text:
Approximate calculations of the product distribution under kinetic control
What is the product distribution when the energy difference between the transition states (∆TS = TS1 – TS2), is at 2 kcal/mol and when it is at 5 kcal/mol (assuming room temperature)?
I know that 1.4 kcal/mol approximately corresponds to factor 10 and that for a difference of, for example, 3 kcal/mol the product distribution would approximately be 100:1 P2:P1.
But I find it hard to know the distribution at a difference of 1, 2, 4, and 5 kcal/mol.
Could anyone please hook me up with some approximate calculations or values and explain to me how to think in order to understand this without having to go through all sorts of equations like Eyring and Ahrrenius?

I am afraid you have to use either the Arrhenius or the Eyring equation to get a rough estimate of the product distribution.
If you assume that the pre-exponential terms in either of these equations are the same or very similar, you will find that
$$ \mathrm{\frac{k_2}{k_1} = \exp \left( \frac{\Delta TS}{RT}\right)} $$
At 298 K, $\mathrm{RT = 0.593} \, \mathrm{kcal.mol^{-1}}$ so
$$ \mathrm{\frac{k_2}{k_1} = \exp \left( \frac{\Delta TS}{0.593}\right)} $$
and substituting in this equation we will obtain the values we are looking for.

I doubt the assumption of the equality of the frequency/ pre-exponent  factor is well justified. It could be for the common transition state, but that is not the case.

@Poutnik  I completely agree with you. Just thought the author of the question wanted to avoid entering on the nitty-gritty of the theory (i.e. a too detailed explanation) and just needed a rule of the thumb rule though it may have plenty of caveats.

Yeah, for educative cases one has often go for the most simplified model ... and then some.

Quick Answer
The relative product concentrations are dependent on $\Delta E_a$, $T$, orders, and reversibility of the reactions. It would be inaccurate to specify relative product concentrations simply based on $\Delta E_a$. For competitive reactions of the same order, the relative product concentrations are a simple ratio of the rate constants. Moreover, reversibility is crucial for thermodynamic and kinetic control.
The following values tabulated from Equation (5) should help you estimate the relative product concentrations $\dfrac{\ce{[P_2]}}{\ce{[P_1]}}$ for competitive same-order non-reversible reactions:

$\Delta E_a \pu{kcal mol^{-1}}$
$\pu{100 K}$
$\pu{200 K}$
$\pu{300 K}$
$\pu{400 K}$
$\pu{500 K}$

1
1.53E2
1.24E1
5.35
3.52
2.74

2
2.35E4
1.53E2
2.86E1
1.24E1
7.49

3
3.61E6
1.89E3
1.53E2
4.36E1
2.05E1

4
5.53E8
3.25E4
8.21E2
1.53E2
5.60E1

5
8.48E10
2.91E5
4.39E3
5.40E2
1.53E2

Sigh! I tried, but there is no generalized simple formula to do this for such a wide range of $\Delta E_a$, even approximately so at a fixed temperature. You will have to go through The Fun Part.
The Fun Part
The information provided just isn't enough to claim any results. The following derivation makes some assumptions, that the two:

reactions have the same order,
products have zero initial concentration,
rate constants follow the Arrhenius equation, and
pre-exponential factors in the Arrhenius equation are equal.

Consider two reactions:
$$
\ce{A ->[$k_1$][$n_1$] P1} \tag{1}
$$
$$
\ce{A ->[$k_2$][$n_2$] P2} \tag{2}
$$
where $k_1$ and $k_2$ and $n_1$ and $n_2$ are the rate constants and orders, respectively, of Reactions (1) and (2). Note that I have assumed that the reactions are not reversible:
$$
\ce{P1 ->[$k\,=\,0$] A}\\
\ce{P2 ->[$k\,=\,0$] A}
$$
Same Order Competitive Reactions
Assuming $n_1=n_2=n$, the rates $R_1$ and $R_2$ for reactions (1) and (2), respectively, are given by:
$$
R_1 = \dfrac{\mathrm{d}\ce{P1}}{\mathrm{d}t} = k_1\ce{[A]^n} \tag{3}\\
$$
$$
R_2 = \dfrac{\mathrm{d}\ce{P2}}{\mathrm{d}t} = k_2\ce{[A]^n} \tag{4}\\
$$
Dividing Equation (4) with Equation (3):
$$
\dfrac{R_2}{R_1} = \dfrac{\dfrac{\mathrm{d}\ce{[P_2]}}{\mathrm{d}t}}{\dfrac{\mathrm{d}\ce{[P_1]}}{\mathrm{d}t}} = \dfrac{\mathrm{d}\ce{[P_2]}}{\mathrm{d}\ce{[P_1]}} = \dfrac{k_2}{k_1}
$$
Assuming initial concentrations of products is zero, upon integration we obtain:
$$
\dfrac{\ce{[P2]}}{\ce{[P1]}} = \dfrac{k_2}{k_1}\\
$$
Arrhenius Approximation
Assume the rate constants follow the Arrhenius equation:
$$
k_1 = A_1\mathrm{e}^{-\dfrac{(E_a)_1}{RT}}\\
k_2 = A_2\mathrm{e}^{-\dfrac{(E_a)_2}{RT}}\\
$$
Assuming $A_1 = A_2$ and with Equation (3):
$$
\dfrac{\ce{[P2]}}{\ce{[P1]}} = \dfrac{k_2}{k_1} = \mathrm{e}^{-\dfrac{\Delta E_a}{RT}} \tag{5}
$$
where $\Delta E_a =(E_a)_2-(E_a)_1$. Notice that we still have a temperature dependence. Plotting $\dfrac{\ce{[P2]}}{\ce{[P1]}}$ with varying $T$ and $\Delta E_a$, we obtain, with $R = \pu{1.987\times 10^{-3} kcal K^{-1} mol^{-1}}$:

Legend

DEa: $\Delta E_a/\pu{kcal mol^{-1}}$
T: $T/\pu{K}$
[P2]/[P1]: $\dfrac{\ce{[P2]}}{\ce{[P1]}}$

Discussion
Higher $\dfrac{\ce{[P2]}}{\ce{[P1]}}$ is obtained at lower $T$ and higher $\Delta E_a$.

I know that 1.4 kcal/mol approximately corresponds to factor $\left( \dfrac{\ce{[P2]}}{\ce{[P1]}} \right)$ 10, 3 kcal/mol the product distribution would approximately be $100:1::\ce{[P_2]}:\ce{[P_1]}$.

From Equation (5), the relative product concentrations are dependent on both $\Delta E_a$ and $T$. It would be inaccurate to specify relative product concentrations simply based on $\Delta E_a$. You can immediately see that your rough estimates are extremely affected by even small changes in $\Delta E_a$ and $T$.
Note: Since reactions are not reversible, there is no thermodynamic or kinetic control here. Relative product concentrations are simply dependent on $\Delta E_a$ and $T$. In other word, reversibility is crucial in establishing thermodynamic and kinetic control.
Further Considerations
Varying the equation for rate constant, orders of the reactions, and reversibility may provide interesting insights, but will require extensive numerical analysis.

I think you must also consider competitive first order kinetics.  In that case the relative yield of the different products $\mathrm{[P_1]:[P_2]:...}$ reflects the ratio of the rate constants $\mathrm{k_1:k_2:...}$. And you can use Arrehnius or Eyring equations to estimate this values from the difference in the activaction energies or TS energies.

@PAEP I hope the edit addresses your comment.

From my point of view the answer is much clearer now. Perhaps you can refer your table of relative product concentrations to equation (5) below.

I believe in the line "Dividing Equation (4) with Equation (5):" should read (3) not (5).

I am posting an alternative answer because none provide a general expression for the product distribution. It is impossible to know it without:

The rate laws that follow both reactions. If you do not know the rate law, to answer the question is impossible.
Kinetic data regarding the Arrhenius equation for both reactions. If you only know the difference of activation energies, to answer the question is impossible. You also need the ratio of the frequency factors, i.e., $A_2/A_1$.
The answer depends on the concentration of the reactant $R$. If the final concentration is unknown, to answer the question is impossible.

1. Reactions The chemical reactions are
\begin{align}
  \ce{R -> P\text{1}} \quad r_1 &= k_1 C_\mathrm{R}^\alpha \tag{1} \\
  \ce{R -> P\text{2}} \quad r_2 &= k_2 C_\mathrm{R}^\beta  \tag{2}
\end{align}
where we assume power laws for both of them.
2. Fractional Yield In multiple reactions, it is useful to define the fractional yield $\varphi_{X,Y}$. This is defined as the rate of generation/consumption of species $X$ with respect to the rate of generation/consumption of species $Y$. This allows for rapid calculations as we show below.
We will consider the case for $P1$. In here, $R$ can be consumed by two reactions, so its rate has two terms. Using Eqs. (1-2) yields
\begin{align}
  \varphi_\mathrm{P1/R} &= \frac{\mathrm{d}C_\mathrm{P1}}{-dC_\mathrm{R}} \tag{3} \\
  \varphi_\mathrm{P1/R} &= \frac{r_\mathrm{1,P}}{-(r_\mathrm{1,R} + r_\mathrm{2,R})} \\
  \varphi_\mathrm{P1/R} &= \frac{k_1 C_\mathrm{R}^\alpha}{k_1 C_\mathrm{R}^\alpha + k_2 C_\mathrm{R}^\beta} \\
  \varphi_\mathrm{P1/R} &= \frac{1}{1 + (k_2/k_1) C_\mathrm{R}^{\beta - \alpha}} \tag{4} \\
\end{align}
Two comments regarding Eq. (4):

Only the ratio of the frequency factors $A_2/A_1$ and the difference between the activation energies $E_\mathrm{a2} - E_\mathrm{a1}$ is needed. Let me call $ k := k_2/k_1$.
Only the difference of partial reaction orders $\alpha - \beta$ is needed. Let me call $\nu := \beta - \alpha $. As an example, I will consider $\nu = 1$.

Now we integrate Eq. (4) by noting the equality with Eq. (3), under the assumption that initially $P1$ was not present
\begin{align}
  \frac{\mathrm{d}C_{P1}}{-\mathrm{d}C_R}
  &= \frac{1}{1 + k C_\mathrm{R}^\nu}  \\
  \int_{0}^{C_\mathrm{P1}} \mathrm{d}C_\mathrm{P1} &= - \int_{C_\mathrm{R0}}^{C_\mathrm{R}} \frac{\mathrm{d}C_\mathrm{R}}{1 + k C_\mathrm{R}} \rightarrow
  \boxed{C_\mathrm{P1}(C_\mathrm{R}) = -\frac{1}{k} \ln\left(\frac{1 + kC_\mathrm{R}}{1 + kC_\mathrm{R0}}\right)} \tag{5}
\end{align}
Now, we could do the same for the fractional yield of $\varphi_\mathrm{P2/R}$. However, we have an easier path looking at Eqs. (1-2). The stoichiometry and the extents of reactions $X_1$ and $X_2$ (in terms of molarity) tells us
\begin{align}
  C_\mathrm{R}  &= C_\mathrm{R0} - X_1 - X_2 \\
  C_\mathrm{P1} &= X_1 \\
  C_\mathrm{P2} &= X_2 \\ \tag{6,7,8}
\end{align}
but
\begin{align}
  \require{cancel}
  &C_\mathrm{R} + C_\mathrm{P1} + C_\mathrm{P2}
  = C_\mathrm{R0} \cancel{- X_1 - X_2} + \cancel{X_1 + X_2} \\
  &C_\mathrm{R} + C_\mathrm{P1} + C_\mathrm{P2} = C_\mathrm{R0} \\
  &C_\mathrm{P2} = C_\mathrm{R0} - C_\mathrm{R} - C_\mathrm{P1} \rightarrow
  \boxed{C_\mathrm{P2}(C_\mathrm{R}) = C_\mathrm{R0} - C_\mathrm{R} +
  \frac{1}{k} \ln\left(\frac{1 + kC_\mathrm{R}}{1 + kC_\mathrm{R0}}\right)} \tag{9}
\end{align}
Eqs. (5) and (9) are the final expressions that give the distribution of the products. If other order of reactions take place, the process is the same, but the primitives are different. An analytical expression always exists provided that $\nu$ is an integer.
Thus, to know the distribution we need:

The initial and final concentration of the reactant $R$.
The value of $k$, that at temperature $T$ is
$$
  k = \frac{k_2}{k_1} =
  \frac{A_2}{A_1} \exp\left(-\frac{E_{a2} - E_{a1}}{RT}\right) \tag{10}
$$

3. Example Eqs (5) and (9) may be plotted as a function of $C_\mathrm{R}$ at a fixed temperature that fixes $k$. I will do an example with an initial concentration of $C_\mathrm{R0} = \pu{1 mol/L}$ with two different values of $k$. This situation is funny, because you should read the graph from right to left because $C_\mathrm{R}$ decreases in that way:

Note how the product distribution changes with $k$:

If $k<1$ (circles) $\text{P1}$ dominates.
If $k>1$ (squares) $\text{P2}$ dominates.

