Calculation of free energy from potential and kinetic energy I am having trouble understanding how (and even whether it's possible) to get free energy from kinetic and potential energy.
I have run molecular dynamics simulations (NPT ensemble, CP2K software), but this is a basic thermodynamics question.  There's a results file called "energy."  The file contains energy-related output: temperature, kinetic and potential energy, and a "conserved" quantity.
(The "conserved quantity" is the sum of the kinetic and potential energies and also the thermostat energies).
From courses a long time ago and reading I have done online, I know that
kinetic energy + potential energy = total energy

I think "total energy" would be the internal energy.  It would be great if someone could please confirm/debunk that--I don't know if the situation is distinct for distinct ensembles, for example.  Whatever the case, I don't know how it relates to the free energy.  I would like to know if I have enough information to calculate the free energy.
If anyone could please point me in the right direction, I would appreciate it.  I can't find anything online and maybe that's because I don't even know where to start looking.

The sum of the potential and kinetic energies is the internal energy, E (or U).
However, you don't necessarily need to know the entropy to calculate the free energy.  If you perform an equilibrium molecular dynamics simulation, you can derive the free energy from the equilibrium constant.
See, for instance:
https://onlinelibrary.wiley.com/doi/full/10.1002/jcc.21776
If you are modeling a constant T and V system, where there is no work done, the equilibrium constant would give you the Helmholtz free energy:
$$\Delta A = - R T ln K^{'}_{eq}$$
where A = E - TS
If you are modeling a constant T and p system, where you have no non-pV-work, the equilibrium constant would give you the Gibbs free energy:
$$\Delta G = - R T ln K^{''}_{eq}$$
where G = H - TS = E + pV - TS
The reason for this is that, at constant T and V, with no work term, minimizing A_sys corresponds to a maximization of S_univ, and thus determines the equilibrium condition. By contrast, at constant T and p, with pV-work only, minimizing G_sys corresponds to a maximization of S_univ, and thus determines the equilibrium condition.
I've marked the equilibrium constants as prime and double-prime to indicate that these would be numerically different.


Why does ammonia react with water to produce hydroxide? My question relates to dissolving $\ce{NH3}$ in water and how it produces $\ce{OH-}$ for the reaction $\ce{NH3 + H2O -> NH4+ + OH-}$?

I know that the $\mathrm{p}K_b$ of $\ce{NH3}$ is $4.8$ and $\mathrm{p}K_a$ of $\ce{NH4+}$ is $9.2$ so $\ce{NH3}$ is a stronger base than $\ce{NH4+}$ is an acid. But why the free $\ce{OH-}$? I am now taking organic chemistry and all I am reading is how strong $\ce{OH-}$ is and how it can attack anything slightly positive. So why is the above reaction favorable towards creating $\ce{OH-}$?

Actually any acid-base reaction that is creating $\ce{OH-}$ as a free species?

The shortest answer is "because the change in free energy is negative," but maybe you want to know how the mechanism works?

Ammonia is a Lewis base, and hydroxide is a stronger Lewis base. However, both are stronger Lewis bases than water is. Pure water always has some amount of "free" $\ce{H+}$ (it's actually hydronium ion) and $\ce{OH-}$ due to self-ionization ($pK_W \approx 14$):

$$
K_W = \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]^2}
$$

When you add ammonia to the system, some of those hydronium ions (which are very strong acids) react with the ammonia to form ammonium. Since the concentration of hydronium decreases, the concentration of hydroxide must increase to maintain equilibrium. Those hydroxide ions are stronger bases than ammmonia is, and so they will "win" the tug-of-war over protons. However, in the meantime, some other hydronium reacts with some other ammonia, and the net result is a dynamic equilibrium in which there is more hydroxide then there would be in pure water.

The same mechanism is responsible for any other acid/base reaction where a base accepts protons from water and leaves hydroxide as a free species, but the net concentration of hydroxide will be different for different strengths of bases.


True muonium energy levels In a true muniom(muon-antimuon atom) do the energy levels match with that of the hydrogen's?

The core is lighter for about 100 times but the muon is 100 times heavier than the electron so it would make sense the energy levels of a true muonium and a hydrogen atom to match?

Or is this wrong?

The theory for a muon-antimoun "atom" is essentially the same as any hydrogen-like atom, just with masses changed.  See this Wikipedia page for the relevant formulae for relativistic (Dirac) and non-relativistic treatments.

The non-relativistic energy levels are given by :

$$E_n =  \frac { \mu c^2 \alpha^2}{2n^2}$$

where $alpha$ is the fine structure constant and $\mu$ is the reduced mass, in this case because the masses are equal, it's $\mu=\frac m 2$ where $m$ is the muon mass.

Note that the muon mass is about $206.8$ times that of the electron and the reduced mass is then $103.4$ times than of the electron making the energy levels for the non-relativistic state about $103.4$ times that of the Hydrogen energy levels.

As the Wikipedia page explain (in more detail that you asked for), a more detailed relativistic calculation using the Dirac equation includes corrections for the spin-orbit interaction.

  The core is lighter for about 100 times but the muon is 100 times heavier than the electron

This is wrong.

Both the muon and antimoun have the same mass, hence the effect of reduced mass being a factor of one-half.

