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(B) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 \ R} [\frac{T_2 - T_1}{T_1 T_2}]$Given: $\frac{K_2}{K_1} = 2$,$T_1 = 300 \

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$53$

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(B) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 \ R} [\frac{T_2 - T_1}{T_1 T_2}]$Given: $\frac{K_2}{K_1} = 2$,$T_1 = 300 \ K$,$T_2 = 310 \ K$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$\log 2 = 0.3$Substituting the values:$0.3 = \frac{E_a}{2.303 	imes 8.3} [\frac{310 - 300}{300 	imes 310}]$$0.3 = \frac{E_a}{19.1149} [\frac{10}{93000}]$$0.3 = \frac{E_a}{19.1149} 	imes 0.00010753$$E_a = \frac{0.3 	imes 19.1149}{0.00010753} \approx 53341 \ J \ mol^{-1} = 53.34 \ kJ \ mol^{-1}$The closest value is $53.33 \ kJ \ mol^{-1}$.

The rate of a first order reaction doubles when the temperature changes from $300 \ K$ to $310 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is
$R=8.3 \ J \ K^{-1} \ mol^{-1}, \log 2=0.3$ (in $.33$)

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The rate of a first order reaction doubles when the temperature changes from $300 \ K$ to $310 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is$R=8.3 \ J \ K^{-1} \ mol^{-1}, \log 2=0.3$ (in $.33$)