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The 2nd order graph shows a bit of a kink, whereas the 1st order graph is smooth.

  1. What is the significance of this kink, and why does it appear in only the 2nd order graph.

  2. What is the significance of the numerical value at this kink? The graph below show it occurs at $[A]=1.5-2$.

enter image description here

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  • $\begingroup$ Can you add an annotation to that upper 2nd-order kinetics plot, to indicate what you're referring to when you say "kink"? I think I know what you mean, but I'd like to be sure. $\endgroup$ – hBy2Py Jan 19 '17 at 4:47
  • $\begingroup$ (Depending on which feature you're asking about, it might just be a plotting glitch.) $\endgroup$ – hBy2Py Jan 19 '17 at 4:58
  • $\begingroup$ The shape is different so it looks like it has a kink, the kink itself isn't important. $\endgroup$ – Kevin Kostlan Jan 19 '17 at 6:49
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    $\begingroup$ @K-Feldspar Ahh, thanks for that edit, that was very helpful. I think there was a small plotting glitch in the prior figure, but it's not in the revised one. ortho's answer is correct to a point, but I think I may be able to provide some further insight, once I get a chance to write up an answer. $\endgroup$ – hBy2Py Jan 19 '17 at 12:05
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Summary

As orthocresol correctly notes, the presence of this kink is simply a feature of the $x$-shifted hyperbola that is the solution to the second-order kinetics problem, and the presence/characteristics of the kink will vary depending on the particular parameters of the problem.

However, this answer aims to demonstrate that:

  1. The kink in the 2nd-order plot occurs due to a rapid change in the slope of the curve from "large" to "small" values.
  2. The kink only occurs in the 2nd-order plot if (a) the initial concentration and rate constant are both "high enough"; and (b) the time scale of interest is "short enough".
  3. A similar kink also occurs with first-order kinetics, again if the rate constant is "large enough" and the time scale is "short enough".

In order to generalize the solutions to the first- and second-order kinetics problems, coming from a chemical engineering background my standard approach is to non-dimensionalize, then solve.

1st-Order Problem

$$ \ce{A ->[k_1] P} \\ ~ \\ {\mathrm dC_\mathrm A\over \mathrm d t} = -k_1 C_\mathrm{A}~;\quad C_\mathrm A(0) = C_\mathrm{A}^\mathrm o $$

2nd-Order Problem

$$ \ce{2A ->[k_2] Q} \\ ~ \\ {\mathrm d C_\mathrm A\over \mathrm d t} = -k_2 C^2_\mathrm{A}~;\quad C_\mathrm{A}(0) = C_\mathrm{A}^\mathrm o $$

(Note that although I'm writing both reactions with $\ce A$ as the reactant, I'm assuming that these reactions do not occur simultaneously.)

To nondimensionalize, you define a new set of variables by dividing each dimensional variable by a scaling factor. Here, it makes the most sense to scale the concentration by the initial value:

$$ \Theta = {C_\mathrm A \over C_\mathrm A^\mathrm o}~,~~\text{or}~~~ C_\mathrm{A} = C_\mathrm{A}^\mathrm{o}\,\Theta $$

The scaling factor for the time doesn't have an immediate, natural definition, so I'll use the generic variable $t^*$:

$$ \tau = {t\over t^*}~,~~\text{or}~~~ t = t^*\,\tau $$

Then, these new variables are substituted into the original equations. The first-order equations become (using subscripts to indicate the order of the reaction under consideration):

$$ {C_\mathrm A^\mathrm o\over t^*} {\mathrm d \Theta_1\over \mathrm d \tau} = - k_1C_\mathrm A^\mathrm o\, \Theta_1 $$

The scaling constants can then be grouped on the RHS of the differential equation:

$$ {\mathrm d \Theta_1\over \mathrm d \tau} = - k_1 t^* \Theta_1 $$

For convenience, define the parameter $\eta = k_1 t^*$, yielding:

$$ {\mathrm d \Theta_1\over \mathrm d \tau} = -\eta\,\Theta_1~;\quad \Theta_1(0)=1 \tag{1} $$

The form of $\eta$ reveals that it is dimensionless, along with all of the other terms in the equation. This approach allows representation of any 1st-order kinetics problem to be represented by a single equation, once appropriate values for the system parameters $k_1$ and $t^*$, and thus $\eta$, are defined.

Non-dimensionalization of the 2nd-order equation proceeds as follows:

$$ {C_\mathrm A^\mathrm o\over t^*} {\mathrm d \Theta_2\over \mathrm d \tau} = - k_2\left(C_\mathrm A^\mathrm o\right)^2\, \Theta_2^2 \\ {\mathrm d \Theta_2\over \mathrm d \tau} = - k_2 t^* C_\mathrm A^\mathrm o\,\Theta_2^2 $$

$$ {\mathrm d \Theta_2\over \mathrm d \tau} = -\kappa\,\Theta_2^2~;\quad\Theta_2(0)=1 \tag{2} $$

The non-dimensional parameter $\kappa = k_2 t^* C_\mathrm{A}^\mathrm{o}$ is defined analogously to $\eta$, but has a different form due to the different reaction kinetics.