If you have some way of controlling the temperature, and enlarging the differences between the distribution of products, you can play with the equations in order to obtain the value of $k$. This means, ending the reaction with essentially $\text{P1}$ or $\text{P2}$. This is only true if you do not activate other reactions when you change temperature, which unfortunately, happens a lot.


text:
Why is fluoroantimonic acid so unique?
Fluoroantimonic acid is the strongest acid because its conjugate base $\ce{SbF6-}$ is exceptionally stable. $\ce{SbF6-}$ is in perfect octahedral symmetry wherein the negative charge is equally shared between the 6 fluorine atoms. Fluorine is the most electronegative element which makes it difficult to pull out its lone pairs (protonation), and the high oxidation state of antimony further reduces the polarizability of fluorine atoms.
What’s I don’t understand is that elements in the same column can form similar structures ($\ce{PF6-}$, $\ce{AsF6-}$, and $\ce{BiF6-}$) except for nitrogen, but what makes $\ce{SbF6-}$ so unique? $\ce{SbF6-}$ is neither the lightest nor the heaviest. The d block contraction and inert pair effect makes the high oxidation states of arsenic and bismuth unusually electronegative, but what is special about antimony? The only weaker conjugate base is $\ce{AuF6-}$ because $\ce{Au^{V}}$ is extremely electronegative, but $\ce{AuF6-}$ is unstable under acidic environment and quickly releases fluorine gas.

+1 //Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

Not that there is  $\ce{[H2F . nHF][SbF6. m SbF5]}$, with n,m >= 0

https://onlinelibrary.wiley.com/doi/10.1002/anie.201308023

Arnold Schwarzenegger is pretty strong, but is he the strongest man alive? Is he unique among all who ever lived or will live? No and no. Same thing with HSbF6. It is not unique.

Is "unique" the correct word here? You may call it "special" if you want. Just because it is a superacid doesn't make it "unique".

As @NilayGhosh states, unique is an absolute. You can't be more unique, or less unique; it means the only one of its kind.

The answer is surprisingly simple, or perhaps elegant.
If we consider the radius ratios of nitrogen(V), phosphorous(V), arsenic(V), antimony(V), and bismuth(V) in VI-coordination with that of fluoride ion in I-coordination (for comparing gas phase acidic behaviour), we obtain the following:

E
Radius Ratio
Value
Octahedral (with ionic bonds)

$\ce{N}$
$\dfrac{r_\ce{N^{+5}}}{r_\ce{F-}}$
0.10
No

$\ce{P}$
$\dfrac{r_\ce{P^{+5}}}{r_\ce{F-}}$
0.29
No

$\ce{As}$
$\dfrac{r_\ce{As^{+5}}}{r_\ce{F-}}$
0.35
No

$\ce{Sb}$
$\dfrac{r_\ce{Sb^{+5}}}{r_\ce{F-}}$
0.46
Yes

$\ce{Bi}$
$\dfrac{r_\ce{Bi^{+5}}}{r_\ce{F-}}$
0.59
Yes

These values are in close agreement with Pauling's values as well and, accordingly, $\ce{SbF6-}$ is the most natural octahedral hexafluoride complex.$^\text{1}$ Bismuth also shows a natural tendency to form octahedral hexafluoride complex, but with a slightly loosely packed structure. Nonetheless, antimony and bismuth are expected to form more stable octahedral hexafluoride complexes $\ce{EF6-}$ and stronger corresponding acids $\ce{HEF6}$ than other group-5 elements. This is supported by:

Fluoride Ion Affinities
Conductivity Measurements

Fluoride Ion Affinities$^\text{2}$
The fluoride ion affinity of $\ce{SbF5(g)}$, $\pu{-506 \pm 63 kJ mol^{-1}}$, is higher than that of $\ce{AsF5(g)}$, $\pu{-421 \pm 22 kJ mol^{-1}}$, suggesting $\ce{HSbF6(g)}$ is more strongly acidic than $\ce{HAsF6(g)}$.
Conductivity Measurements$^\text{3}$
The order of acidity of group-5 pentafluorides, based on conductivity measurements is:
$$
\ce{SbF5>BiF5>AsF5>PF5}
$$
Suggesting the following order of acidity.
$$
\ce{HSbF6>HBiF6>HAsF6>HPF6}
$$
References

Clifford, A. F., Beachell, H.C., and Jack W.M. (1967). The hydrogen fluoride solvent system—I A qualitative survey of acids. J. Inorg. Nucl. Chem., 5, 57. 10.1016/0022-1902(57)80081-5.
Jenkins, H. D. B., Roobottom, H. K., and Passmore, J. (2003). Estimation of Enthalpy Data for Reactions Involving Gas Phase Ions Utilizing Lattice Potential Energies:  Fluoride Ion Affinities (FIA) and pF- Values of mSbF5(l) and mSbF5(g) (m = 1, 2, 3), AsF5(g), AsF5·SO2(c). Standard Enthalpies of Formation:  ΔfH°(SbmF5m+1-,g) (m = 1, 2, 3), ΔfH°(AsF6-,g), and ΔfH°(NF4+,g). Inorg. Chem., 42(9), 2886–2893. 10.1021/ic0206544.
Gillespie, R. J., Ouchi, K., and Pez, G. P. (1969). Fluorosulfuric acid solvent system. VI. Solutions of phosphorus, arsenic, bismuth, and niobium pentafluorides and titanium tetrafluoride. Inorg. Chem., 8(1), 63–65. 10.1021/ic50071a015.

@OscarLanzi the table is accurate. Even I had the same doubt. What happens, based on theoretical studies, is that the (Mulliken) charges on P is actually close to 2, increasing the radius ratio to accommodate the octahedral geometry, and as we go down the group the (Mulliken) charges and atomic radii increase, reducing the inter-fluoride repulsions and stabilizing the anions on the whole. The radius ratio is to show the most 'natural' geometries for the elements in (formal) V oxidation states.

In a way, the highly electronegative fluorine has to provide lesser electron density to the central atom as we move down the group to achieve an octahedral geometry.

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text:
Jablonski diagram and absorption
I was reading about Jablonski diagrams and how they are structured. Bold lines would indicate ground electronic levels, and thin lines indicate the associated vibrational levels. Absorption only happens between the different electron energy (orbital) levels, however any additional associated vibrational level can be occupied (for example, transition from S(0) ground state to the third vibrational state of S(1)).

Why is this the case though? Couldn't a transition happen between the vibrational levels of an energy level (for example, from ground state S(0) to third vibrational state of S(0)) or does an absoprtion always change the electronic level?

That's a good depiction. Where did you get it, is it free?

Yes I found it on: https://www.edinst.com/de/blog/jablonski-diagram/

An electronic transition e.g. from ground level $S_0$ to $S_1$ requires much more energy as between ground level $S_0$ to energetically closer vibrational levels (without changes the electronic state).
Let's assume only one photon is "in charge" to provide the relevant excitation.*  The energy light provides is not a continuous, but a discrete quantity; literally, the photons of electromagnetic radiation carry energy in quanta.  Because the energetic difference between the electronic states $S_0$ and $S_1$ are farther away, than the about the second (only vibrational transition), you typically observe electronic excitations of molecules with (visible) light (e.g., perception of colors, UV-Vis spectroscopy). To induce changes in vibrational states, you however typically use infrared radiation (as in IR spectroscopy); in comparison to visible light, this is a radiation of longer wavelength.
addition to answer a comment:

Spectra can be either absorption spectra, because you observe how much incident radiation intensity (of varying wavelength, or energy) is attenuated by the sample.  This uptake of energy (excitation) can yield the release of of energy (relaxation) for example by radiation of similar, or different wavelength; in some instances launch chemical reactions (e.g., radical photochlorination and photobromination).
Or, the spectrum is an emission spectrum.  An example is the emission map in fluorescence spectroscopy; a sample is excited by radiation of a fixed wavelength, you monitor the wavelength dependent intensity of light "broadcast" by the sample.

A molecule typically can undergo electronic, vibrational, and rotational transitions.  Most often, the gap between two electronic states (e.g. $S_0 \rightarrow S_1$, thick black lines in the figure below) is much larger than the one between two vibrational ($\nu_0 \rightarrow \nu_1$, thin black lines below) belonging to the same electronic state.  Transitions between adjacent rotational states (not shown below) are even closer, energetically speaking.  As already symbolized in the illustration used in your question, the states are all «there» (though not always equally well accessible, as transitions follow selection rules). Simultaneously, there is partially overlap; see e.g. some vibrational states of $S_0$ could be energetically equivalent to vibrational states of $S_1$, or $S_2$.

For simplicity of the representation, the straight blue arrows depart from $(S_0, \nu = 0)$ to higher $S_1$ and some $\nu$ (i.e., not only to $\nu = 0$). It equally is possible for an electronic transition to depart from an other vibrational state than $\nu = 0$:

(image credit to LibreTexts Chemistry)
This contributes to the observation of UV-Vis absorption bands extending multiple nm width (contrasting to the line spectra e.g. in atomic absorption spectroscopy), especially with growing complexity of the investigated molecule (many energetically similar vibrations) at non-zero temperature, surrounded by neighboring molecules of solvent, or solid matrix.

* Second-harmonic generation is an example where two-photon absorptions may occur.

Thank you that makes sense, as IR photons have low energy. One question though. So basically suppose if I would want to excite from S(0) to S(1), which is an electronic transition which needs a photon of 495 nm. But in theory, I could use 490 nm so I get the excitation from ground state S(0) to S(1), but in the third vibrational mode? Is that why it's called an absorption spectrum, because I can also use photons which are slightly higher in energy than for the simple electronic transition?

If you stay in the ground state then an infra-red or vibrational transition could populate, say, $v=3$ from $v=0$ always depending on selection rules. When the transition happens to an excited state then this is an electronic transition and typically needs more energy i.e. visible/uv light, rather than infra-red as in the ground state. The electronic transition also has selection rules but now the excited state is different to the ground state.
You can also have transition in an excited state, at $S_1, (v=0 \to v=3)$ but only using time resolved infra-red methods as excited states always decay back to the ground state.
(btw. The diagram shows states, not orbitals, each state is the total energy of the electrons in each orbital, with the ground state set to zero.)
Edit: I would also add that your diagram shows the situation in solution, in the vapour/gas phase vibrational relaxation can be completely absent then fluorescence can occur from the initially excited level to the ground state vibrational levels.

Because in solution, fluorescence is rare, rather de-excitation because of collision with solvent molecules?

Fluorescence is not rare in solution at all.

In solution fluorescence comes primarily from the lowest vibrational level, as energy is lost via collisions usually far faster than the fluorescence decay rate constant. This need not be the case in the gas phase as vibrational relaxation can be far slower or effectively absent. Fluorescence yield can be v. low in some types of molecules due to intersystem crossing forming triplets but is often large, particularly in aromatics and dyes.