The reduced mass of two particles is given by :

$$\mu = \frac {m_1m_2}{m_1+m_2}$$

Strictly speaking there is no "core" as both the muon and antimoun are of the same mass whereas in Hydrogen the electron is of much smaller mass than the proton.


Clarifications about acid-base theory Considering the Arrhenius theory of acids and bases:

Is every existing compound either an acid or a base?
Why is a compound such as $ \ce {NH_3}$ considered a base, if it doesn't even have an oxygen atom and a base is supposed to release $ \ce {OH-}$ in water?
If $X$ is a base according to Arrhenius theory, will $X$ still be a base according to the other two theories (and the same if $X$ were an acid)?

According to the Brønsted-Lowry theory:

There are no absolute acid and bases, but only acid-base reaction in which one compound acts as a base and the other as an acid. So, is every reaction of two compounds an acid-base one?
Except for water, are there any other substances which in some reactions act as bases while in others as acids?

A couple of thoughts:

There are mainly 3 different definitions of acids and bases:

The Arrhenius formulation: Acid = proton donor in water,  Based = hydroxide ion donor in water
The Brønsted-Lowry formulation: Acid = potential proton donor, Base = potential proton acceptor
The Lewis formulation: Acid = Electron pair acceptor, Base = Electron pair donor

The Lewis definition is the most general, followed by the BL and finally by the Arrhenius definition.

You are right, for example HCl is considered a strong acid in water, but it is very weakly acidic in glacial acetic acid. If you follow the Lewis definition a large number of reactions will be considered acid base reactions. An exception are free-radical based reactions

There are many compounds that are amphoteric, but it really also depends on the definition you use.


What is the definition of porous materials? Is there a consensus on scientific definition of porous materials? There are many definitions out there which do not seem accurate enough:

from wikipedia (accessed July 2017):
A porous medium or a porous material is a material containing pores (voids).

-problem : Then almost all materials are porous, since there is some sort of void in everything.

from Merriam Webster (website accessed July 2017):
Containing pores, permeable to fluids, capable of being penetrated.

-problem : A piece of tubing would be porous since fluid passes through it.

At what scale do we consider the pores to qualify a material as porous? How much does the permeability matter in a formal definition? Is there a convention on defining pores and porous materials that is agreed upon by scientific organizations like IUPAC or ASTM etc?

I have used the following definition in my presentations:

"Porosity is where a material has accessible voids which are permeable to liquids or gases".

Another definition:

"a  solid with pores, i.e. cavities, channels or interstices, which are deeper than they are wide."

IUPAC also defines different types of porosity based on the diameters of the pores:

Micropores: d < 2 nm
Mesopores: 2 < d < 50 nm
Macropores: d > 50 nm

M. Thommes, K. Kaneko, A. V. Neimark, J. P. Olivier, F. Rodriguez-Reinoso, J. Rouquerol, K. S. W. Sing, Physisorption of gases, with special reference to the evaluation of surface area and pore size distribution (IUPAC Technical Report), Pure Appl. Chem. 2015, 87, 1051.


Diazotisation of aniline derivatives I want to know if N-methyl aniline $(\ce{PhNHCH3})$ can undergo diazotisation under suitable conditions and in the presence of $\ce{NaNO2}/\ce{HCl}.$

The mechanism involves loss of two $\ce{H+}$ during the formation of the diazo compound, but if we have one $\ce{CH3}$ and one $\ce{H}$ instead of two $\ce{H}$ (as we have for aniline), will the $\ce{CH3+}$ act as a good leaving group? Is diazotisation of $\ce{PhNHCH3}$ feasible?

Secondary amines react with nitrous acid to give N-nitroso compounds reference here

N-Methylaniline will give N-Methyl-N-nitrosoaniline as shown below


Reaction between anisole and concentrated HI in excess Question:

My Attempt:  In most of the ethers reaction of HI in excess yields alkane iodide products. But in case of anisole two scenarios are possible, I believe:

The HI breaks bond between methyl and oxygen leading to formation of phenol and methyliodide.
The phenol formed further reacts to form phenyliodide.

But would HI be strong enough to break partial double bond in phenol?

The correct answer here should be C. $\ce{I-}$ is an excellent nucleophile, it would attack the $\ce{C-O}$ bond in the methly group because phenoxide is a good leaving group as it is resonance stabilized and quite polarizable. So, the products are phenol and methyliodide. As for the excess of conc. $\ce{HI}$, it is just there so that some other nucleophile like $\ce{OH-}$ doesn't grab the limelight. I don't think it is energetically favourable to remove hydroxide from phenol to make it iodobenzene.


Active transport of Ca2+ ions against chemical and electrostatic gradients, assisted by ATP hydrolisis I have a homework question that I am not able to solve. No solution is available, so I was hoping someone here could help. The question is this:

  Calculate the maximal concentration ratio that can be obtained for
  transport of Ca$^{2+}$ ions if two ions are transported per ATP hydrolyzed
  at 310 K, and they are transported to a potential that is 70 mV higher
  at the receiving side. Assume that the $\Delta G$ for hydrolysis of ATP = -45 > kJ\mol.

