[单选题]
The rate constant of a reaction at $298K$ is $3\times 10^{-3}$$s^{-1}$, while the rate constant at$308K$is$6\times10^{-3}$$s^{-1}$
What is the activation energy of this reaction, in$kJ$ $mol^{-1}$?
$ln$$\frac{k_1}{k_2}$=$\frac{E_a}{R}\times(\frac{1}{T_2}-\frac{1}{T_1})$;$R$=8.31$JK^{-1}mol^{-1}$
A. $\frac{8.31\times ln(2)}{(1000\times(\frac{1}{298}-\frac{1}{308})}$
B. $\frac{8.31\times ln(0.5)}{(1000\times(\frac{1}{298}-\frac{1}{308})}$
C. $\frac{1000\times ln(2)}{(8.31\times(\frac{1}{298}-\frac{1}{308})}$
D. $\frac{1000\times ln(0.5)}{(8.31\times(\frac{1}{298}-\frac{1}{308})}$
参考答案:
A
本题详细解析:
A
The$ln(2)$term appears because$k_1$ is double$k_2$, and the factor of 1000 appears in the denominator to convert$J mol^{-1}$to $kJ mol^{-1}$
Option B is almost correct, but the temperature terms$\frac{1}{298}$and$\frac{1}{308}$would need to be swapped to give a positive value for the activation energy.
The question is in the “hard” or “very hard” category since students must be able to identify the correct terms in the question for use in the equation, rearrange the equation correctly, and also perform a unit conversion.
