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[综合交流] 能垒和温度的关系

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各位老师,
       最近的一篇文章遇到了一点麻烦,对于审稿人的意见我不知道该如何回复合适。贴出审稿人的意见和我的疑问,请各位老师帮忙出出主意,这里先行谢过!我先简单说明一下我的实验与计算。实验是对葡萄糖在300,400,500℃加热20s,通过GC/MS检测产物,其中有一个产物产率在1%左右(以反应物计),我在气相条件计算该产物的形成路径,计算级别是B3LYP/6-31+G(d,p)。两个审稿人都在质疑我的实验与计算不匹配的问题。审稿人1认为我的产物不是主产物(他说的没错,不是主产物),审稿人2认为根据公式和我算出的能垒计算出的k与20s的反应时间差距很大。
       我的问题是:
       1.我该怎么回复审稿人的意见
       2. 我该怎么改进我的工作,这只是我的第二篇文章,以后还想做的更多更深入的工作。
       3. 根据阿累尼乌斯公式的假设,活化能和温度不相关,在300-500℃下这个产物都能生成,说明其实多高的反应能垒都是可以克服的,温度只是在改变反应速率,温度越高,反应速率越大,温度越低,反应速率越小,所以低温+长时间和高温+短时间可以得到相同的结果,这种理解对么?
审稿人1的意见:
Most importantly, all the reaction pathways leading the formation of hydroxyacetone involves elementary steps that require activation energy of more than 287 kJ/mol, as shown in Fig. 9, Fig. 11, and Fig. 12. However, Within the fast pyrolysis timescale of seconds as reported by numerous experimental studies in the literature (20s), kinetics suggest that a reaction pathway requires an activation energy of more than 287 kJ/mol temperature of 500 °C. In other works, reaction pathways requiring such high energy barrier can not be important or competitive during fast pyrolysis.
审稿人2的意见:
The study is very systematic and interesting, and for sure deserve publication in JAAP, but a crucial point should be solved before publication. Using the equation ln (k/T) = 23.76 – ΔG/RT to compute approximate rate for the reaction, and putting inside the equation the calculated energy barriers (Fig. 9), the resulting rates are really higher than the timescale of the experiment (20 seconds).The study is very systematic and interesting, and for sure deserve publication in JAAP, but a crucial point should be solved before publication. Using the equation ln (k/T) = 23.76 – ΔG/RT to compute approximate rate for the reaction, and putting inside the equation the calculated energy barriers (Fig. 9), the resulting rates are really higher than the timescale of the experiment (20 seconds).



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发表于 Post on 2017-1-20 04:54:35 | 只看该作者 Only view this author
所以低温+长时间和高温+短时间可以得到相同的结果,这种理解对么?

反应路径不止一条,温度变了每条路径的速率都会变,各个产物的比例也会变,所以不能这么说。

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 楼主 Author| 发表于 Post on 2017-1-26 11:04:08 来自手机 | 只看该作者 Only view this author
agent99 发表于 2017-1-20 04:54
所以低温+长时间和高温+短时间可以得到相同的结果,这种理解对么?

反应路径不止一条,温度变了每条路径 ...

谢谢老师指点,关于我提的前两个问题,您有什么建议么?

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发表于 Post on 2017-1-26 19:25:50 | 只看该作者 Only view this author
The reactions (or products) might be controlled kinetically or thermodynamically.

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发表于 Post on 2017-1-26 19:32:10 | 只看该作者 Only view this author
Since the activation energy is so high that the minor product is not important in short time as pointed out by the reviewer. But one may see more of the products with a long reaction time if the free energy is favored among all the products.

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