K$.Using the Arrhenius equation: $\log(K_2 / K_1) = \frac{E_a}{2.303 \, R} \left[ \frac{1}{T_1} - \frac{1}{T_2} ight]$.Substituting the values: $\log(10) = \frac{209000}{2.303 	imes 8.314} \left[ \frac{1}{300} - \frac{1}{T_2} ight]$.$1 = 10923.6 	imes \left[ 0.003333 - \frac{1}{T_2} ight]$.$0.0000915 = 0.003333 - \frac{1}{T_2}$.$\frac{1}{T_2} = 0.003333 - 0.0000915 = 0.0032415$.$T_2 = 308.5 \, K$.$X = 308.5 - 273 = 35.5^{\circ} C \approx 35^{\circ} C$.

$A$ reaction has an activation energy of $209 \, kJ \, mol^{-1}$. The rate increases $10$-fold when the temperature is increased from $27^{\circ} C$ to $X^{\circ} C$. The temperature $X$ is closest to
[Gas constant,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$ ]

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$A$ reaction has an activation energy of $209 \, kJ \, mol^{-1}$. The rate increases $10$-fold when the temperature is increased from $27^{\circ} C$ to $X^{\circ} C$. The temperature $X$ is closest to[Gas constant,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$ ]