每月彙整: 2018 年 9 月

樹莓派、樹莓派之學習、樹莓派之教育

STEM 隨筆︰鬼月談化學︰☲ 麗 《光明》

絕句‧杜甫

兩個黄鸝鳴翠柳，
一行白鷺上青天。
窗含西嶺千秋雪，
門泊東吳萬里船。

詩聖杜甫七言絕句有四首

堂西長筍别開門，塹北行椒卻背村。
梅熟許同朱老吃，松高擬對阮生論。

欲作魚梁雲複湍，因驚四月雨聲寒。
青溪先有蛟龍窟，竹石如山不敢安。

兩個黄鸝鳴翠柳，一行白鷺上青天。
窗含西嶺千秋雪，門泊東吳萬里船。

藥條藥甲潤青青，色過棕亭入草亭。
苗滿空山慚取譽，根居隙地怯成形。

，何必特舉其三說？

難悟實相金剛義！娑婆世界有情多，純量造化色菩提☆★
……

不空強空成頑空，空空不空有時盡，一念大悲觀自在，人間哪處非紅塵！！？？

─── 《GOPIGO 小汽車︰格點圖像算術《彩色世界》詩說》

 

『離』與『麗』一音之轉乎？故耳假借為『麗』也！

《易經》第三十卦‧《䷝ 離》

離：利貞，亨。 畜牝牛，吉。

彖曰：離，麗也﹔日月麗乎天，百谷草木麗乎土，重明以麗乎正，乃化成天下。 柔麗乎中正，故亨﹔是以畜牝牛吉也。

象曰：明兩作離，大人以繼明照于四方。

初九：履錯然，敬之無咎。
象曰：履錯之敬，以辟咎也。

六二：黃離，元吉。
象曰：黃離元吉，得中道也。

九三：日昃之離，不鼓缶而歌，則大耋之嗟，凶。
象曰：日昃之離，何可久也。

九四：突如其來如，焚如，死如，棄如。
象曰：突如其來如，無所容也。

六五：出涕沱若，戚嗟若，吉。
象曰：六五之吉，離王公也。

上九：王用出征，有嘉折首，獲其匪丑，無咎。
象曰：王用出征，以正邦也。

 

因其在『天』為『日』；在『身』為『目』；在『心』為『火』耶？？

若考之『甲骨』  離，證以《說文解字》：

離，黃倉庚也。鳴則蠶生。从隹离聲。

實恐是『捕獲』一隻『美麗』的『黃鸝鳥』呦！！

或該說︰

沒有『光明』，哪來『美麗』的呢？？！！

沒有『眼睛』，怎麼『欣賞』的呀！！？？

大概暗示『天之道』︰

『正』『反』相生；『順』『逆』相成。

的吧☆★

於是用『新觀點』閱讀

Opposed reactions

A pair of forward and reverse reactions may occur simultaneously with comparable speeds. For example, A and B react into X and Y and vice versa (s, t, u, and v are the stoichiometric coefficients):

\displaystyle {\ce {{{\mathit {s}}A}+{\mathit {t}}B<=>{{\mathit {u}}X}+{{\mathit {v}}Y}}}

The reaction rate expression for the above reactions (assuming each one is elementary) can be expressed as:

\displaystyle r={k_{1}[{\ce {A}}]^{s}[{\ce {B}}]^{t}}-{k_{2}[{\ce {X}}]^{u}[{\ce {Y}}]^{v}}

where: k1 is the rate coefficient for the reaction that consumes A and B; k2 is the rate coefficient for the backwards reaction, which consumes X and Y and produces A and B.

The constants k1 and k2 are related to the equilibrium coefficient for the reaction (K) by the following relationship (set r=0 in balance):

\displaystyle {k_{1}[{\ce {A}}]^{s}[{\ce {B}}]^{t}=k_{2}[{\ce {X}}]^{u}[{\ce {Y}}]^{v}}
\displaystyle K={\frac {[{\ce {X}}]^{u}[{\ce {Y}}]^{v}}{[{\ce {A}}]^{s}[{\ce {B}}]^{t}}}={\frac {k_{1}}{k_{2}}}

Simple example

Concentration of A (A0 = 0.25 mole/l) and B versus time reaching equilibrium kf = 2 min−1 and kr = 1 min−1

In a simple equilibrium between two species:

\displaystyle {\ce {A <=> B}}

Where the reactions starts with an initial concentration of A, \displaystyle {\ce {[A]_0}} , with an initial concentration of 0 for B at time t=0.

Then the constant K at equilibrium is expressed as:

\displaystyle K\ {\stackrel {\mathrm {def} }{=}}\ {\frac {k_{f}}{k_{b}}}={\frac {\left[{\ce {B}}\right]_{e}}{\left[{\ce {A}}\right]_{e}}}

Where \displaystyle [{\ce {A}}]_{e} and \displaystyle [{\ce {B}}]_{e} are the concentrations of A and B at equilibrium, respectively.