Relevant formulas:

\begin{align}
\Delta G_{\text{electrostatic gradient}} &= ZF\Delta \Psi\\
\Delta G_{\text{chemical gradient}} &= RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}}\\
\Delta G_{\text{hydrolysis of ATP}} &= -45000~\mathrm{J/mol}\\
\end{align}

Moving two positive $\ce{Ca^2+}$ ions against both a concentration gradient and an electrostatic gradient implies that $G_{\text{chemical gradient}}$ and $\Delta G_{\text{electrostatic gradient}} > 1$. The negative value for $\Delta G_{\text{hydrolysis of ATP}}$ means that hydrolysis of ATP is a spontaneous process, or a "work-yielding" process. The energy released from hydrolysing ATP is essentially what drives the transport process. Since we are transporting "4 charges" per ATP molecule, $\Delta G_{\text{chemical gradient}}$ is multiplied by 2. Also (I presume) the concentration of $\ce{Ca^2+}$ ions is larger on the inside of the cell than on the outside, implying that the concentration ratio is larger than 1.

The combined expression for $\Delta G$ for the unassisted transport is

$$\Delta G = RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}} + 2ZF\Delta \Psi$$

Rearranging this yields

$$\frac{C_\mathrm{in}}{C_\mathrm{out}} = \exp\left\{\frac{\Delta G - 2ZF\Delta\Psi}{RT}\right\}$$

I am not sure what to plug in for $\Delta G$, but it needs to account for the push from ATP. I believe this is done by summing the two free energies affecting the transport, ergo the sum of $\Delta G_{\text{electrostatic gradient}}$ and $\Delta G_{\text{hydrolysis of ATP}}$. However, this gives me a concentration ratio much smaller than 1.

Can anyone see where I am incorrect?

At equilibrium, the entire process (transport of $2~\ce{Ca^2+}$ ions and hydrolysis of $1~\text{ATP}$) involves no change in G.

\begin{align}
0 &=\Delta{G_\mathrm{total}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+}}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+},\text{ concentration gradient}}} + 2\Delta{G_{\ce{Ca^2+},\text{ electrostatic}}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}} + 2ZF\Delta \Psi\\
\end{align}

So basically I'm saying you made two mistakes, one a sign error and one a factor of 2 error.


Reactivate dried up ink of ballpoint pens According to this question, ballpoint pens start writing intermittently when the ink is drying up. In other words, the ink is becoming more viscous and does not flow any more.

Are there ways to make it flow again?

As ink typically consists of non-polar dyes dissolved in a non-polar liquid, I would recommend using a straightened paper clip and a drop of acetone to clear out the dried ink clogging up the ink cartridge. This works by agitating the dried ink, as well as by replenishing the ink's solvent. Just be careful to not make too much of a mess!

Another method I have seen work before is heating the end of the pen under a flame, which melts the ink clogging the tip. For this method, my advice would be to not hold the pen in one place for too long, lest you risk melting the plastic.


Marker dissipates on White board paint What causes white board markers to dissipate when writing on dry erase painted surfaces. It almost immediately start to pull together and seems as if it disappears

The issue at hand is competition between the intermolecular forces that pull the solvent together and the forces that cause the solvent to adhere to the surface. In this case, the solvent's interactions with itself are greater, causing the ink to bead. Since the amount of ink that is actually drawn out when writing is very small, the beads of ink are tiny and very hard to see.


can we use con HCl zn amalgam to reduce alcohol? all says that we can't use zn con HCl to reduce alcohol. can anyone explain this with the mechanism or with explanations

Here is the mechanism for reduction of aldehyde/ketone by clemmenson reduction.

Note that alcohol is formed as intermediate which further reduces. Considering that alcohol can be reduced by Zn/HCl


What is the product of the chemical reaction between phenol and ferric chloride? The chemical reaction between phenol and ferric chloride is a test for the presence of phenol. They react with each other to produce a violet complex. However, the reaction is given differently in different links:

$$\ce{3ArOH + FeCl3 → Fe(OAr)3 + 3HCl}$$

source 1 , source 2
$$\ce{6 ArOH + FeCl3 -> [Fe(OAr)6]^3- + 3H+ + 3HCl}$$
source

I guess the 2nd reaction is the appropriate one because it shows the formation of a complex, $\ce{[Fe(OAr)6]^3-}$. How could  $\ce{Fe(OAr)3}$ be considered a complex? It should act like a ionic salt like $\ce{Na^+C6H5O^-}$  or $\ce{NaOAc}$.

So, what is the correct product of the reaction; $\ce{[Fe(OAr)6]^3-}$ or $\ce{Fe(OAr)3}$?

After doing some looking around, and going back to check what I thought I knew about coordination complexes, I think I understand why you're seeing two equations. I first recommend reading the pages on Chemguide that talk about coordination chemistry, particularly the first half of this page on acidity, and the pages on reactions with hydroxide and ammonia. Both were helpful to me in understanding an investigation I'm doing on copper (II) complexes, but the same ideas apply to 3+ compounds.

The important concepts to keep in mind are that coordination compounds (in situations like this, at least) will either undergo acid-base reactions or ligand-exchange reactions. Acid-base reactions result in neutralization and a precipitate, ligand-exchanges cause a color change. Phenol is a relatively strong acid (stronger than alcohol at least), and so is ferric chloride. I had to look around but this page says at the very bottom that all oxygen-containing compounds act as bases in the presence of Lewis acids (like ferric chloride).