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text:
Change of equilibrium constant with respect to temperature
Suppose we have an arbitrary chemical reaction $A+B\rightleftharpoons  2C+D$ and its equilibrium constant at two temperatures $T_{1},T_{2}$ are $k_{1},k_{2}$. We can relate them as
$$\log\frac{k_{2}}{k_{1}}=\frac{\Delta H}{2.303R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)\ \ \ \ -(i)$$
I am unclear on the significance of $\Delta H$ in the above equation.

Does $\Delta H$ represent the enthalpy change of the reaction at any given temperature or is it the difference of reaction enthalpies at the two different temperatures (given by $\Delta H_{2}-\Delta H_{1}=C_{p,reaction}\Delta T$).

I tried to go back to the Arrhenius equation to understand this. Arrhenius' equation gives a relation for the forward rate constant $k_{f}$ of the reaction with its activation energy $E_{a}$
$$k_{f}=Ae^{-\frac{E_{a}}{RT}}$$
and we take $k_{eq}=\frac{k_{f}}{k_{b}}$. However, it doesn't improve my understanding on the subject because we have to assume $\Delta H$ to be constant with temperature to get equation (i).

There is convention to keep TD constants as K while kinetic constants as k, what comes handy if mixing them up. //  See also Van 't Hoff equation $$\frac{d}{dT} \ln K_\mathrm{eq} = \frac{\Delta_r H^\ominus}{RT^2},$$ of which is your equation above the integral form.  The equation implies $\Delta H_r$ is constant on the interval $T_1 - T_2$.

The activation energy is that energy needed to reach the transition state (top of the hill as it were ) between products and reactants  and is not the same as $\Delta H$ which is the enthalpy change from reactants to products at equilibrium so you cannot mix them or get one from the other.

There is error in you formula, there is just /R, not /(RT)  // Easy check will tell you that both equation sides have to be unitless, what is not possible with the extra T.

There is no need to invoke the Arrhenius equation to understand it. I will give the various form of that equation so you are out of doubts.
The significance is clear when you go back to the differential form, this is, van't Hoff's equation
$$ \boxed{\frac{\mathrm{d}\ln K}{\mathrm{d}T} = \frac{\Delta_\mathrm{r} H^\circ(T)}{RT^2}} \tag{1}$$
in words: $\Delta_\mathrm{r} H^\circ$ is the enthalpy of reaction when $1$ mol of $A$ in its standard state at temperature $T$ reacts with $1$ mol of $B$ in its standard state at temperature $T$, to yield $2$ mol of $C$ in its standard state at temperature $T$ and $1$ mol of $D$ in its standard state at temperature $T$.
The following points are true:

The value of $\Delta_\mathrm{r} H^\circ$ depends on temperature.
The integration of Eq. (1) assuming its constant value is (there is a typo in your equation)
$$ \ln\left[\frac{K(T_2)}{K(T_1)}\right] = -\frac{\Delta_\mathrm{r} H^\circ}{R}
   \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \tag{2}$$
this integration is only carried on exams or easy exercises on books. However, if you are dealing with a new reaction, it is a nice way to get an estimate of $K(T_2)$. Its validity is simple: the estimation will be worse as the distance from $T_1$ to $T_2$ is large. It is also an excellent guess when the reaction is not that exothermic neither endothermic, so large temperature changes are needed to shake the value of $\Delta_\mathrm{r} H^\circ(T)$.
The dependence of $\Delta_\mathrm{r} H^\circ$ can be obtained by doing an additional integration
\begin{align}
  \frac{\mathrm{d}\Delta_\mathrm{r} H^\circ}{\mathrm{d}T} &=
  \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) \\
  \int_{\Delta_\mathrm{r} H^\circ(T_1)}^{\Delta_\mathrm{r} H^\circ(T_2)}
  \mathrm{d}\Delta_\mathrm{r} H^\circ &=
  \int_{T_1}^{T_2} \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) \; \mathrm{d}T \\
  \Delta_\mathrm{r} H^\circ(T_2) &= \Delta_\mathrm{r} H^\circ(T_1) +
  \int_{T_1}^{T_2} \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) \; \mathrm{d}T \tag{3} \\
\end{align}
where in your reaction
\begin{align}
  \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) &=
  \sum_\mathrm{j} C_\mathrm{p,j}^\circ(T) \\
  \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) &= 2C_\mathrm{p,C}^\circ(T) +
  C_\mathrm{p,D}^\circ(T) - C_\mathrm{p,A}^\circ(T) - C_\mathrm{p,B}^\circ(T) \tag{4}
\end{align}
In Eq. (3), typically you have access (at the back of the books) for the value $ \Delta_\mathrm{r} H^\circ(T_1) $ at $T_1 = \pu{298.15 K}$. Thus, if you get heat capacity data for your substances $A$, $B$, $C$, and $D$ as a function of temperature, you can combine Eqs. (3) and (4) and just integrate.
You have to respect the heat capacites in Eq. (4). If $A$ is a liquid, then the heat capacity must be that of a real liquid for the pressure of $\pu{1 bar}$. If it is a gas, then the heat capacity must be that of an ideal gas for the pressure of $\pu{1 bar}$.
When you do the full integration, but not for a temperature $T_2$ but arbitrary temperature $T$, then you can go back to Eq. (1). Once you integrate it again, you have the "exact" value of $K(T_2)$, in the sense that you get a more exact value the better your data is.
The worst assumption is $\Delta_\mathrm{r} H^\circ = \text{constant}$. If you feel strong, you perform the full integration. An intermediate level is to consider the functionality of temperature, but to assume that the heat capacities of $A$, $B$, $C$, and $D$ are constant in $T \in [T_1,T_2$]. Eq. (3) becomes
$$ \Delta_\mathrm{r} H^\circ(T_2) = \Delta_\mathrm{r} H^\circ(T_1) +
   \Delta_\mathrm{r} C_\mathrm{p}^\circ (T_2 - T_1) \tag{5} $$
Eq. (5) is the one you wrote. Typically, $ C_\mathrm{p,j}^\circ$ is given also at the temperature $T = \pu{298.15 K}$ at the back of the books. Again the reasoning is the same, larger the range, worse the estimation. Again, you can combine Eq. (5) and Eq. (1) to obtain an estimate of $K(T_2)$. This calculation often happens, generally for non-mainstream compounds where maybe you only have the heat capacity at $T = \pu{298.15 K}$.

The equation you wrote assumes that the standard heat of reaction $\Delta H^0$ over the temperature range from T1 to T2 is nearly constant and that A, B, C, and D are ideal gases.   If the standard heat of reaction over the temperature range is not nearly constant, an integration over the temperature range must be applied.

It does  not seem to be assuming an ideal gas. Are you sure you are not confusing it with Clausius–Clapeyron equation of the similar pattern ? That one does.

@Poutnik. Sorry.  You are right.

The reason equilibria change with temperature is heat, energy in general, is involved with the reaction. At equilibrium forward and reverse reactions have equal rates and usually different rate constants and activation energies. A Temperature increase raises the energy of the products and reactants, each in different amounts depending on their heat capacities and overall chemical potential energy. This involves chemical study. Generally the lower energy products increase in energy faster so raising the T favors the endothermic reaction direction. Delta H or U does change with T so in detailed study must be considered. For small temperature changes the change in Delta H is usually small.
This is best explained in any good PChem text.


text:
Have I named this compound correctly using IUPAC naming conventions?
I think it is 3-[3-[2-[[(2R)-2-(3-chlorophenyl)-2-hydroxyethyl]amino]ethylamino]-R-phenyl]-4-Ar. But I'm not sure and would really appreciate some help


text:
In a reversible reaction, how can endothermic and exothermic processes happen simultaneously?
A reversible reaction has both the forward and the backward reaction occurring at the same time. I struggle to understand how this is logically possible. One reaction pathway is always endothermic, while the other reaction pathway is exothermic. Hence under the same conditions, how is it theoretically possible for both reaction pathways to occur at the same time?

Being exothermic is neither required neither sufficient condition for the thermodynamic tendency to undergo the reaction.

Once there is a product molecule there is a chance that it can be converted back into reactant. Of course at the start of the reaction there is v. little product so v. little reactant can be re-formed, but each individual reactant molecule still has the same chance of forming a product or a (different) chance of a product molecule reforming a reactant. At equilibrium the rate of forwards and backwards rates are equal.
The reason an endothermic reaction (positive enthalpy change) occurs can be due to entropy (making the free energy negative, $A\to 2B, \Delta S^\text{o} \gt 0$ ) but there is also a Boltzmann distribution of energies in all molecules so some amount of product will be formed even if energetically 'uphill' wrt. reactant. The activation energies will primarily determine the rate constants and so equilibrium constant and so amounts of reactant and product.

"Once there is a product molecule there is a chance that it can be converted back into reactant" this is the basis of the principle of microscopic reversibility

This is a consequence of being taught incomplete truths as absolutes and blindly believing it. At first Chemical reactions are ABSOLUTE; They happen! A bit later there are complications, possible side reactions, CO as well as CO2 is formed in combustion, ice below zero Celsius does not melt![unless pressure is applied]. The final straw! Reactions can go in reverse!! They can reach equilibrium and both forward and reverse reactions are happening simultaneously!! It is time to learn that everything we have been taught or think we know is but a snippet of the truth.
Perhaps a better way is to start with simple equilibria with a loose understanding of the kinetic theory of matter and the First and Second Laws of Thermodynamics: Molecules are in motion, Energy is conserved, Entropy generally increases and all are important.  Molecules must be moving fast enough, Energy must be transferred and entropy must increase [somewhere]. At equilibrium ΔG = 0 = ΔH - TΔS. ΔH and ΔS have the same sign. In one direction the reaction is energy driven in the other it is entropy driven. When ΔH and ΔS are of different sign or one is zero the reaction cannot be reversed. Think of a thermonuclear bomb, expansion of a [ideal] gas into a vacuum, racemization of an enantiomer, respiration, photosynthesis.
A reversible reaction can reach equilibrium from either direction. The mechanism is the same for each direction. Since rates are a function of the motion of molecules and concentrations and the activation energies, what is changing at constant temperature is the concentrations of reactants in each direction. Eventually concentrations are reached where the rates in both directions are the same, equilibrium. The position is determined by the two activation energies that are incorporated into the rate constants.

Typically when we think about thermodynamics, we're not describing an individual molecule but rather the whole system. So thinking about this just in terms of endothermic and exothermic reactions is not sufficient. It's a matter of kinetics, occasionally two molecules will collide with enough energy and in the correct orientation, such that some of the energy of the collision will be used to create the less energetically favorable molecule.
If you're trying to explain this with thermodynamics, in spontaneous reactions the energy required to overcome the activation energy of the reverse reaction comes from the reaction energy of the forward reaction. Whereas, in non-spontaneous reactions the energy required comes from the surroundings.

You have the idea; at equilibrium that is about what happens locally. However, without a constant source of energy to maintain T, Temperature will change altering the equilibrium conditions. This is effected by blackbody radiation. In the lab or reactor a constant temperature bath or oven is used to maintain temperature. At equilibrium the reaction rates are equal and DeltaG = zero. Ponder that for each reaction. How is it possible for an endo and exothermic reaction to have the same rate?

I don't think that both reactions happen at the same time. It goes one way and then the other. One could associate that with a prey-predator model, where the preys are the reactants and the predator the products. If there too much preys the predators will thrive (right to left reaction) and vice-versa. The reaction react to the way that is the more favorable. Two quantities that can help you study such reactions are the Equilibrium constant along side the reaction quotient. These quantities can tell you which of the two pathways is the more favorable given the current state of your system.
There will be some inaccuracies since I am not a chemist.

Every chemical equilibrium, that looks statical, has dynamic nature on molecular level, with opposite reactions going both ways at equal rates.

No, as soon as a product molecule is formed there is a chance that it well reform reactant.

