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标题: [其他] 问个关于单位的问题----gmol 是什么的单位 [打印本页]
作者: 北京人在亚琛    时间: 2007-7-28 08:53     标题: 问个关于单位的问题----gmol 是什么的单位

不是g/mol
gmol没见过，哪位给指点一下
还有，我想查化学反应的速率，哪里有相关的数据库？
比如反应 n+o2-->no+o 的反应速率k ?
作者: vivian_leigh    时间: 2007-7-28 09:48

gmol 不是 SI-Einheit。
1 gmol 表示 6.023×10^23 个分子，和mol大概是定义上有区别吧。
z.B

1 gmol NaOH = 1 gmol * 40 g/gmol = 40g



[ 本帖最后由 vivian_leigh 于 2007-7-28 10:55 编辑 ]
作者: NeuMond    时间: 2007-7-28 10:04

原帖由 北京人在亚琛 于 2007-7-28 09:53 发表
不是g/mol
gmol没见过，哪位给指点一下
还有，我想查化学反应的速率，哪里有相关的数据库？
比如反应 n+o2-->no+o 的反应速率k ?

不会是Gmol吧, 很大的单位, 1G等于十的九次方.
反应速率取决与温度, 反应物浓度, 不一定能直接查到, 也许可以通过查反应激活能间接算出来.
作者: 北京人在亚琛    时间: 2007-7-28 10:51

原帖由 vivian_leigh 于 2007-7-28 10:48 发表
gmol 不是 SI-Einheit。
1 gmol 表示 6.023×10^23 个分子，和mol大概是定义上有区别吧。
z.B

1 gmol NaOH = 1 gmol * 40 g/gmol = 40g


还是没太明白,6.023×10^23 个分子不就是1mol分子么?!! 如果不说具体物质的话 gmol 和 g 之间怎么换算呢 我现在遇到的单位是m^3 /gmols  
和活化能的单位j/kgmol 怎么换算呢?
作者: 北京人在亚琛    时间: 2007-7-28 10:53

原帖由 NeuMond 于 2007-7-28 11:04 发表

不会是Gmol吧, 很大的单位, 1G等于十的九次方.
反应速率取决与温度, 反应物浓度, 不一定能直接查到, 也许可以通过查反应激活能间接算出来.

对，我就是说关于 Arrhenius-Gleichung的那些参数上哪里去找??
作者: cherubyupan    时间: 2007-7-28 12:03

Overview

In short, the Arrhenius equation is an expression that shows the dependence of the rate constant k of chemical reactions on the temperature T (in Kelvin) and activation energy Ea, as shown below:.[3]
.
where A is the pre-exponential factor or simply the prefactor and R is the gas constant. The units of the pre-exponential factor are identical to those of the rate constant and will vary depending on the order of the reaction. If the reaction is first order it has the units s-1, and for that reason it is often called the frequency factor or attempt frequency of the reaction. When the activation energy is given in molecular units, instead of molar units, e.g. joules per molecule instead of joules per mol, the Boltzmann constant is used instead of the gas constant. It can be seen that either increasing the temperature or decreasing the activation energy (for example through the use of catalysts) will result in an increase in rate of reaction.
Given the small temperature range in which kinetic studies are carried, it is reasonable to approximate the activation energy as being independent of temperature. Similarly, under a wide range of practical conditions, the weak temperature dependence of the pre-exponential factor is negligible compared to the temperature dependence of the  factor; except in the case of "barrierless" diffusion-limited reactions, in which case the pre-exponential factor is dominant and is directly observable.
Some authors define a modified Arrhenius equation,[4] that makes explicit the temperature dependence of the pre-exponential factor. If one allows arbitrary temperature dependence of the prefactor, the Arrhenius description becomes overcomplete, and the inverse problem (i.e. determining the prefactor and activation energy from experimental data) becomes singular. It has been pointed out that "it is not feasible to establish, on the basis of temperature studies of the rate constant, whether the predicted T1/2 dependence of the pre-exponential factor is observed experimentally".[2].. but if additional evidence is available, from theory and/or from experiment (such as density dependence), there is no obstacle to incisive tests of the Arrhenius law.
Taking the natural logarithm of the Arrhenius equation yields:
.
So, when a reaction has a rate constant which obeys the Arrhenius equation, a plot of ln(k) versus T -1 gives a straight line, whose slope and intercept can be used to determine Ea and A. This procedure has become so common in experimental chemical kinetics that practitioners have taken to using it to define the activation energy for a reaction. That is the activation energy is defined to be (-R) times the slope of a plot of ln(k) vs. (1/T):