As an aside, note that in Eqs. $(1)$ and $(2)$ all of the information about the scale of each problem has been wrapped into $\eta$ and $\kappa$, respectively. Taking $\kappa$ as an example, if $C_\mathrm{A}^\mathrm{o}$ is 'large enough' relative to $t^*$ and $k_2$, then $\kappa$ will be large and the reaction will be "fast". Similarly, for a given $C_\mathrm{A}^\mathrm{o}$ and $k_2$, if you extend your time range of interest long enough (viz., increase $t^*$), it will make the reaction term 'seem faster' compared to a shorter time scale.


The solutions to the above differential equations are:

$$ \Theta_1(\tau) = e^{-\eta\,\tau} \tag{3} $$

$$ \Theta_2(\tau) = {1\over 1 + \kappa\tau} \tag{4} $$

The derivations of the above are standard enough that I won't go into them in detail.

It will be useful later to have explicit expressions for the rates as functions of $\tau$. Substituting Eq. $(3)$ into Eq. $(1)$:

$$ {\mathrm d \Theta_1 \over \mathrm d \tau} = -\eta\, e^{-\eta\, \tau} \tag{5} $$

And, substituting Eq. $(4)$ into Eq. $(2)$:

$$ {\mathrm d \Theta_2 \over \mathrm d \tau} = -{\kappa\over \left(1 + \kappa\tau\right)^2} \tag{6} $$


Now (finally!) to start tying these equations to the plots in the original question.

First, to demonstrate the utility of the scaled approach, consider the lower plot in the original question, where numerical values for the rate constant and initial concentration are given: $k_2 = 2~{1\over \mathrm{\mu M\, s}}$ (I'm assuming a typo on the units here) and $C_\mathrm{A}^\mathrm{o} = 10~\mathrm{\mu M}$. Further, the time scale has implicitly been defined by the choice of the domain of the abscissa: $t^* = 2~\mathrm{s}$. This leads to a value of $\kappa = 40$, which yields the following plot:

2nd order, kappa=40

As can be seen, the visual appearance of the profile is essentially identical to that in the OP's plot.

Anticipating subsequent work, I'm actually going to work from here with a slightly higher value, $\kappa=54$:

2nd order, kappa=54

The highlighted band in the figure marks the range where the (non-dimensional) slope decreases from $4$ to $0.25$, and falls right on the "kink" noted by the OP. Qualitatively, it marks the transition region from the "high slope" to the "low slope" portion of the curve. It is the narrowness of this transition region $\left(\Delta\tau \ll 1 \right)$ that gives the visual impression of the "kink" in the curve.

There are a couple of ways one could compare this curve to a 1st-order curve. Naturally, one cannot define a 1st-order curve that lies exactly on the 2nd-order curve, as the two functional forms are different. However, after taking as given that $\Theta_1(0) = \Theta_2(0) = 1$, I see two intuitive ways of selecting a 'comparable' 1st-order curve:

  1. Constrain the initial slopes to be equal [cf. Eqs. $(5)$ and $(6)$]:

$$ \left.{\mathrm d \Theta_1 \over \mathrm d \tau}\right|_{\tau=0} = \left.{\mathrm d \Theta_2 \over \mathrm d \tau}\right|_{\tau=0} \quad\longrightarrow\quad \eta = \kappa $$

  1. Constrain the 'final values' (here, at $\tau = 1$) to be equal [cf. Eqs. $(3)$ and $(4)$]:

$$ e^{-\eta} = {1\over 1+\kappa} \quad\text{or}\quad \eta = \ln{\left(1+\kappa\right)} $$

For Case #1, I'll plot $\eta = 54$ along with $\kappa = 54$:

Both curves, kappa = eta = 54

As can be seen, in actuality, one can introduce a "kink" into a 1st-order curve, given high enough a value of $\eta$. In fact, when the initial slopes of the non-dimensional curves are made to be equal, the kink in the 1st-order curve is more dramatic than that in the 2nd-order curve!

On the other hand, Case #2 looks much more like that depicted in the original question, without a visible kink in the 1st-order curve. Plotting for $\kappa=54$ and $\eta = \ln{\left(\kappa+1\right)} \approx 4$ gives:

Both curves, kappa = 54, eta = 4

(This is is why I picked $\kappa = 54$ above, so that the value for $\eta$ here would be approximately an integer value.)

In this case, the slope transition region for the 1st-order curve is broad $\left(\Delta\tau\sim 1\right)$, leading to no perceptible kink.

Further, it’s possible to plot a second-order curve where no kink is present. Taking $\kappa = 4$, comparable with the above $\eta = 4$ curve, gives the following:

2nd-order only, kappa=4

As with the first-order chart with $\eta = 4$, the broader transition region above corresponds to the lack of an observable kink in the curve.


So: To kink, or not to kink? The scale is the question!

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  1. There's no significance to the shape, that's simply how any graph of the form $y = 1/(c+x)$ looks like ($c$ being any constant you like).

  2. Even if the supposed "kink" has some significance, there isn't any significance to the number, especially since the number depends on your system anyway (it's dependent on the value of your rate constant as well as the initial concentration).

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