Not really a chance, the rate is k2[products] where  Keq= k1/k2


text:
Can I put both Epsom Salts and Lectric soda into a bath
I would like to get the benefits of absorbing Magnesium by using either Epsom salts (magnesium sulphate) or magnesium chloride and the benefit that lectric soda (sodium carbonate) has on sore, tired muscles. Can either or both forms of magnesium and lectric soda be put in a bath together and remain in their original forms to act as they do individually or will they react and form something that won’t do anything or worse do something undesirable?

Mixing solutions of magnesium sulphate or chloride and sodium carbonate is effective way to produce precipitate of magnesium carbonate, known also as the mineral magnesite.
When filtered out, you would have solution of sodium sulphate (Glauber salt) or sodium chloride (table salt), with eventual excess of one of the used salts.
I doubt it is what you want.

Addition: $\ce{MgCO3}$ has a low solubility in water (property box on Wikipedia reports  $\pu{0.0139 g}/\pu{100 mL}$ (at $\pu{25 ^\circ{}C}$); most of it will be not in solution, and in consequence, not accessible for anything (maybe "precipitate" is foreign to the OP). Even if, the substance had to travel across layers of skin (much of collagen) prior to reach the muscle. More likely: the warmth of the bath soothes some of the sensation of sore (overstretched / ruptured) muscle fibers if warming up (e.g., prior to a run) was skipped.

@Buttonwood I see. I wrote it maybe under false impression "precipitate" is a common word, not limited to chemistry or science.

I agree (depending on language) "precipitate" is not constrained to chemistry, or science.  In comparison of an English general purpose dictionary and Robert, a French one, however, only the later explicitly includes the note of low solubility (though uncertain if by chance by the editors' choice, or by use [of related words] equally in other circumstances, than chemistry), photo taken.

@Buttonwood I cannot even without looking remember other suitable verb for solids excluding themselves from solutions. // It reminds me a user objecting against adjective "laconic",  suggesting an alternative I have not known it's meaning, not even ever seeing such a word. (and forgetting it immediately :-)


text:
How to Calculate Entropy Changes at Non-Standard Conditions
I am trying to calculate the values of $\Delta H$ and $T\Delta S$ for the reaction taking place at $\pu{800 ^\circ C}$
$$
\ce{CO2 + H2O + 2CH4 -> 3CO + 5H2}
$$
I calculated $\Delta H$ to be $\pu{\approx 480 kJ mol^{-1}}$ using Hess’ Law, i.e., by taking the reactants down to $\pu{ 25  ^\circ C}$ and calculating the enthalpy change, performing the reaction at standard conditions, and then taking the products back up to $\pu{800 ^\circ C}$.
However, I am having trouble with $T\Delta S$. Can I take the same approach as I did for the enthalpy change, taking the reactants/products between $\pu{25 ^\circ C}$-$\pu{800 ^\circ C}$ and calculating the entropy change at $\pu{25 ^\circ C}$?  Then multiplying by the temperature in Kelvin?
I tried using the Shomate equation expression below for the reactants/products (taken from NIST), where A-G are Shomate equation coefficients; however, I’m not getting the correct answer of $\pu{\approx -80 kJ mol^{-1}}$, so unsure if this is the correct approach:
$$
\begin{align}
\Delta S = &A \ln{(T_2-T_1)} + B(T_2-T_1) + \dfrac{C}{2}(T_2^2-T_1^2)\\ &+ \dfrac{D}{3}(T_2^3-T_1^3) - \dfrac{E}{2(T_2^2-T_1^2)} + G
\end{align}
$$

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

As you can evaluate integral  Cp.dT for  enthalpy, you can do the same for Cp.dT/T for entropy.

Are you using absolute temperatures in your entropy calculation?

You maybe applying the Shomate equation incorrectly:
$$
\ln{T_2} - \ln{T_1} = \ln{\dfrac{T_2}{T_1}}
$$
$$
\dfrac{1}{T_2^2} - \dfrac{1}{T_1^2} = \dfrac{T_1^2-T_2^2}{T_1^2T_2^2}
$$
The correct application of the Shomate equation would be:
$$
\begin{align}
S ^\circ  &= A \ln{(T)} + B(T) + \dfrac{C}{2}(T^2)\\ &+ \dfrac{D}{3}(T^3) - \dfrac{E}{2(T^2)} + G\\
\implies \Delta S ^\circ &= A \ln{\dfrac{T_2}{T_1}} + B(T_2-T_1) + \dfrac{C}{2}(T_2^2-T_1^2)\\
&+ \dfrac{D}{3}(T_2^3-T_1^3) - \dfrac{E}{2}\left( \dfrac{T_1^2-T_2^2}{T_1^2T_2^2}\right) + G
\end{align}
$$

How to Calculate Entropy Changes at Non-Standard Conditions

Do note that conditions are still standard, and not non-standard, as indicated by the $\circ$ superscript in the above equations.

Thank you very much for your help ananta!  I am still slightly confused however.  If the Shomate equation is only applicable for standard conditions, how is it applicable to calculate the entropy change between two temperatures?  Thanks again.

@KarlNow yes! Temperature can change; however, pressure must be kept constant at $\pu{1 bar}$.

@KarlNow another factor to consider is the phase change of water. I am not sure if the Shomate equation is valid for liquid phase of water. I could help you out if you share your calculations as well.

After working through the problem I've figured out the issue now!  Thanks very much for all your help ananta!!

@KarlNow happy to help :)


text:
Question about solubility of silver halide
In many japanese chemistry books for highschool students, it is written that AgF is soluble in water because compared to other silver halides ionic nature of Ag-F bond is greater.
I don't think this is valid explanation, because what determines solubility must be free energy of the process not the nature of the bond.
Am I thinking the right way?

https://chemistry.stackexchange.com/questions/97868/why-do-we-use-different-arguments-for-determining-the-strength-of-hydracids-and

Also see: https://chemistry.stackexchange.com/questions/6238/which-silver-halide-is-the-most-soluble

Thank you. I didn't know about HSAB theory well. So F and Ag is hard and soft pair.

You are certainly correct. Whether something has more ionic or more covalent bonding does not in general correlate directly with solubility in water. Among alkaline earth fluorides (which all have network bonding and not discrete molecules) beryllium fluoride has the most solubility in water — and the most covalent character.
When push comes to shove, we really have no reliable way to predict the solubility of salts in water a priori.


text:
NIST SRMs: Determination of the Validity Period
The National Institute of Standards and Technology (NIST) produces Standard Reference Materials (SRMs) of basically anything. That also means that things like peanut butter are among those SRMs. SRMs come with a Certificate of Analysis (CoA) and have a validity period attached.
Given that NIST makes sure to know almost everything about their SRMs, how do they determine the validity period? Complex matrices can have difficult to estimate long-term influences on all the compounds in the sample. If they are running stability studies, when do they start those and for how long do they run?

Search Accelerated Stability Tests for Certified Reference Materials. Also read the poster on "Certification process for organic certified reference materials (CRM) using high-performance quantitative 1H-NMR (HP-qNMR®): special focus on stability studies." It is available on ResearchGate for free.


text:
Colour composition of Bromine during diffusion?
Consider the diffusion of Bromine Gas within a tube filled with air.
After a prolonged period of time, is the colour composition uniform?
From my understanding, Bromine is a denser gas. Hence, on average, the Bromine gas is more concentrated at the bottom of the flask. There is always diffusion, even against gravity, so I think the colour would be relatively uniform. However, the bottom of the tube would still be slightly darker. ( May not be noticeable ).
I've read that the colour composition is entirely uniform. I can see where this comes from. I suspect we consider the particles to be too small to have any gravitational influence. However, if we were to have a longer tube would there be a slightly darker portion at the bottom of the tube? Do we consider the colour composition to be exactly uniform because the tube is too small to have any noticeable seperation of colour?
This problem has puzzled me a bit. Is my intuition right? That is, there is a slight seperation of colour, even though it may not be noticeable?

In a tall tube set upright, the density will change with height provided that the gas is not stirred. i.e. no convection occurs, so with that condition your intuition is correct. As the density (or pressure) changes so does the optical density ( colour). The relative density changes according to the Boltzmann distribution as
$$\displaystyle \frac{\rho}{\rho_0}=\exp(-gMh/(RT)) $$
where $\rho_o$ is the density at $ h=0$, where $h$ is the height in metres where you want to calculate the density, $g$, acceleration due to gravity(m/s$^2$), $M$, molecular weight (kg/mol), $R$, the gas constant (J/mol/K), and $T$ the temperature in Kelvin.
Edit.
The equation above can be derived in an intuitive way from the ideal gas law and the fact that each molecule experiences a force under the effect of gravity as well as thermal motion. Using the ideal gas law is not a restriction as we can always choose a low concentration so that it accurately describes the gas/ vapour behaviour.
In the cylinder of gas of base area $A$ select at layer of thickness $\Delta Z$ and so a volume $V=A\Delta z$. The number of molecules $n$ in this volume is found using the ideal gas law ($pV=nk_BT$) and is
$$ \displaystyle n=\frac{p\,A}{k_BT}\Delta z$$
where $k_B$ is the Boltzmann constant. The force on a molecule is $f=mg$ so that there is a pressure change across layer $\Delta z$ of $\Delta p=p_{(z+\Delta z)}-p_z$. Equating forces (pressure is force/area) then  $A\Delta p  = -n\,mg$ (negative as the force acts downwards) then rearranging gives
$$\displaystyle \Delta p= -\frac{n\,mg}{A}=-\frac{p\,mg}{k_BT}\Delta z$$
and next make the $\Delta$ very small and in the limit into differentials so that
$$\displaystyle dp=-p\frac{mg}{k_BT}dz$$
which integrates to the equation above where molar units are used.

in theory this is true. But this is misleading. What is the density difference in, say, a 1m tall tube that fits in a typical lab? While, were the tube 1km tall, the difference might be detectable, is it detectable in anything that fits in a typical lab?

I agree in principle, the OP can calculate this for themselves what the difference can be. (In the lab its probable that convection would dominate even if utterly tiny differences in optical density could be measured. ) In the distant past I have done demonstrations using small air-agitated polystyrene spheres that can demonstrate the effect of gravity on gas density.

You'd need a good glassworker to actually make it, but you could envisage a metre-long vacuum flask with the gas mix inside, to massively reduce thermal inputs that would drive convection.  Silvered apart from a series of small windows through which to observe (e.g. shine a very weak laser that will be absorbed by the bromine) and measure the absorption as a function of height.  Heater coils round the inner tube would provide even heating, and they could be switched off once it was all gaseous - or really on it having a non-zero vapour pressure

I will humbly disagree with the solution of my fellow trying to justify the opposite answer.

[OP] After a prolonged period of time, is the colour composition uniform?

Under the sole condition of diffusion, a solution made up of several substances can never achieve a uniform composition even at $t \rightarrow \infty$. Moreover, gravity does not influence the composition profile, it only ensures that the fluid does not "fall" by specifying an adequate pressure at every point.
We will consider for simplicity the diffusion evolution along the $z$-axis, so we can imagine that the $x$ and $y$ directions are of infinite lenght than that of the $z$ axis. In this scenario, the velocity field can only have the $z$ direction, and since air doesn't flow in the $x$ or $y$ direction, is only a function of $z$, i.e. $\mathbf{v}(z) = v_z(z)\mathbf{\hat{z}}$. The image is the following:

1. Mass Balance The continuity equation is
\begin{align}
  \frac{\partial \rho}{\partial t} + \mathbf{\nabla} \cdot \mathbf{v} = 0 \tag{1}
\end{align}
but a prolonged period of time traduces into equating the time derivative to zero, that is, we want the solutions for $t \rightarrow \infty $ or the steady-state solutions. Eq. (1) says
\begin{align}
  \mathbf{\nabla} \cdot \mathbf{v} = 0 \rightarrow
  \boxed{\frac{\mathrm{d}v_z}{\mathrm{d}z} = 0} \tag{2}
\end{align}
Thus, the velocity field at $t \rightarrow \infty $ is just a constant for every point. Here we need to make a decision:

If we consider $v_z \neq 0$, then we cannot stop at $z = L$, because the air will continuously go up. This can be circumvented by considering an ideal solution, of semi-infinite diffusion, where $L \rightarrow \infty$. This is an interesting solution often studied in transport phenomena but not our case.
If we want the tube of a finite size then $v_z = 0$, which is the condition that we will take.