$\ce{3ArOH + FeCl3 -> Fe(OAr)3 + 3HCl}$ is the acid-base reaction, and you get a ferric phenolate salt as a precipitate.

$\ce{6ArOH + FeCl3 -> [Fe(OAr)6]^3- + 3H+ + 3HCl}$ is the ligand-exchange. It's not immediately obvious, but you can rewrite the two reactions a different way and it becomes clear.

$$\ce{3ArOH + [Fe(H2O)6]^3+ -> Fe(H2O)3(OAr)3 + 3HCl}$$

$$\ce{6ArOH + [Fe(H2O)6]^3+ -> [Fe(OAr)6]^3- + 6H2O + 3H+ + 3HCl}$$

Those three $\ce{H+}$ ions are from the excess phenol. The ligand-exchange occurs only when the phenol is in excess and more water molecules can be replaced, making the iron compound ionic and soluble again.

So depending on the context, you'll see the one with three or the one with six. Both are correct, but $\ce{[Fe(OC6H5)6]^3-}$ is the new complex and $\ce{Fe(OAr)3}$ is an intermediate product.

Hope that helps.


Which d-orbitals split according to CFT? According to CFT, the degenerate d-orbitals split into $eg$ and $t_2 g$ groups. Are these the outermost nd orbitals which are vacant(4-d in case of 4th period) or the penultimate filled (n-1)d orbitals?

If they are the (n-1)d orbitals, how can they be degenerate if they are not completely filled? If they are (vacant) nd orbitals, how do they split into $eg$ and $t_2 g$ groups if they have no electrons to be repelled by the incoming ligands (CFT is based purely on electrostatic forces)?

The $d$ orbitals that split are the $(n-1)$ $d$-orbitals because they are the relevant ones that do chemistry (because they have the electrons).

If the orbitals are degenerate, it doesn't matter whether they're filled or not, the energy level is the same for all the degenerate states. So it also doesn't matter if they're filled or not, the energy states will split up according to CFT.

The compound only "feels" this raising or lowering of energy depending on how many electrons are filled in into the different orbitals after splitting.

To maybe clarify some confusion: The existence of an orbital doesn't mean that the electron has to be in there, it is only a mathematical construct.


Formation of Calcite from Calcium carbonate So, in lab we are forming the different polymorphs of calcium carbonate. It required us to age the sample that was prepared for the formation of calcite for one week. This sample mixture was suspended in water. Whilst the other samples that were prepared for vaterite and aragonite were immediately washed/filtered.
Why do you need to age a calcite sample for a week?

It's my understanding that the production of the calcite takes a decent amount of time as there are 2 stages being undertaken for the crystallization of the amorphous calcium carbonate (ACC). The first of these is relatively fast whilst the second stage is the one that you want to be looking at as it is roughly 10 times slower. This is the stage which looks at a dissolution and precipitation mechanism.


Cation-anion radius ratio of manganese difluoride How to determine the cation-anion radius ratio for $\ce{MnF2}?$

I tried to calculate the $r_\mathrm{c}/r_\mathrm{a}$ ratio using the relationships:
$$2r_\mathrm{c} = c$$

and

$$(2r_\mathrm{c} + 4r_\mathrm{a})^2 = 2a^2,$$

but that gives $r_\mathrm{c}/r_\mathrm{a} > 1$ with the given $a$ and $c$ parameters, which is impossible. $\ce{MnF2}$ has a rutile structure, so the predicted ratio should be between $0.414$ and $0.732.$

You assumed that the cations touch along the $c$ direction.  They don't.  In most ionic compounds you have only anion to cation contact and, therefore, any attempt to determine ionic radii from lattice parameters gives only the sum of the two radii, not the individual values.  There are exceptions such as lithium iodide, in which the lattice parameters enable determination of an ionic radius for iodide, but then they carry no information about the lithium ion.

In principle, you could use the data on lithium iodide, where there is iodide-iodide contact, to get a radius for just the iodide ion.  Then you turn to manganese(II) iodide, where anions contact cations, and get the sum of the manganese and iodide radii.  Then subtract the iodide radius to get radius for manganese ion.  Use that plus the lattice information for manganese(II) fluoride to get the fluoride radius.  It all works perfectly -- if the radius of a given ion is the same in all its ionic compounds.  Maybe it is to sufficient accuracy, maybe it isn't.


How (in)flammable is "isopropyl alcohol" really? I have an electric device which frequently gets very oily (it's a vaporizer for herbs). It came with a few individually packaged small "tissues" drenched in "isopropyl alcohol". I've used up them all now to clean the insides.

After each cleaning session, I wait a few minutes before I put in new herbs and heat it up again (to 365-420 degrees F). Every time, it feels scary, because "isopropyl alcohol" is said to be "extremely" flammable. I feel like it's going to explode every time, but it luckily never does. Since they shipped those with the product, and didn't mention a word about this in the "manual", I assume it's safe.

But how can it be safe? Is it not hot enough? Does it evaporate so quickly as to never actually be left in there when I heat it up?

