For a chemical reaction at 27^{circ} C, the a | vedclass

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(C) From the Arrhenius equation:$\ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} \left[ \frac{1}{T_{1}} - \frac{1}{T_{2}} \right]$Given $T_{1} = 27 + 273 =

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$e$

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(C) From the Arrhenius equation:$\ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} \left[ \frac{1}{T_{1}} - \frac{1}{T_{2}} ight]$Given $T_{1} = 27 + 273 = 300 \ K$, $T_{2} = 327 + 273 = 600 \ K$, and $E_{a} = 600 R$.Substituting the values:$\ln \frac{k_{2}}{k_{1}} = \frac{600 R}{R} \left[ \frac{1}{300} - \frac{1}{600} ight]$$\ln \frac{k_{2}}{k_{1}} = 600 \left[ \frac{2 - 1}{600} ight]$$\ln \frac{k_{2}}{k_{1}} = 600 	imes \frac{1}{600} = 1$Therefore, $\frac{k_{2}}{k_{1}} = e^{1} = e$.

For a chemical reaction at $27^{\circ} C$, the activation energy is $600 R$. The ratio of the rate constants at $327^{\circ} C$ to that of at $27^{\circ} C$ will be

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