2. Momentum Balance  The equation of motion in the case for a newtonian fluid is
\begin{align}
  \rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot
  \mathbf{\nabla}\mathbf{v}\right) &= -\mathbf{\nabla}P + \mu \nabla^2\mathbf{v} + \rho \mathbf{g} \tag{3} \\
\end{align}
where $\mu$ is the viscosity of the air+bromine solution. In a prolonged period of time, employing Eq. (2), and remembering that the gravity points down in our coordinate system
\begin{align}
  \rho(v_z \mathbf{\hat{z}}) \cdot \left(\underbrace{\frac{\mathrm{d}v_z}{\mathrm{d}z}}_{=0}\mathbf{\hat{z}}\right)
   &= -\frac{\mathrm{d}P}{\mathrm{d}z}\mathbf{\hat{z}} + \mu \underbrace{\frac{\mathrm{d}^2 v_z}{\mathrm{d}z^2}}_{=0}\mathbf{\hat{z}} +
  \rho (-g\mathbf{\hat{z}}) \\
  0 &= -\left(\frac{\mathrm{d}P}{\mathrm{d}z} + \rho g\right)\mathbf{\hat{z}} \rightarrow
  \boxed{\frac{\mathrm{d}P}{\mathrm{d}z} = -\rho g} \tag{4}  \\
\end{align}
Eq. (4) establishes the pressure profile in order to balance the weight of the fluid, that can be solved once stating a boundary condition. This is typically the pressure at a boundary, for example, if open to the atmosphere that would be $P(z=L) = P_\text{atm}$. This profile is a linear function and has its maximum at the base of the tube.
3. Mass Balance for Bromine The continuity equation for $\ce{Br2}$ under the absence of sources or sinks (like chemical reactions), and considering Fick's law to relate the flux and the concentration, is
$$
  \frac{\partial C_\ce{Br2}}{\partial t} + \mathbf{v} \cdot
  \mathbf{\nabla}C_\ce{Br2} = D_\ce{Br2,air} \mathbf{\nabla}^2 C_\ce{Br2}
  \tag{5}
$$
where $D_\ce{Br2,air}$ is the diffusion coefficient of $\ce{Br2}$ in air. Again, for steady state conditions Eq. (5) yields
\begin{align}
  (v_z \mathbf{\hat{z}}) \cdot \left(\frac{\mathrm{d}C_\ce{Br2}}{\mathrm{d}z}\mathbf{\hat{z}}\right)
  &= D_\ce{Br2,air} \frac{\mathrm{d}^2 C_\ce{Br2}}{\mathrm{d}z^2} \\
  v_z \frac{\mathrm{d}C_\ce{Br2}}{\mathrm{d}z}
  &= D_\ce{Br2,air} \frac{\mathrm{d}^2 C_\ce{Br2}}{\mathrm{d}z^2} \rightarrow
  \boxed{\frac{\mathrm{d}^2 C_\ce{Br2}}{\mathrm{d}z^2} - \frac{v_z}{D_\ce{Br2,air}} \frac{\mathrm{d}C_\ce{Br2}}{\mathrm{d}z} = 0} \tag{6}
\end{align}
Eq. (6) establishes the concentration of bromine, for $t \rightarrow \infty$, which is independent of gravity, once two boundary conditions are stated. For our case of $v_z = 0$, Eq. (6) states a linear profile, with a maximum at the base of the tube, and a minimum at the top of the tube. Therefore we have proven that the concentration of $\ce{Br2}$ is not uniform in the tube.
4. Conclusion The lower the diffusion coefficient, the more difference of concentration there will be between the bottom and the top. This is the reason why we mix coffee with a spoon every morning, why we use a magnetic stirrer for titration, for the variables of interest to be independent of position. From a mathematical point of view, those actions are done to avoid this type of behaviour, where we have an undesired dependency of position in the concentration of a substance.
If $\ce{Br2}$ diffuses very easily in air, then we may consider that $C_\ce{Br2} \approx \text{constant}$ in the tube, because we may not be able to notice. However, in this scenario that we are studying, and strictly prohibiting to shake the tube, a difference of concentration and hence of colour is inevitable.

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text:
Uncertainty for repeated measurements
I am having some trouble calculating the percent uncertainty of obtained cell potential. My cell potentials were 1.788, 1.787, and 1.792 volts with the uncertainty of voltmeter $\pm$ 0.001 volts. Here was my attempt:
The average cell potential:
\begin{align*}
    \frac{1.788+1.787+1.792}{3}=1.789
\end{align*}
Percent uncertainty when measurements are repeated was found by the following formula:
\begin{align*}
    \text{% $E_{cell}$ uncertainty}=\frac{\frac{\text{range}}{2}}{\text{mean}}\times 100\%.
\end{align*}
Therefore:
\begin{align*}
    \text{% $E_{cell}$ uncertainty}=\frac{\frac{1.792-1.787}{2}}{1.789}\times 100\%\approx\text{0.1\% (1 s.f.)}.
\end{align*}
But, my calculation doesn't involve voltmeter's uncertainty of $\pm$ 0.001 V. What should I do?
If I were to use standard error, then,
Standard deviation:
\begin{align}
\sqrt{(1.788-1.789)^2+(1.787-1.789)^2+(1.792-1.789)^2}=0.002
\end{align}
Standard error of average measurement:
\begin{align}
\frac{0.002}{\sqrt{3}}=0.0012
\end{align}
Total uncertainty:
\begin{align}
0.0012 + 0.001 = 0.0022
\end{align}
Total percentage uncertainty:
\begin{align}
\frac{0.0022}{1.789}\times100=0.1\%
\end{align}

I'm sorry, I didn't get it. May I ask you to 'answer the question' instead of 'add a comment' for a more detailed explanation?

where did sqrt(12) come from?
what is meant by s^2_1meas,Corr = s^2_1meas + s^_device? I am not familiar with this formatting.

As you have few measurements you need to use Student's t test because so few samples may not represent a normal distribution. With $t=n-1=2$, calculate $\pm t;s/\sqrt(3)$. At $90$% confidence $t=2.9$ and 4.3 at $95$%. If this is bigger than your instrumental error you can ignore the latter.

@porphyrin, thanks for the comment! Can you explain it to me some more?

@Poutnik, thanks for the link! I'll check it out!

The details of t distributions can be found in any good stats book. Meanwhile you can just use the formula I have given. There is some detail in chapter 13 of the ebook https://chemistry-maths-book.com/intro.html

Imagine you measured 10 values:

6 times 1.790 V
2 times 1.789 V
2 times 1.791 V

Estimation of standard deviation of 1 measurement is $$s_\mathrm{meas} = \sqrt{\frac {\sum_{i=1}^{n} {(x_i - \bar{x})^2}}{n-1}} = \\
\sqrt {\frac{2 \cdot (1.789-1.790)^2 + 2 \cdot (1.791-1.790)^2 }{ 10-1}} \pu{V} = \\
\sqrt {\frac{\pu{4e-6}}{ 9}} \pu{V}  =\pu{0.0006667 V} \approx \pu{0.0007} V $$
With the voltmeter resolution $\ce{0.001 V}$, the related estimation of the standard deviation for the uniform distribution is $s_\mathrm{resol} = \frac{\pu{0.001 V}}{\sqrt{12}}=\pu{0.0002887} V \approx \pu{0.0003 V}$
The overall corrected estimation of the measurement standard deviations is:
$$s_\mathrm{meas, corr} = \sqrt{s^2_\mathrm{meas} + s^2_\mathrm{resol}}=\sqrt{0.0007^2 + 0.0003^2} V \approx \pu{0.0008 V}$$

For you case, you need first to do the Student test of   values that are off, as porphyrin suggested. Note that 3 values is not exactly many to work with.
For a small number of measurements,it is often recommended to estimate the standard deviation from the data range: $s = R \cdot \frac{1}{d_n}$.
|  n | 1/d_n |
|----+-------|
|  2 | 0.886 |
|  3 | 0.591 |
|  4 | 0.486 |
|  5 | 0.430 |
|  6 | 0.395 |
|  7 | 0.370 |
|  8 | 0.351 |
|  9 | 0.337 |
| 10 | 0.325 |

For the measurement data $\pu{1.788 V}$, $\pu{1.787 V}$, and $\pu{1.792 V}$,
$$s(n=3)_\mathrm{meas} \approx (\pu{1.792 V} - \pu{1.787 V} ) \cdot 0.591 = \pu{0.002955 V} \approx \ce{0.003 V}$$
$$s_\mathrm{meas, corr} = \sqrt{s^2_\mathrm{meas} + s^2_\mathrm{resol}}=\sqrt{0.003^2 + 0.0003^2} V \approx \pu{0.003 V}$$

Note that the reason for keeping $s$ with just the single valid digit is this:  (Standard deviation of standard deviation)

$$\frac{\mathrm{SD}(s)}{s} =  \frac{\Gamma( \frac{n-1}{2} )}{ \Gamma(n/2) } \cdot \sqrt{\frac{n-1}{2} - \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2 } \approx \frac{1}{\sqrt{2(n-1)}}$$

For $n=51$,  $\frac{\mathrm{SD}(s)}{s} \approx 0.1$
what well illustrates why uncertainties with two valid digits are rare and must be well justified and supported by much of data.

It helped me so much!! Thank you!!

Consider then accepting the answer, unless you need further clarification.


text:
Hydrogen Isotopes and Bronsted Lowry Acid
a Bronsted Lowry Acid is defined as an acid that donates a proton. However it's actually a Hydrogen Ion, since hydrogen only has 1 proton and 1 electron, a positive Hydrogen Ion would have no electron and therefor only be a proton which is why we call it that. However Isotopes exist:
deuterium has 1 neutron and tritium has 2, my question is: Can an (Bronsted Lowry) acid donate a hydrogen Isotope? Is the naming just out of convenience or very specific? Surely an acid can donate a deuterium ion, or perhaps there's a reason these isotypes are different?

Deuterium works essentially the similar way as hydrogen but only slower. For example, in benzyne mechanism, the reaction slows down if hydrogen is replaced with deuterium showing isotope effect.

BL theory have been created 8 years before deuterium discovery. It was left as is probably for simplicity as D relative abundance is about 0.015%. Whoever works with deuterated acids is aware of this formulation inaccuracy.

Note that some formulations say "hydrogen cation".

Just remember that titles and names do not matter for elements or atoms. As Shakespeare wrote "What's in a name? That which we call a rose, by any other word would smell as sweet." An element is to be identified by the number of protons ($Z$) not by the number of neutrons or electrons. Isotopes have very similar chemical properties, but their reaction rates are different because of different bond strengths. Say, you have HCl gas (dissolved in $\ce{H2O}$) and DCl gas (dissolved in $\ce{H2O}$), both liquids will still behave like very strong Bronsted-Lowry acids. If you were to add a base like NaOH and NaOD, in each case, one will get water ($\ce{H2O}$ and $\ce{D2O}$) and salt as products. Heavy water is still chemically "water".