I also bought a separate, very expensive bottle of what I assume is "high-quality" isopropyl alcohol and poured it into a glass container with a firm plastic lid, which I keep in a wardrobe. Luckily, it doesn't seem to evaporate into nothing, like used to happen with other alcohol which I previously attempted to store in the same container. After a couple of weeks, half of it had disappeared! That doesn't happen with "isopropyl" alcohol, apparently, but why doesn't it?

As you can hear, I wonder both how (in)flammable it really is, and what makes it not evaporate inside the box, but does evaporate very quickly (apparently) when not restricted by a closed container.

The Flash Point of a liquid is the minimum temperature at which a liquid gives off vapor within a test vessel in sufficient concentration to form an ignitable mixture with air near the surface of the liquid. The flash point is therefore dependent on the boiling point and consequent vapor pressure of the liquid (Source of Vapor Pressure):

$$
\begin{array}{l|c c c}
\text{Flammable Liquid} & \text{Boiling Point at $\pu{1 atm}$} & \text{Flash Point} & \text{Vapor pressure at $\pu{25 ^\circ C}$} \\
\hline
\text{Diethyl ether} & \pu{35 ^\circ C} & \pu{-45 ^\circ C} & \pu{532.7 mmHg} \\
n\text{-Pentane} & \pu{36 ^\circ C} & \pu{-40 ^\circ C} & \pu{434.3 mmHg} \\
\text{Hexane} & \pu{69 ^\circ C} & \pu{-26 ^\circ C} & \pu{151.2 mmHg} \\
\text{Ethyl acetate} & \pu{77.1 ^\circ C} & \pu{-4 ^\circ C} & \pu{94.7 mmHg} \\
\text{Methanol} & \pu{65 ^\circ C} & \pu{11 ^\circ C} & \pu{126.89mmHg} \\
\text{Ethanol} & \pu{78 ^\circ C} & \pu{12 ^\circ C} & \pu{58.8 mmHg} \\\text{Isopropanol} & \pu{82.2 ^\circ C} & \pu{12 ^\circ C} & \pu{43.6 mmHg} \\
\text{Water} & \pu{100 ^\circ C} & \text{N/A} & \pu{23.7 mmHg} \\
\end{array}
$$

By definition, any liquid with a flash point less than $\pu{37.8 ^\circ C}$ ($\pu{100 ^\circ F}$) is considered to be a flammable liquid. Any liquid with a flash point between $\pu{100-200 ^\circ F}$ is considered combustible. In general, the relative hazard of a flammable liquid increases as the flash point decreases.

However, OP's isopropyl wipes are soaked with about 70% isopropanol and DI water (the best combination for better sanitation). For example, see these wipes on market. As a consequence, the flash point ($\approx \pu{18 ^\circ C}$) is increase on these wipes. Also, since vaporizer has good ventilation system (I assumed), these vapors never would get to saturation point in addition to combination of water and isopropanol vapors. That would reduce the flammability of vapors even with a spark (here are heating coils, I assume). That's probably why the user manual did not warn fire hazard.

OP's question about evaporation from an open bottle: As you can see in the given table, vapor pressure of ethanol and isopropanol is quite high in room temperature ($\approx \pu{25 ^\circ C}$) compared to that of water. If bottle is open, the vapor escapes without ever getting saturated like wipes in closed box. Thus, alcohol in the bottle keep evaporate and, as a result, volume in the bottle reduces with time. On the other hand, the closed space in the box of wipes eventually get saturated with alcohol/water vapors, and evaporation stops at that point.


Why do different H+ in polyprotic acids have different K? Why do different $\ce{H}$ atoms in a polyprotic acid have different values of $\pu{K_a}$?

All $\ce{H}$ atoms (which can show acidic nature) have similar properties within the molecule. It does not matter which one reacts. For example, in $\ce{H3PO4}$, all the $\ce{H}$ atoms can be dissociated from the molecule to act as acids. All of them are attached to an $\ce{O}$ atom, which is attached to the central $\ce{P}$ atom. If all the properties are same, then why is $\pu{K_a}$ different?

`

You are correct that all of the $\ce{H}$'s in $\ce{H3PO4}$ are equivalent.

However, once the first H is removed, the remaining two $\ce{H}$'s are no longer equivalent to the $\ce{H}$'s in the original $\ce{H3PO4}$. After the initial deprotination, the molecule gains an overall negative charge making it more difficult to remove the second $\ce{H}$.  Then when the second $\ce{H}$ is removed the molecule gains a second negative charge making it even more difficult to remove the final $\ce{H}$.  Thus, the $K_a$ values for the originally identical $\ce{H}$'s change each time the molecule is deprotinated.


Where is the probability of finding an electron in 1s orbital maximum? So the way I understand the question is that it asks us at what r (distance from nucleus) and at what angles is the probability going to be maximum. The way I see probability of finding at a point in a meaningful sense is, just like in 1D probability of finding particle at a point is taken as probability it is in dx region beside it, we find probability of finding it in a small volume near the point of interest.
Now the probability density of 1s orbital times dV in polar coordinates will end up containing a sine term (since small volume in polar cooditnates contains sine of angle with z axis). Thus the probability is dependent on not just r but also angle with z axis, which isn't expected for such a spherically symmetric orbital. I am confused here.