The concentration of A at time t, \displaystyle [{\ce {A}}]_{t} , is related to the concentration of B at time t, \displaystyle [{\ce {B}}]_{t} , by the equilibrium reaction equation:

\displaystyle {\ce {[A]_{\mathit {t}}=[A]_0 - [B]_{\mathit {t}}}}

Note that the term \displaystyle {\ce {[B]_0}} is not present because, in this simple example, the initial concentration of B is 0.

This applies even when time t is at infinity; i.e., equilibrium has been reached:

\displaystyle {\ce {[A]_{\mathit {e}}=[A]_0 - [B]_{\mathit {e}}}}

then it follows, by the definition of K, that

\displaystyle [{\ce {B}}]_{e}=x={\frac {k_{f}}{k_{f}+k_{b}}}{\ce {[A]_0}}

and, therefore,

\displaystyle \ [{\ce {A}}]_{e}={\ce {[A]_0}}-x={\frac {k_{b}}{k_{f}+k_{b}}}{\ce {[A]_0}}

These equations allow us to uncouple the system of differential equations, and allow us to solve for the concentration of A alone.

The reaction equation, given previously as:

\displaystyle r={k_{1}[{\ce {A}}]^{s}[{\ce {B}}]^{t}}-{k_{2}[{\ce {X}}]^{u}[{\ce {Y}}]^{v}}
\displaystyle -{\frac {d[{\ce {A}}]}{dt}}={k_{f}[{\ce {A}}]_{t}}-{k_{b}[{\ce {B}}]_{t}}

The derivative is negative because this is the rate of the reaction going from A to B, and therefore the concentration of A is decreasing. To simplify annotation, let x be \displaystyle [{\ce {A}}]_{t} , the concentration of A at time t. Let \displaystyle x_{e} be the concentration of A at equilibrium. Then:

\displaystyle {\begin{aligned}-{\frac {d[{\ce {A}}]}{dt}}&={k_{f}[{\ce {A}}]_{t}}-{k_{b}[{\ce {B}}]_{t}}\\-{\frac {dx}{dt}}&={k_{f}x}-{k_{b}[{\ce {B}}]_{t}}\\&={k_{f}x}-{k_{b}({\ce {[A]_0}}-x)}\\&={(k_{f}+k_{b})x}-{k_{b}{\ce {[A]_0}}}\end{aligned}}

Since:

\displaystyle k_{f}+k_{b}={k_{b}{\frac {{\ce {[A]_0}}}{x_{e}}}}

The reaction rate becomes:

\displaystyle \ {\frac {dx}{dt}}={\frac {k_{b}{\ce {[A]_0}}}{x_{e}}}(x_{e}-x)

which results in:

\displaystyle \ln \left({\frac {{\ce {[A]_0}}-[{\ce {A}}]_{e}}{[{\ce {A}}]_{t}-[{\ce {A}}]_{e}}}\right)=(k_{f}+k_{b})t

A plot of the negative natural logarithm of the concentration of A in time minus the concentration at equilibrium versus time t gives a straight line with slope kf + kb. By measurement of Ae and Be the values of K and the two reaction rate constants will be known.[20]

Generalization of simple example

If the concentration at the time t = 0 is different from above, the simplifications above are invalid, and a system of differential equations must be solved. However, this system can also be solved exactly to yield the following generalized expressions:

\displaystyle \left[{\ce {A}}\right]={\ce {[A]_0}}{\frac {1}{k_{f}+k_{b}}}\left(k_{b}+k_{f}e^{-\left(k_{f}+k_{b}\right)t}\right)+{\ce {[B]_0}}{\frac {k_{b}}{k_{f}+k_{b}}}\left(1-e^{-\left(k_{f}+k_{b}\right)t}\right)
\displaystyle \left[{\ce {B}}\right]={\ce {[A]_0}}{\frac {k_{f}}{k_{f}+k_{b}}}\left(1-e^{-\left(k_{f}+k_{b}\right)t}\right)+{\ce {[B]_0}}{\frac {1}{k_{f}+k_{b}}}\left(k_{f}+k_{b}e^{-\left(k_{f}+k_{b}\right)t}\right)

When the equilibrium constant is close to unity and the reaction rates very fast for instance in conformational analysis of molecules, other methods are required for the determination of rate constants for instance by complete lineshape analysis in NMR spectroscopy.