In its common usage "water" is not definitive. A quick glance thru the Wikipedia tome on water [read it, the entry on deuterium depleted water, and on Heavy water] affords an amazing array of properties, etc., of H2O with no mention of isotopic [or isomeric, yes H2O has "isomers"] composition. I do not know if the data presented are for isotopically pure H2O or the natural isotopic distribution. Since natural water is mostly H2O its properties are very similar to but not identical to pure H2O. Natural water also differs depending on the source; there is more D in the oceans than in water vapor, fresh water, or glacial ice [I do not know about sea ice].
Natural water devoid of non-water impurities is a mix of many compounds: H2O, HDO, D2O, H2O[17] and its D analogues. H2O[18]and its D analogues culminating in the super heavy D2O[18]; there is also the mix of ortho and para-H2O and the trace of tritium compounds. We are fortunate that this can all be almost ignored and left to the specialists to worry about.
These, excepting the HD and o,p compounds that can interconvert, are distinct chemicals with distinct, while similar, chemical and physical properties. The differences manifest in kinetic isotope effects and in the compositions of products in reactions that do not reach equilibrium or completion. D2O is a slightly weaker acid or base than H2O. Mixes act as mixtures.
Isotopes are similar chemically but can be very different in nuclear and spectroscopic properties. Low temperature behavior of Helium[3] and Helium[4] is an example; They are almost different elements.

What is the reason for the downvote?


text:
Small amount of sulfuric acid down drain - questions
Bought some drain cleaner from my local hardware store, the guy at the counter said its powerful stuff that barbers use to clear hair from drains. Said to not put my head above the drain while using it or breathe it in (standard for most household chemicals), open a window and use about 150ml, let it sit 5-10 mins then rinse with cold water, simple enough.
Go to use the stuff today on a tub/shower drain. I'd just finished a shower and got dressed, so there could have been some residual lukewarm water in the drain. As soon as I tipped the stuff into the drain, it bubbled violently and created some reddish-orange stains around the drain (since rinsed off) and some dark grey staining on the drain ring itself. Frantic googling revealed that it was sulfuric acid.
I started running cold water from the tap about 15 seconds after this reaction (googling while water was running) since that was the advice in the first place and rinsed the inside of the tub with cold water numerous times from a bucket and with detachable shower head, rinsing even the walls next to the shower that were most likely not even in contact with the bubbling drain (though there was some steaming and spitting, so something could have spattered).
the Window was open and I didn't smell anything during or after. I left the cold water run for about 5 minutes after the initial use, and checked on the thing and run water a few more times and aside from the dark grey staining, everything seems to be normal. It was sulfuric acid it turns out. If I had to guess - I would say I got maybe somewhere between 1-2 ounces/30-60ml (if that) out of the bottle.  The reaction was so immediate that I stopped right away.
I live in a foreign country and can't read all labeling which is why I got the full download from the gent at the hardware store. Wasn't warned to expect bubbling/boiling so make sure all the water was cleared from the drain or wear PPE so guessing that's what caused the initial violent reaction. I did not use gloves or glasses - very stupid I know, but I would not have used the solution at all if I'd known.  I did not get any on my skin but rinsed my hands many times regardless.  I also threw some baking soda down after I'd rinsed the tub out many times and run the water for a very long time (that was probably at least 60-90 mins after initial use).  Putting the soda down did not generate any reaction aside from absorbing water.  nonetheless I scooped it out after about 45 mins bagged it a put it in the trash outside.
I called poison control about potential exposure on skin/inhalation - they said if it were there, I'd know by now (approx 45 - 60 mins after initial use), but based on the situation its likely that there was no exposure. I even changed and washed my clothes and my towel and bathmats that were nearby just in case (as was recommendation on many chem company websites/MSDS)
I won't be using any more of the acid, but is there anything else I should do to clean/clear the acid that was used? I immediately capped the bottle and set it down on the tile near the tub - but if there were a drop or two that happened on the tile or floor, what danger could that bring.
I'm freaking out a bit about how I was no naive and did not take the time to translate the label.  Wondering if there are any future dangers to consider if its left in small amounts on the surfaces around the drain? most of the MSDS are geared toward severe spills/exposure/inhalation rather than smaller applications.

Well, maybe you didn't notice that drain cleaners are all very aggressive by design - they are supposed to react with stuff in the drain.

I am aware - I just didnt expect such an immediate and strong reaction and now understand this product isnt really for household applications unless in the hands of a qualified professional.

I wouldn't say so, NaOH is hardly less dangerous. You just need to be careful and definitely not panic. You only saw a reaction that was supposed to happen inside the plumbing - where it wouldn't be visible.

An MSDS will tell you what safety precautions might be necessary with respect to handling and disposal. Is your question what reaction occurred? Personal medical questions are not accepted on the forum. If you develop symptoms from use of the product contact a medical practitioner.

It's odd that $\ce{H2SO4}$ was used as a drain cleaner, since it attacks metal pipes, and is not effective at saponifying fats. Double-check that the cleaner was not $\ce{NaOH}$ solution.
That said, $\ce{H2SO4}$ is readily soluble in water, and, once diluted, was safely neutralized with baking soda.
However, a concern is that plumbing might have been damaged. Check for leaks, and possibly, use a small camera to check for holes in the drain pipe, or where flange meets the drain and tub or base.

Apparently there does exist a drain cleaner, in the USA, that is concentrated sulfuric acid. Enough said …

It was sulphuric acid, I translated the label and it said "based on sulphuric acid" but does not give a chemical solution or % concentration, but I'm guessing based on my digging its around 95%.  I could not find an MSDS for this exact product online, only similar solutions.  I can access my tub plumbing from below and I see that after running water for a very long time, there is a known drip but thats all.  Its PVC piping  but may just call a plumber out to double check anyway. Thanks

This is in europe, but I did find this common and available for extreme blockages and industrial applications.  to be frank, I can't even believe they sold it to me at a regular shop

The most surprising phenomena was the "violent bubbling"  and the "dark grey staining" that happened when you used your stuff for the first time. This is due to the stuff that was previously deposited in the drain. It is not due to sulfuric acid alone, or caustic soda alone.  What was this drain used for, previously ?

its a household shower drain - that's all. occasionally used to pour other standard household cleaners down, but by that I mean CLR or floor cleaner for mopping. That was several months ago at the most recent. I did read on a plumbing forum that its pretty common for the acid to discolor nickel finish by turning it black.  The bubbling, I am only guessing was because of the water left in the trap and soap/hair that could have been there.  Also said this on the plumbing forum.

I also found this answer on SE - seems much more of the product was used in that case.  https://chemistry.stackexchange.com/questions/39502/sulfuric-acid-fumes-from-cleaning-shower-drain#:~:text=Sulphuric%20acid%20is%20not%20toxic,will%20be%20no%20further%20problem.


text:
How does the anti isomer below have a plane of symmetry? The R group and the H end up in opposite sides if you divide in half
If we add substituents R to this structure, we can have two diastereoisomers with the two R groups on the same side (syn) of the flat ring or on opposite (anti) sides. Although the plane of the paper is no longer a plane of symmetry, neither isomer is chiral as the other plane bisects the substituents and is still a plane of symmetry.
How does the anti-diastereoisomer have a plane of symmetry?

The plane of symmetry contains both R's and both H's, so they don't move when reflected.

Exactly like it is drawn in the picture...

The two $\ce{CHR}$ groups at opposite corners of the ring are co-planar, and both isomers are symmetric about this plane.
Yes, seeing a molecule in three dimensions when you have a paper drawing is difficult. In my day at school the recommendation was to use a kit to build a molecular model, which you could rotate to get different views. Today we can use computer images.

So, what is wrong this time? (Other than getting burned by Autocorrect, which forced an early-morning edit?)

Both molecules have a plane of symmetry. The $\ce{R}$ groups lie on the plane of symmetry. Try visualizing it this way:

Unless the $\ce{R}$ groups are chiral, both the molecules can exist in an achiral conformation.
Also, likely, the ring is not planer because $\pu{120 ^\circ}$ is not the natural angle for the $\ce{N}$ atoms in (somewhat) $\mathrm{sp^3}$ hybridization.

Amide nitrogens are very much planar.

@orthocresol 'very much' being the keywords, the six-membered ring is strained in the planer configuration. Based on geometrical optimization, non-planarity adds stability.


text:
What is the best way to draw chemical structures in LaTeX?
I was just wondering what the best option is for getting chemical structures/reactions in LaTeX. Should I just draw the molecule in ChemDraw/ChemSketch then export it to LaTeX or is there a particular set of packages that would be better in the long run?
Thanks!

Can you make chemical structure diagrams in LaTeX?


text:
HPTLC Densitogram - Not Showing Other Secondary Metabolites
I have run a HPTLC analysis of methanolic extract of Gymnema Sylvestre R.Br. plant in order to determine the concentration of active ingredient (Gymnemic acid using Gymnemagenin as standard). Sample run was outsourced to a third party. The densitogram of standard and sample are presented. Is it possible that other secondary metabolites do not appear as peak in the densitogram of the plant extract. In the result of the sample only the peak of the gymnemic acid appears while no other significant peak is witnessed indicating presence of other metabolites?.

For readers, read about the anti-sweetener Gymnemic acid


text:
What if kc=1 will reactants concentration equals to product concentration
I read a book, it says if kc=1 it mean reactant concentration equals to product concentration at equilibrium. But I'm wondering that in kc expression we dont just use concentration of an element we raise them to their stoicheometric coefficient(power). Can somebody explain me about this

You reasoning is correct, $K_c = 1$ doesn't necessarily mean the reactant and product concentrations are equal.
Consider a general reaction:
$$
\sum_ir_i\ce{R}_i \ce{->} \sum_jp_j\ce{P}_j
$$
The equilibrium constant is given by:
$$
K_c = \dfrac{\prod_{j}[\ce{P}_j]_\text{eq}^{p_j}}{\prod_i[{\ce{R}_{i}}]_\text{eq}^{r_i}} \tag{1}\\
\text{if }K_c=1 \text{ then } \dfrac{\prod_{j}[\ce{P}_j]_\text{eq}^{p_j}}{\prod_i[{\ce{R}_{i}}]_\text{eq}^{r_i}} = 1
$$
There is a whole range of concentrations where this can happen.
Example
Consider a simple reaction:
$$
\begin{align}
\ce{$a$A} &\ce{->} \ce{$b$B}\\
K_c &= \dfrac{\ce{[B]}^b}{\ce{[A]}^a}\\
\text{if }K_c=1 &\text{ then } \ce{[A]}^a=\ce{[B]}^b
\end{align}
$$
I have made an interactive plot where you can play around with different values of $a$ and $b$:

Although concentrations need not be equal, there is special case. From equation (1):
$$
\text{if } [\ce{R}_i] = 1 \text{ and } [\ce{P}_j] = 1 \text{ for all }i,j\\
\text{then }K_c = 1
$$
This has been highlighted in the interactive plot as well, where all graphs pass through the point $(1,1)$.


text:
The isotopic effect in chemical kinetics
I conducted a chemical experiment (degree in chemical engineering) to understand the isotopic effect in chemical kinetics. In the experiment, I reacted isopropanol and D-isopropanol (D stands for deuterium) with chromic acid (H2CrO4).
In order to measure the rate of the reaction - we used a spectrophotometer, to measure the absorption of the Cr(+6) ion as a function of time. As expected - the absorption of the reaction solutions diminished over time, as more Cr(+6) reacted and turned into Cr(+3).
The graphs in the experiment are attached here:

Now, I need to somehow find the connection between the rate equations (written in the pictures) and the Arrhenius equation.
If we write the Arrhenius equation for the reactions (with isopropanol and D-isopropanol) we get:
K(iso) = A*exp(-Ea_iso/RT)
K(D-iso) = A*exp(-Ea_Diso/RT)
thus:
K(iso) / K(D-iso) = exp(-dEa/RT)
where:
dEa = Ea_iso - Ea_Diso
I need to find the ratio K(iso) / K(D-iso), given the rate equations in the graphs above. How can I do that?
Thanks in advance

There would be A(iso) and A(D-iso), the latter lower. // Do you consider isotope exchange before oxidation? like IPrOD + H+ <=> IPrOH + D+?

I'm not sure I understand your question. Can you explain it again?

Do you ask whether the coefficient A is different for the reactions? In the analysis we assume that A is equal for both reactions.

The first part is the statement, not a question. IMHO, the isotopic effect on frequency factor is more significant than on activation energy.