Of course one could say that since this is spherically symmetric, probability depends only on r and thus use only radial probability distribution and maximise it. But why is the above way which seems more general giving an apparently different answer.

Sorry if the above part isn't clear or is messy. Just to clarify things, I am not asking a "how to find" question. One way to look at my question is that if instead of maximising radial probability distribution, if I evaluate probabilities in small volume around points and maximise it, my answer should match with the one I get from maximising radial probability distribution. Consequently, I should get all points on sphere with Bohr radius but on maximising probability in small volume around a point, I get points at Bohr radius and on XY plane only (since for maxima, sine must be 1 meaning perpendicular to z axis).

You are right in saying that an s orbital is spherically symmetric, hence the probability density must not depend on the angles. Also, you are right in saying that $dV$ in polar coordinates contains $\sin\theta$. So what? You don't have to multiply or divide anything by that $dV$ (nor do you need to even know what it is). Old man Schrödinger took care of that for you. $|\psi^2|$ already gives you the probability density. Just as expected, it is a function of $r$ and not of the angles.


Why does slaked lime water lose alkalinity over time? When I throw 3 handfuls of white lime $\ce{Ca(OH)_2}$ into 60 gallons of tap water, the initial pH value is 11.4, but unknowingly it changes to pH 7.

What is the chemistry behind it?

How long does it take to completely change to neutral pH at open space?

I put lime in the water to pasteurize straw for mushrooms, and I need to know how long and how to make neutral pH faster so that I can pour the neutral water in the soil, so it won't ruin the soil.

If you dissolve slaked lime in water it will absorb carbon dioxide from the atmosphere over time.
Carbon dioxide will react with hydroxide ions from the slaked lime forming mainly hydrocarbonate ions. Hydrocarbonate ions will equilibrate with water and at about pH < 8 carbon dioxide will slowly be liberated from the solution. After some time, no significant amounts of hydroxide ions will be left from the slaked lime and the pH of the solution will close into neutral pH.

Well, what will happen to the calcium ions from the slaked lime? You will probably find them as an insoluble deposit at the end. This is because carbonate ions will also be formed during the process, forming calcium carbonate of very low water solubility.


Why is "Spontaneous" a part of 2nd law wording? The way I understand the law, any process is implausible if the total entropy of the universe does not increase. Why is "Spontaneous" even a part of the wording, what does it mean?

Is it perhaps a part of the probabilistic theory of thermodynamics, in which it is not "impossible" for heat to flow from cold to hot, simply very very unlikely?

In chemical thermodynamics we are interested in the amount of work that can be done by a reaction as it proceeds towards equilibrium and whether a reaction will spontaneously proceed towards products or, if not, how much work is needed to bring it about. Spontaneous is used because experimentally reactions and processes occur without us having to do anything to bring them about, and this needed an explanation.

The second law can be stated as$^1$  'Spontaneous changes are those which, if carried out under the proper conditions can be made to do work. If carried out reversibly they yield the maximum amount of work. In natural processes the maximum amount of work is never obtained.' This is also a general definition of equilibrium. A shorter definition of the second law is 'heat does not spontaneously flow from from a cold to a hot body'.

If there are differences in intensive properties then a spontaneous change is always that which eliminates the difference. Thus if there is a (perfect) piston with different pressures on either side, it will move to equalise the pressure. Similarly if there is a temperature difference between two bodies in contact with one another then 'heat flows' from the warmer to the colder. Similarly, a concentration difference in a solution will equalise. However, this cannot be used to predict the direction of a chemical reaction in which there may already be constant pressure, temperature or concentrations.

In a purely mechanical system, the direction of spontaneous change seems obvious and is one in which the potential energy is lowered to a minimum. It was thought at one time that the same would apply to chemical systems, i.e that the direction of spontaneous change was one that minimises the heat content of the system. But experiment shows that this is not the case, there are, for example, spontaneous endothermic reactions$^2$, so that another criteria is needed to predict the direction of a spontaneous reaction.  This means understanding why for example, a gas expands to fill a container even though no loss of energy occurs on doing this. The missing idea is the entropy$^3$.

It used to be thought, before the second law was properly understood, that the maxiumum work a chemical reaction could do was to be found by converting all the heat of reaction into work. This was not bourne out by experiment and it is the free energy $\Delta G$ that gives the maxmum amount of work. Gibbs was the first to understood that the maximum amount of work is given by the first law unless some heat is taken in or given out to the surroundings. Thus the work done may be greater than the heat of reaction.

The second law requires that there be an increase in entropy during any actual process. Thus $dS_{sys} + dS_{surr} \ge 0$ for the system and surroundings. The sum of the two entropy production processes is always a positive quantity, but which can be zero in the case of a reversible process.

(Your speculation about statistical mechanics is partly correct, it is possible for entropy to 'run backwards' if only a very small number of molecules is considered and then only for a very short time. In 'normal' lab chemistry the number of molecules involved is vast say, $10^{18}$ so deviations from the average are minute.)

$^1 $ E. B. Smith, 'Basic Chemical Thermodynamics'.

