The activation energy for a reaction is 9.0 , | vedclass

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(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 	imes R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$Given: $E_a = 9.0 \, kca

Vedclass · Accepted Answer

$65$

Vedclass · Answer

(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 	imes R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$Given: $E_a = 9.0 \, kcal/mol = 9000 \, cal/mol$,$R = 1.987 \, cal \, K^{-1} \, mol^{-1} \approx 2 \, cal \, K^{-1} \, mol^{-1}$,$T_1 = 298 \, K$,$T_2 = 308 \, K$.$\log \frac{K_2}{K_1} = \frac{9000}{2.303 	imes 2} \left[ \frac{308 - 298}{308 	imes 298} ight]$$\log \frac{K_2}{K_1} = \frac{9000}{4.606} 	imes \frac{10}{91784} \approx 1954.19 	imes 0.0001089 \approx 0.2129$$\frac{K_2}{K_1} = 	ext{antilog}(0.2129) \approx 1.632$Percentage increase = $\frac{K_2 - K_1}{K_1} 	imes 100 = (1.632 - 1) 	imes 100 = 63.2 \% \approx 63 \%$.The closest option is $65 \%$.

The activation energy for a reaction is $9.0 \, kcal/mol$. The increase in the rate constant when its temperature is increased from $298 \, K$ to $308 \, K$ is $......... \, \%$.

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