You have the two rate constants as shown in your data so you can calculate what you want. Normally to get at activation energies  measurements at different temperature are made.  I'm assuming you use D8 propanol.

To Include or not to Include
The constants in your fitted equation make your kinetics different than first order kinetics. Which is why, if we consider the constatnts, this problem cannot be solved.
Explanation: Because the reactions are following different order kinetics, the data provided is not enough to determine the activation energies of the two reactions. More data needs to be retrieved to determine the activation energies.
If we represent $Ab$ as absorption, then:
$$
\ln{Ab_\text{iso}} = \ln{(0.5682\times \mathrm{e}^{-0.04958\times t}+0.007084)}\\
\ln{Ab_\text{D-iso}} = \ln{(0.6474\times \mathrm{e}^{-0.02649\times t}-0.005161)}\\
$$

Now, $Ab_\text{D-iso}$ seems to be following first order kinetics, because the graph is linear. However, $Ab_\text{iso}$ doesn't seem to follow first order kinetics. This makes our analysis somewhat trickier.
However, if you do want to include the constants, I have exactly outlined the steps needed to solve the question, which is attached at the end.

Not Including the Constants
If we consider the constants as an error term, your equations become:
$$
Ab_\text{iso} = 0.5682\times \mathrm{e}^{-0.04958\times t}\\
Ab_\text{D-iso} = 0.6474\times \mathrm{e}^{-0.02649\times t}\\
$$
These equations follow first order kinetics exactly:
$$
\text{first order rate law}\\
\ce{[A]} = \ce{[A]}_\circ\mathrm{e}^{-kt}
$$
Since $\ce{[A]} \propto Ab$:
$$
Ab = Ab_\circ\mathrm{e}^{-kT}
$$
By comparing the equations, we obtain:
$$
Ab_\text{iso} = (Ab_\text{iso})_\circ\mathrm{e}^{-k_\text{iso}t}\\
k_\text{iso} = 0.04958 \\
Ab_\text{D-iso} = (Ab_\text{D-iso})_\circ\mathrm{e}^{-k_\text{D-iso}t}\\
k_\text{D-iso} = 0.02649 \\
$$
Using the same method outlined by you, at room temperature:
$$
\ln{\dfrac{k_\text{D-iso}}{k_\text{iso}}}\approx -0.62682\\
\boxed{\Delta E_\mathrm{a} \approx \pu{1.527 kJ mol^{-1}}}
$$

Including the Constants
Let us assume the order of reaction with D-isopropanol ($\ce{R1}$) is one and that with isopropanol $\ce{R_2}$ is $n$.
$$
\ce{R1 -> P1}; \text{first order}\\
\ce{R2 -> P2}; n^\text{th}\text{ order}
$$
The rate equations are given by:
$$
\ce{[R1]} = \ce{[R1]}_\circ \mathrm{e}^{-k_1t}\\
\ce{[R2]} = \dfrac{1}{\left( k_2t - \dfrac{1}{\ce{[R2]}_\circ^{n+1}}\right)^{n+1}}
$$
Assuming $Ab_{\ce{[X]}} = c\ce{[X]}$
$$
Ab_\text{D-iso} = (Ab_\text{D-iso})_\circ \mathrm{e}^{-k_1t}\\
cAb_\text{iso} = \dfrac{1}{\left( k_2t - \dfrac{1}{c^{n+1}(Ab_\text{iso})_\circ^{n+1}}\right)^{n+1}}
$$
We have arrived at two equations with six unknown parameters: $(Ab_\text{D-iso})_\circ$, $(Ab_\text{iso})_\circ$, $k_1$, $k_2$ ,$c$, and $n$. This is unsolvable.
You will need to make a few adjustments and require more data:
Steps to Determine Activation Energies

Ensure initial concentrations of all reactants are equal and known, reducing two unknown parameters. These can also be approximated as the initial values of your data:

$$
Ab_\text{D-iso} \approx 0.642 \mathrm{e}^{-k_1t}\\
cAb_\text{iso} \approx \dfrac{1}{\left( k_2t - \dfrac{1}{c^{n+1}0.575^{n+1}}\right)^{n+1}}
$$
We still need to reduce two more unknown parameters, which are $n$ and $c$.

Find the value of $c$ by preparing samples of different concentration and measuring the absorbance, reducing one unknown parameter.
Determine the value of $n$ using the differential rate law, reducing one unknown parameter.

That should reduce the variables to two and the equations will be solvable for $k_1$ and $k_2$. Then, you can use the Arrhenius approximation to calculate activation energies.

This is nonsense. The equations that you are treating as his data are best fit first-order curves, so they are by definition first-order. The reason you get non-linearity is because you're including the non-zero y-intercept from the fitted curve.  And the $k$ values that you already have from his non-linear least squares fit are likely to be much more accurate than those that you extracted from a linear fit to the transformed points.

@Andrew the only information was the fitted equation, I had to assume it to be correct, and from the given data, there is no way of knowing if the reaction is following first order perfectly. If it was so perfect the logarithm graphs would be perfectly linear. What if the fitted equation is actually correct and kinetics do change under the isotope effect, as seen in the logarithm graph. Nonetheless, I did answer the question assuming first order kinetics as well, what more can I do here? Anyways, I hardly expect you to reply, so this is probably the end of our 'discussion.'

I'm not sure why you do not expect a reply, as I have replied to your previous questions in other answers. To your question 'What if the fitted equation is actually correct and kinetics do change under the isotope effect, as seen in the logarithm graph." If the equation is correct, the reaction is first-order. That's the point. The equation is a fit that assumes first order kinetics. If you removed the y-intercept which your linearization did not account for, you would necessarily have a perfect straight line, since you are "fitting" a first order curve to a first-order equation.

@Andrew sorry about that, I have been receiving comments recently and when I try to discuss, I hear no reply. Thing is, the equation for first order kinetics does not have a constant after the exponential term, do we assume that the fitting is incorrect? If I had the original data, I would have checked for it, but I only had the equation, so I decided to work with it. I did reverse-engineered the data, differential analysis is usually more accurate than integral analysis, and fitting integral data adds another source of error... so I understand the constant term may just be that.

@Andrew I wonder if it is okay to call it the $y$-intercept here, since it is the value when the graph approaches an asymptote as $t \to \infty$.

yes, I'm using "y-intercept" sloppily here, since "asymptotic as t goes to infinity" is wordier. A non-zero asymptote can arise experimentally in many ways, for example if the instrument is not zeroed properly if the product has a small absorbance at the wavelength of interest. It is up to the experimenter to determine if it is better to fit with a constant or without. If you want to use a linear transformation that does not account for that constant, you can simply remove it, but keeping it and not accounting for it in your transformation is incorrect.

Furthermore, I do not understand the point of transforming his first-order fit equation in order to extract rate constants that are already available in his equation. Since you are working with the equation rather than the raw data, there is no possibility that you will somehow get more accurate values. And more generally fitting first-order data to a linear equation is less accurate than fitting directly to a curve (though it is possible that the original fit was done via linearization.)

I understand. The constant could be an error, and it could arise from various sources. Instead of reverse engineering the data, I should've neglected the constants in the original equation, and I will make that correction; although it is probably a good thing that we have analysis with and without neglecting the constants. Probably should be OP's call whether to ignore the constants or not.

"Probably should be OP's call whether to ignore the constants or not." Right. And they chose to include them. So we don't "have analysis with. . .neglecting the constants.", because that can't be done without the original data. Your "analysis" is just incorrect math.

Let us continue this discussion in chat.

Best practice might arguably be to fit the two pieces of data simultaneously, allowing for the two different substrates to have different rate constants and different pre-exponential factors, but forcing them to have the same offset i.e. y-intercept.  If the y-intercept is to account for experimental imperfections, presumably many of these imperfections would be exactly the same in each run.


text:
What is the pH of acetone in water?
Reaction of Acetone with Water
Acetone in water can react in three ways:

The different reactions change the concentrations of $\ce{H+}$ and, consequently the pH, differently. My question is what is the pH of acetone in water? Is it greater than (basic), less than (acidic), or equal to (neutral) seven.
Context
Labeled Oxygen Exchange
In case of labeled water, exchange does occur. Since I couldn't find a source for this, I have outlined three pathways, that I could think of, for the same.
Neutral Pathways for Exchange

Since the pathways contains either both acidic and basic or only neutral intermediates, acetone in water should be, more or less, neutral ($\mathrm{pH} \approx 7$).
Basic Pathway for Exchange
Exchange can also occur via another pathway.

Only basic intermediates are involved in this pathway suggesting acetone in water should be slightly basic ($\mathrm{pH} > 7$).
Acidic, Basic, or Neutral?
Since, in the three pathways, I could come up with were either neutral or basic solutions, I am thinking acetone in water should be either neutral ($\mathrm{pH} = 7$) or basic ($\mathrm{pH} > 7$). However, going back to the first figure, I cannot discard the possibility of an acidic ($\mathrm{pH} < 7$) solution.

Is water in water "Acidic, Basic, or Neutral?" Same thing. You should reword it to ask about specific equilibrium constants - hydration, pKa of diol and of protonated aceton, instead of something as vague as "neutral".

@Mithoron water is known to have equal concentrations of $\ce{H+}$ and $\ce{HO-}$ which is why we call it neutral and, having a pH of 7 (at Room Temp.), we have decided to make that the standard for neutrality. Since water can react with both acidic and basic moieties, we call it amphoteric. So, water is an amphoteric neutral moiety. It is very common for ketones to show nucleophilic addition at $\alpha$-carbon, and electrophilic addition at $\ce{O}$, I am asking which reaction is favored, and if we measure the pH of a solution of acetone and water, would it be greater or less than seven.

"we have decided to make that the standard for neutrality", "we call it amphoteric" who is "we"? Water is just both acidic and basic, like virtually everything. If you want to edit according to my pointers than rethink and edit it. If not than my downvote stays. That would be all.

Adding a cosolvent changes equilibrium constants so a substance that is neither acid nor base still can change the pH of a solution: https://pubs.rsc.org/en/content/articlepdf/1994/p2/p29940000467?casa_token=4PTAh4KTeNcAAAAA:uTD5mDGCnWHqwXtBGPdD4CsqJi7Hci95R0CIEWwN8z1CXqD54bKxfe6z5BC103Q2jCzgslyuNPeDUe4

The problem stated in the question can be reduced to a problem of labeling something as acidic, basic or neutral. All human devised labels break down under one condition or the other. This is why there are plenty of definitions of acids and bases.
Concurrently, then people assume that acid means the pH is < 7 and neutral means pH=7, and base means pH>7 in water. This label also breaks down. As mentioned in a recent question, atoms, molecules, solutions do not care what humans call them or how people label them hydrogen isotopes and Bronsted-Lowry , they will behave the way they have been designed to. Recall the ambiguous issue of labelling chemical vs. physical changes.

Now, if we were to dissect your thesis, conventional pH is strictly defined for aqueous solutions of substances which partially or completely ionize in water and maintain this equilibrium. Acetone, certainly does not ionize in water in a traditional electrolyte sense (no molecular rearrangement). So the question of acetone being acidic, neutral or basic in water does not arise. Alternatively, one better refine the thought process to asking if I add a small amount of acetone or any other water miscible ketone (without another functional group) to water, will the aqueous pH change if we measure it with a pH probe (glass electrode)? Even then, electrochemical purists will raise their eyebrows because acetone is not a conventional electrolyte.

Next issue with this thought process is what percentage of acetone are we talking about 1% mole ratio, 60% mole ratio acetone:water? In that case, the former solution is aqueous, and the latter the solution is more organic. The experimental measurement of pH and its interpretation becomes murky in organic solvents.

If you are solely thinking about isotope exchange, in that case, the total mass balance of free protons will not change, and the bulk solution pH of acetone (again measured with a glass electrode) should remain the same. In short, if you add a small amount of acetone in water, the pH will not change measurably.
If you are more interested, search how the pKa of acetone or any other ketone is measured in water. The experiment would certainly be interesting and worth knowing.