$^2$ e.g. $\ce{Ba(OH)2(8H2O)_{(s)} + NH4SCN_{(s)} +heat \rightarrow Ba(SCN)2 + 2NH3 + 10H2O}$

$^3 $ The second law is needed because the first law offers too much latitude compared to experimental observations; by the first law heat could spontaneously flow from a cold to a hot body or a broken glass could spontaneously reform itself.


How do the def2-SV, def2-SV(P), and def2-SVP basis sets differ? I am trying to calculate the energy of complex molecules with DFT and have a question about the Ahlrichs (Karlsruhe) def2 basis sets.

What is the difference  between the def2-SV, def2-SV(P), and def2-SVP?
Do they only differ in whether the basis set includes split valence polarization or not?

Def2-SVP has the polarisation functions on all atoms, -SV(P) does not have these functions on the hydrogen atoms and -SV does without them.
In future, you can check on the EMSL Basis Set Exchange It wouldn't give you an answer explicitly, but you can view the basis set for different atoms and see how the different sets vary.
A side note - If your 'complex molecules' aren't too large, consider the larger Def2-TZVP basis set. Errors at the Def2-SVP level can be several kcal which may significantly impact your study. As an alternative, consider the PC-n family of basis sets from Frank Jensen. If the Def2-TZVP set is too large, the double-zeta PC-1 might work well.


Can one predict the entropy change for aqueous phase reactions without calculation? I have seen several sources that claim aqueous solutions always have more entropy than pure liquids or solids, however, some aqueous ions have negative standard entropy.  This opens the possibility of solubility reactions with negative standard entropy change.  For example,
$$\begin{align}
\ce{NaF (s) &<=> Na+ (aq) + F- (aq)}&&\Delta S^{\circ} =-6.26 \ \mathrm{J/(mol\cdot K)}\\
\ce{AlCl3 (s) &<=> Al^3+ (aq) + 3 Cl- (aq)}&&\Delta S^{\circ} =-261.5 \ \mathrm{J/(mol\cdot K)}\\
\ce{AlF3 (s) &<=> Al^3+ (aq) + 3 F- (aq)}&&\Delta S^{\circ} =-429.58 \ \mathrm{J/(mol\cdot K)}
\end{align}$$

I have used thermodynamic data from this source.

So it seems to me that it is impossible to easily predict whether the entropy change for a reaction involving an aqueous phase is either positive or negative without calculation (unless a gas phase is involved, in which case the aqueous phase is moot)

Am I missing something, or are all the sources that claim an aqueous phase always has more entropy than a liquid phase oversimplifying the situation or just plain wrong?

I brought up the example of aluminum chloride to a chemistry instructor who then said that it is possible for molecular compounds to have negative entropy change when dissolved, but never for ionic compounds.  Yet both sodium fluoride and aluminum fluoride are ionic, and seem to have a negative entropy change when dissolved.

I understand why the standard entropies are negative, and about solvation shells. What I'm asking is whether the change in entropy can be predicted at a glance instead of looking up values and calculating as I have done.  Many textbooks claim all of these reactions should have a positive $\Delta S^{\circ}$ simply because "aqueous solutions have more entropy than solids or liquids." The instructor I talked to claimed that $\ce{AlCl3}$ will have negative $\Delta S^{\circ}$ because it is a molecular solid whereas $\ce{NaF}$ and $\ce{AlF3}$ should have positive $\Delta S^{\circ}$ because they are ionic solids. Are the textbooks and this instructor incorrect?

Some ions when dissolved in water may form a solvation shell that orders the solvent molecules in a way that the entropy decrease through this ordering outweighs the entropy increase through the ions leaving the crystal structure. Then you would have an overall decrease in entropy when dissolving the salt.

Small ions with high charge have a tendency for this.


Meaning of Molecular Oxygen Does molecular oxygen just mean the oxygen molecule? Or could it refer to a certain state of the molecule (eg. gas or aqueous)?

Molecular Oxygen means that it's bonded to another Oxygen (O2). Atomic oxygen would just give you O.


What could these letters "S" in red circles mean in a biochemical diagram? What could be the meaning of the red circles with letters S in them in the diagram below? I searched in the text but could not find.
From "Role of TREK-1 in Health and Disease, Focus on the Central Nervous System", Figure 2:

It almost certainly refers to the amino acid (residue) serine. The numbers (300, 333, 351) refer to the residue numbers. The COOH at the end signifies the C-terminus of the protein (subunit).
The black P's next to it are phosphate groups; both protein kinase A and C ("PKA" and "PKC") are involved in control of protein activity precisely via phosphorylation of serine residues in proteins (amongst others).

https://en.wikipedia.org/wiki/Protein_kinase_A
https://en.wikipedia.org/wiki/Protein_kinase_C


Gibbs free energy of formation vs. Gibbs free energy of reaction? Consider the following reaction:

$$\ce{A(s) + B(s) <=> AB(s)}.$$

I know there might be kinetic limitations but I am only interested in thermodynamic relations.

In general, the Gibbs free energy of reaction reads:

$$\Delta G = \Delta G^⦵ + RT\ln{\prod a_i^{\nu_i}}$$

and for the given reaction:

$\Delta G = \Delta G^⦵ + RT\ln{\frac{a_\ce{AB}}{a_\ce{A}\cdot a_\ce{B}}} \label{eqn:1}\tag{1}$

Since all components are pure solid substances, all activities equal 1 and therefore, $\Delta G = \Delta G^⦵$.