 

然後依樣畫葫蘆『動態模擬』︰

 

似有若無間︰

Consecutive reactions

If the rate constants for the following reaction are \displaystyle k_{1} and \displaystyle k_{2} ; \displaystyle {\ce {A -> B -> C}} , then the rate equation is:

For reactant A: \displaystyle {\frac {d[{\ce {A}}]}{dt}}=-k_{1}[{\ce {A}}]
For reactant B: \displaystyle {\frac {d[{\ce {B}}]}{dt}}=k_{1}[{\ce {A}}]-k_{2}[{\ce {B}}]
For product C: \displaystyle {\frac {d[{\ce {C}}]}{dt}}=k_{2}[{\ce {B}}]

With the individual concentrations scaled by the total population of reactants to become probabilities, linear systems of differential equations such as these can be formulated as a master equation. The differential equations can be solved analytically and the integrated rate equations are

\displaystyle [{\ce {A}}]={\ce {[A]_0}}e^{-k_{1}t}
\displaystyle \left[{\ce {B}}\right]={\begin{cases}{\ce {[A]_0}}{\frac {k_{1}}{k_{2}-k_{1}}}\left(e^{-k_{1}t}-e^{-k_{2}t}\right)+{\ce {[B]_0}}e^{-k_{2}t}&k_{1}\neq k_{2}\\{\ce {[A]_0}}k_{1}te^{-k_{1}t}+{\ce {[B]_0}}e^{-k_{1}t}&{\text{otherwise}}\\\end{cases}}
\displaystyle \left[{\ce {C}}\right]={\begin{cases}{\ce {[A]_0}}\left(1+{\frac {k_{1}e^{-k_{2}t}-k_{2}e^{-k_{1}t}}{k_{2}-k_{1}}}\right)+{\ce {[B]_0}}\left(1-e^{-k_{2}t}\right)+{\ce {[C]_0}}&k_{1}\neq k_{2}\\{\ce {[A]_0}}\left(1-e^{-k_{1}t}-k_{1}te^{-k_{1}t}\right)+{\ce {[B]_0}}\left(1-e^{-k_{1}t}\right)+{\ce {[C]_0}}&{\text{otherwise}}\\\end{cases}}

The steady state approximation leads to very similar results in an easier way.

Parallel or competitive reactions

Time course of two first order, competitive reactions with differing rate constants.

When a substance reacts simultaneously to give two different products, a parallel or competitive reaction is said to take place.

Two first order reactions

\displaystyle {\ce {A -> B}} and \displaystyle {\ce {A -> C}} , with constants \displaystyle k_{1} and \displaystyle k_{2} and rate equations \displaystyle -{\frac {d[{\ce {A}}]}{dt}}=(k_{1}+k_{2})[{\ce {A}}] ; \displaystyle {\frac {d[{{\ce {B}}}]}{dt}}=k_{1}[{{\ce {A}}}] and \displaystyle {\frac {d[{\ce {C}}]}{dt}}=k_{2}[{\ce {A}}]

The integrated rate equations are then  \displaystyle \ [{\ce {A}}]={\ce {[A]_0}}e^{-(k_{1}+k_{2})t} ; \displaystyle [{\ce {B}}]={\frac {k_{1}}{k_{1}+k_{2}}}{\ce {[A]_0}}(1-e^{-(k_{1}+k_{2})t}) and \displaystyle [{\ce {C}}]={\frac {k_{2}}{k_{1}+k_{2}}}{\ce {[A]_0}}(1-e^{-(k_{1}+k_{2})t}) .

One important relationship in this case is \displaystyle {\frac {{\ce {[B]}}}{{\ce {[C]}}}}={\frac {k_{1}}{k_{2}}}

One first order and one second order reaction[21]

This can be the case when studying a bimolecular reaction and a simultaneous hydrolysis (which can be treated as pseudo order one) takes place: the hydrolysis complicates the study of the reaction kinetics, because some reactant is being “spent” in a parallel reaction. For example, A reacts with R to give our product C, but meanwhile the hydrolysis reaction takes away an amount of A to give B, a byproduct: \displaystyle {\ce {{A}+ H2O -> B}} and \displaystyle {\ce {{A}+ R -> C}} . The rate equations are: \displaystyle {\frac {d[{\ce {B}}]}{dt}}=k_{1}{\ce {[A][H2O]}}=k_{1}'[{\ce {A}}] and \displaystyle {\frac {d[{\ce {C}}]}{dt}}=k_{2}{\ce {[A][R]}} . Where \displaystyle k_{1}' is the pseudo first order constant.

The integrated rate equation for the main product [C] is \displaystyle {\ce {[C]=[R]_0}}\left[1-e^{-{\frac {k_{2}}{k_{1}'}}{\ce {[A]_0}}(1-e^{-k_{1}'t})}\right] , which is equivalent to \displaystyle \ln {\frac {{\ce {[R]_0}}}{{\ce {[R]_0-[C]}}}}={\frac {k_{2}{\ce {[A]_0}}}{k_{1}'}}(1-e^{-k_{1}'t}) . Concentration of B is related to that of C through \displaystyle [{\ce {B}}]=-{\frac {k_{1}'}{k_{2}}}ln\left(1-{\frac {\ce {[C]}}{[R]_0}}\right)

The integrated equations were analytically obtained but during the process it was assumed that \displaystyle {\ce {{[A]_0}-[C]\approx [A]_0}} therefeore, previous equation for [C] can only be used for low concentrations of [C] compared to [A]0

 

彷彿了解了 chempy 之用法︰

 

就是方便深入『化學』的『工具』哩◎