[edit]Kinetic theories interpretation of Arrhenius equation

Arrhenius argued that in order for reactants to be transformed into products, they first needed to acquire a minimum amount of energy, called the activation energy Ea. At an absolute temperature T, the fraction of molecules that have a kinetic energy greater than Ea can be calculated from the Maxwell-Boltzmann distribution of statistical mechanics, and turns out to be proportional to . The concept of activation energy explains the exponential nature of the relationship, and in one way or another, it is present in all kinetic theories:
[edit]Collision theory
Main article: Collision theory
One example comes from the "collision theory" of chemical reactions, developed by Max Trautz and William Lewis in the years 1916-18. In this theory, molecules are supposed to react if they collide with a relative kinetic energy along their line-of-centers that exceeds Ea This leads to an expression very similar to the Arrhenius equation, with the difference that the preexponential factor "A" is not constant but instead is proportional to the square root of temperature. This reflects the fact that the overall rate of all collisions, reactive or not, is proportional to the average molecular speed which in turn is proportional to T1/2. In practice, the square root temperature dependence of the pre-exponential factor is usually very slow compared to the exponential dependence associated with Ea, to the point that some think it can not be experimentally proven.
[edit]Transition state theory
Another Arrhenius-like expression appears in the "transition state theory" of chemical reactions, formulated by Wigner, Eyring, Polanyi and Evans in the 1930s. This takes various forms, but one of the most common is

where ΔG? is the Gibbs free energy of activation, kB is Boltzmann's constant, and h is Planck's constant.
At first sight this looks like an exponential multiplied by a factor that is linear in temperature. However, one must remember that free energy is itself a temperature dependent quantity. The free energy of activation includes an entropy term, which is multiplied by the absolute temperature, as well as an enthalpy term. Both of them depend on temperature, and when all of the details are worked out one ends up with an expression that again takes the form of an Arrhenius exponential multiplied by a slowly varying function of T. The precise form of the temperature dependence depends upon the reaction, and can be calculated using formulas from statistical mechanics involving the partition functions of the reactants and of the activated complex.

[ 本帖最后由 cherubyupan 于 2007-7-28 13:05 编辑 ]
作者: cherubyupan    时间: 2007-7-28 12:04

http://en.wikipedia.org/wiki/Arrhenius_equation
作者: 诚实可靠小郎君~    时间: 2007-7-28 12:09

扯乳发彪阿
作者: 北京人在亚琛    时间: 2007-7-28 22:08

我跟期待关于  gmol 的详细解答~~~~
作者: hetty    时间: 2007-7-29 15:20

个人觉得，会不会是两个单位相乘，g（克）和mol（摩尔），所以写在一起了？
作者: cherubyupan    时间: 2007-7-29 15:26

gmol order gmol-1?
作者: cherubyupan    时间: 2007-7-29 15:28

3.                Choose a basis of 100g of mixture for the calculations. The, molar amount of each component is:
          Methyl chloroformate: 50 g/(94.5 g/mol) = 0.529
          Methyl chloride: 50 g/(50.5 g/mol) = 00.99 gmol
The number of gmol of each element is:
          C: (0.529 gmol) (2) + (0.99 gmol) (1) = 2.05 gmol
          H: (0.529 gmol) (3) + (0.99 gmol) (3) = 4.56 gmol
          Cl: (0.529 gmol) (1) + (0.99 gmol) (1) = 1.52 gmol
          O: (0.529 gmol) (2)                             = 1.06 gmol
          Total:                                                 = 9.19 gmol
          The ultimate analysis on a gmol basis (i.e., composition by gmol% of each element) is:
         
          C: (2.05 gmol/9.19 gmol)(100%) = 22.3%
          H: (4.56 gmol/9.19 gmol)(100%) = 49.6%
          Cl: (1.52 gmol/9.19 gmol) (100%) = 16.5%
          O: (1.06 gmol/9.19 gmol)(100%) = 11.5%
作者: cherubyupan    时间: 2007-7-29 15:29

alles klar了伐？:naughty:
作者: 北京人在亚琛    时间: 2007-7-30 20:04

谢谢楼上那么详细的解答，不过说实话还是没明白~~~
50 g/(94.5 g/mol) = 0.529
这个的单位计算 g * mol / g 结果不应该是 mol 么？为啥还有个g !?!?!?!?
作者: doudouwss    时间: 2007-8-1 09:10

原帖由 北京人在亚琛 于 2007-7-28 11:51 发表

还是没太明白,6.023×10^23 个分子不就是1mol分子么?!! 如果不说具体物质的话 gmol 和 g 之间怎么换算呢 我现在遇到的单位是m^3 /gmols  
和活化能的单位j/kgmol 怎么换算呢?






我觉得你的问题就提得不对，你只是想找出m^3 /gmols  和活化能的单位j/kgmol 只见怎么换算吧，那你就不能简单的光看gmol，而是要整体来看。



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