One can take acetone with pKa about 19 as any other very weak acid. The fact it is solvent may not be relevant, we consider diluted solutions as for any other substances where properties of the solvent are assumed not to be altered.

pKa of protonated acetone is reportedly -7.5 and I remember ketone protonization in context of superacids.

I am actually curious how do people experimentally determine pKa of extremely weak acids or bases let us say acetone with a pKa of 19 in water! This must be either thermodynamic measurements or spectroscopic calculations. Yes, indeed superacids can protonate almost everything!

The exchange with H2O[18] is either acid or base catalyzed because of the amphoteric nature of the carbonyl group. You will have to research the rates in pure water-acetone mixes. Your diagrams seem to be confusing you as much as they are confusing me.
Acetone in water is a less polar solvent. Two simple reasons [to me at least] First: the molecule exists in the keto not the enol form, no very polar and acidic OH bond; and Second: the addition of acetone lowers the water activity in the solution. This means any mechanism involving ionic intermediates such as ketal or enol formation should be slower in pure water-acetone solution.
In water:  H2O = H+ + OH-. Keq[H2O] = [H+][OH-].   If Keq remains constant [this requires research], as the water concentration is lowered the H3O+ concentration will be lower. The response of the glass electrode could be tested by addition of measured amounts of strong acid or base in the solvent mixture.
My prediction: pH will increase with additions of acetone. Mettlar-Toledo has a white paper that tries to explain the use of glass pH electrodes in homogeneous organic-water solutions. It is simplistic and empirical. They assign the effects to changes in the ion activities not the change in water activity and discuss many technical caveats. It is difficult to make accurate absolute measurements in such solutions.


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Why does Alcohol-Dehydrogenase prefer ethanol over methanol?
So I've had this experiment at university, where we tried using (yeast) ADH on different alcohols and measuring which one gets turned over the fastest. We also learned that ADH prefers shorter alcohols over longer ones. Now, that would mean the methanol to methanal reaction should've been the fastest, but it didn't react whatsoever. Our lab supervisor reassured me that that's okay, but forgot to tell me the reason why it's the case.

Not an answer, but I want to comment that is extremely common in chemistry that the first/lightest/shortest of a line of homologues falls out. Fluorine/halogenes, lithium/alkali metals, formaldehyde/aldehydes/ketones, ...

Not a chemical explanation, but evolutionary observation: alcohols in organisms tend to be two carbon (i.e., ethanol) or multiples of two, so enzymes that worked efficiently with ethanol helped organisms survive. Admittedly, not in any way an answer as to how the enzyme works.

The alcohol dehydrogenases are dimeric proteins, with catalytically active $\ce{Zn^2+ }$ ions. These catalytic $\ce{Zn^2+ }$ ions have distorted tetrahedral geometry, coordinated to one histidine (His-67) and two cysteine residues (Cys-46 and Cys-174) of the enzyme (Ref.1). The catalytic site of alcohol dehydrogenase is shown in left hand site of following image (part $\bf{a}$) and its mechanism of oxidation of primary ord secondary alcohols to corresponding aldehyde or ketones, respectively are shown in part $\bf{b}$ (Ref.1 & 2):

Accordingly, the role of zinc in the dehydrogenation reaction is to promote deprotonation of the alcohol, thereby enhancing hydride transfer from the zinc alkoxide intermediate. Keep in mind that alcohol dehydrogenases are exquisitely stereo specific; and by binding their substrate via a three-point attachment site, which can enables the enzyme to distinguish between prochiral methylene protons in ethanol (Ref. 1 & 3):

All three hydrogens are equal thus enzyme cannot distinguish them for specific sites for pro-S- and/or methyl (of ethanol) or R-group of other alcohols. It is clear that  the shapes of the binding sites are distinguish  for pro-S- and methyl (of ethanol) or R-group of other alcohols. The binding site for methyl and R-group of other alcohols may be stereoselective for methyl group, but large enough to bind other R-groups as well according to values given in Table 7 of your reference. Yet, it may be too large to bind the hydrogen of methanol because the hydrogen would be too small for necessary tight binding.
References:

Robert R. Crichton, "Chapter 12: Zinc: Lewis Acid and Gene Regulator," In Biological Inorganic Chemistry: A New Introduction to Molecular Structure and Function; Second Edition, Elsevier B.V.: Amsterdam, The Netherlands, 2012, pp. 229-246 (ISBN: 978-0-444-53782-9).
Gerard Parkin, "Synthetic Analogues Relevant to the Structure and Function of Zinc Enzymes," Chem. Rev. 2004, 104(2), 699–768 (DOI: https://doi.org/10.1021/cr0206263).
Donald Voet and Judith G. Voet, In Biochemistry; 3rd edition, John Wiley & Sons: Chichester, NY, 2004, p. 1360 (ISBN-10: 978-0-471-19350-0).


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why does -CHO hold higher priority over -CH2OH according to CIP rule
i am unable to find proper resources to learn CIP rule at a high school level withut delving deeper into terms i cant understand and hence the basic confusion.if i am correct one has to first compare the atomic wieght of groups down the chain until a tie is broken
so in $\ce{-CHO}$ and $\ce{-CH2OH}$ , first we see that both compound have $\ce{C}$ attached
then $\ce{-CHO}$ has $\ce{=O}$ while $\ce{-CH2OH}$ has $\ce{-O}$.However , acc to CIP rule i only need to chech the mass number and not the double bond until i need to break a tie.so both compounds are attached to $\ce{O}$.
then $\ce{-CHO}$ is attached to nothing while $\ce{-CH2OH}$ is attached to $\ce{H}$,hence as they are of dissimilar lengths $\ce{-CH2OH}$ should have the higher priority.
what is wrong with my approach

You can't ignore the double bond.

@IvanNeretin But isn't the double bond used only in case of Ties.that is when all the atoms in both chains have same atomic number?

Here is a resource.  Look at the 2013 rulles for an introduction. In your case, (O,O,H) > (O,H,H).

Karan: Structure 1 not only has the issue that you have addressed, the priority of -CHO vs. -$\ce{CH2OH}$, but also deals with double bond geometry and the configuration at $\ce{C2}$. The Cahn-Ingold-Prelog (CIP) system handles multiple bonds by creating duplicate atoms shown in red in structure 2. For the C=C bond, $\ce{C3}$ is attached to duplicate atom $\ce{C4}$ while $\ce{C4}$ is attached to duplicate atom $\ce{C3}$. The same logic is applied to the C=O bond. All duplicate atoms have the same atomic number (at. no.) as their non-duplicate (original) atoms and each one is bonded to three atoms of at. no. zero. All atoms, even oxygen, are treated as though they were tetravalent.
The digraph is a two-dimensional (circle) representation of the three-dimensional (sphere) aldehyde 1 as one proceeds outward from $\ce{C2}$ to determine the priorities of the surrounding groups. The spheres are numbered in green. In sphere 2 hydrogen has the lowest priority and $\ce{C1}$, $\ce{C3}$ and $\ce{C6}$ are tied. Proceeding to sphere 3, we have $\ce{C1}${O,O,H}, $\ce{C3}${4,4,H} and $\ce{C6}${O,H,H}. Within the braces the atoms are arranged in descending order of at. no. and a one-to-one comparison is made. Clearly, the order of priorities is $\ce{C1}$>$\ce{C6}$>$\ce{C3}$>H.
Just as $\ce{C6}$ is in front of the five-carbon chain, the hydrogen at $\ce{C2}$ is behind the chain. To assign a configuration, point the thumb of your right hand down the $\ce{C2}$-H bond. Now your fingers will point in from $\ce{C1}$ to $\ce{C6}$ to $\ce{C3}$. Accordingly, the configuration at $\ce{C2}$ is "R" (Rectus).
At $\ce{C3}$ and $\ce{C4}$, C>H. The "E" (Entgegen) configuration applies.


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Do transition metals only carry d-electrons as valence electrons until group 11?
as I was revisiting my lecture notes in inorganic chemistry, I noticed that we treated the transition metals as elements where the energetic order of 4s ad 3d orbitals is reversed.
For example starting with scandium, it's valence electron configuration would be 3d3, manganese's would be 3d7 , with the 4s orbitals only starting to fill with copper, 3d10 4s1. Basically, the d-orbitals would be filled completely before an s-electron would be filled.
It wouldn't make a difference which orbitals (4s or 3d) start to fill first, if we're only looking at transition metals with oxidation states of +II and higher, but I revisited a problem in an old inorganic chemistry II exam. The d-electron configuration of iridium(I) was to be determined (in Vaska's complex).
The correct answer given by the university was 5d8. However, iridium (0) is 5d7 6s2, so iridium(I) should be 5d7 6s1, or is this incorrect?

The spectrum of singly-ionized iridium shows a mixture of electron configurations, see this paper (or just the abstract if you don't have access).

The answer is most probably in the view that when there are 8 electrons in $5d$ the exchange energy increases and hence the stability increases as compared to the $5d^7 6s^1$ configuration. The extra electron from the $6s$ moves to $5d$ to achieve more stability by increasing the exchange energy.


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Why SCF-MO methods fail when treating bond-breaking processes?
I'm currently studying the EVB(Empirical Valence Bond) approach. While reading on this article (https://pubs.acs.org/doi/10.1021/ja00224a011), the author mentioned about the usage of SCF-MO formalism on describing bond-breaking/formation processes.

We could use an SCF-MO formalism to determine the energy of the system as a function of the coordinates of the X, Y, and C atoms over the course of the substitution reaction. However, there are bonds broken in the reaction, and it is well-known that the SCF-MO methods fail when treating bond-breaking processes.[38] Thus, we must use a formalism that will effectively treat bond breaking. Any method allowing for more than one Slater determinant (configuration) in the wave function will work. Some possible choices are the Cl, MCSCF, and VB methods, with all of these approaches being exactly equivalent in the limit of a complete basis set of configurations.

I first attempted to track the reference 38 to understand why SCF-MO methods fail when treating bond-breaking processes but it was a QC textbook written by Slater, and published a very past day (Slater, J. C. Quantum Theory of Molecules and Solids-, McGraw- Hill: New York, 1963; Vol. 1), which was inaccessible to me. Can I ask the reason why SCF-MO methods fail when treating bond-breaking processes? It seems like those are related to the number of Slater determinant used in computation but it would be highly appreciated if anyone could add some detailed explanation on here. Suggesting any reference that pointed out the failure of using SCF-MO methods on bond-breaking processes would be highly informative to me as well!


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Lewis Structure of the Guanidinium Ion
In a problem, I was asked to find the Lewis structure of the guanidinium ion $\ce{C(NH2)3}^{+1}$. I followed the following steps which led me to an incorrect structure and I was hoping someone could help me find my mistake.

The total number of valence electrons is given by $4+3(5+2(1)) +1 = 26.$
The total number of electrons with each atom having an octet is $8+3(8+2(2)) = 44.$
The number of bonding electrons is thus $44-26 = 18.$
I drew the following skeletal structure which includes $9$ bonds (corresponding to $18$ bonding electrons):

I have $26 - 18 = 8$ non-bonding electrons remaining, which I assigned as lone pairs to the three nitrogens and carbon:

However, this structure is clearly incorrect because carbon rarely forms three bonds. The correct structure (including resonance) is

There's no lone pair on carbon...

related https://chemistry.stackexchange.com/questions/13593/protonation-of-guanidine

Hint: Start with guanidine $\ce{HN=C(NH2)2}$

You missed a sign. When you incorporate the charge into your electron count, you subtract positive charges and add negative charges because an electron is negatively charged.
$4+[3×(5+2)]\color{red}{+}1=26$
$4+[3×(5+2)]\color{blue}{-}1=24$ ✔
With the correct electron count you should now be able to render a double bond from the carbon to one of the nitrogen atoms. In reality the pi bond this represents is delocalized through the whole ion, providing an extra stabilization that makes guanidine about as strongly basic as sodium hydroxide.