If $A$ and $B$ are elements, I found the following formula:

$$\Delta G_\mathrm{f}^⦵ = N\cdot RT\ln{\left(a_\ce{A}^{x_\ce{A}}a_\ce{B}^{x_\ce{B}}\right)} \label{eqn:2}\tag{2}$$

where $N$ is the total number of atoms and $x_\ce{A}$ and $x_\ce{B}$ are the mole fractions. To me, this latter formula implies, that the activities of $\ce{A}$ and $\ce{B}$ inside the product $\ce{AB}$ are of interest.

I have two questions:

What is the origin of the second formula $\eqref{eqn:2}$? In almost every source, the first type of Gibbs energy is used and I was not able to find any derivation. Also I think in principle, equation $\eqref{eqn:1}$ and $\eqref{eqn:2}$ should produce the same, but it looks very different!?
How does it come, that in one case the activity of the whole product $\ce{AB}$ is important and in another the single activities of the components of the product?

--------------------Update--------------------

Based on comments I'll try to reformulate my problem to make it more clear.

As I have learned, there should be no difference between the reaction product AB and the mixture of A and B with the corresponding stoichiometry (1:1). Is this valid in general or just for reactions like this (cf. my water-example in the comments)?

The transformation of equation (2) (in NightWriter's answer) shows, that it is an identity of the Gibbs free energy of mixing but in terms of moles of compound AB rather than in terms of moles of mixture.

If equation (1) describes the reaction (i.e. formation) of AB and equation (2) the mixing of A and B, and there is no difference between a product of reaction and the corresponding mixture of the two components to form that product, there must be some mathematical relation between equation (1) and equation (2). Furthermore, I am not certain, whether the standard-state symbol in equation (2) is correct or not.

After I read the answer and the comments again, I concluded the following: equation (1) describes the formation of AB and equation (2) the mixture of A and B at arbitrary compositions. So, if I set $x_A = x_B =0.5$, it should be possible to replace $\Delta G^⦵$ in equation (1) by $\Delta G_f^⦵ = N\cdot RT\ln{a_A^{0.5}a_B^{0.5}}$. $N$ must be chosen properly, since the two expressions are related to different things (mixture and compound). Is this correct?

What is the origin of the second formula (2)?

Equation (2) evidently refers to the free energy of formation (at standard pressure) of a solid solution of A and B, in which case neither element is in its standard state as a product (so that $a_i\neq 1$). You can rewrite expression (2) as follows:

$$\Delta G_\mathrm{f}^\circ = nRT\ln{\left(a_\ce{A}^{x_\ce{A}}a_\ce{B}^{x_\ce{B}}\right)} = RT\ln{\left(a_\ce{A}^{x_\ce{A}}a_\ce{B}^{x_\ce{B}}\right)^n}= RT\ln{\left(a_\ce{A}^{nx_\ce{A}}a_\ce{B}^{nx_\ce{B}}\right)}= RT\ln{\left(a_\ce{A}^{n_\ce{A}}a_\ce{B}^{n_\ce{B}}\right)}$$

$$ = n_\ce{A}RT\ln{a_\ce{A}+n_\ce{B}RT\ln{a_\ce{B}}}$$

where $n$ is the total number of moles ($n=n_A+n_B$).

Since the activity of the elements in pure solutions is 1 we can finally write

$$\Delta G_\mathrm{f}^\circ = n_\ce{A}\Delta G_{mA}+n_\ce{B}\Delta G_{mB}$$
where $\Delta G_{mi} = G_{mi}-G_{mi}^\circ= n_\ce{i}RT\ln{\left(\frac{a_\ce{i}}{a_\ce{i}^\circ}\right)}$.

However the most useful way to express this is as follows:

$$\Delta G_\mathrm{f}^\circ = (n_\ce{A}G_{mA}+n_\ce{A}G_{mB})-(n_\ce{B}G_{mA}^\circ+n_\ce{B}G_{mB}^\circ)$$

In words, the free energy change corresponds to mixing of $n_\ce{A}$ moles of pure A and $n_\ce{B}$ moles of pure B (the pure components having activity $a_i^\circ=1$) to form a mixture where the components have activities $a_i$.

  How does it come, that in one case the activity of the whole product AB is important and in another the single activities of the components of the product?

Equation (1) refers to a molar free energy of formation of $\ce{AB}$ from reagents $A$ and $B$, all under the same constant (P,T, composition) conditions, whereas (2) refers to the free energy of formation of a solid solution of $n_A$ moles of A and $n_B$ moles of B from pure components. Equation (1) refers to combination of A and B at a 1:1 mole ratio, or equivalently reaction to form 1 mole of $\ce{AB}$ from $n_A=n_B=\pu{1 mol}$. Reaction (2) refers to mixture of A and B at any arbitrary ratio or total number of moles. Therefore equation (2) is in a way more general. Also, equation (1) refers to a differential process (transformation to form 1 mole of product under contant conditions) whereas (2) refers to an integral (mixing) process.

  For the given reaction:
  $$\Delta G = \Delta G^⦵ + RT\ln{\frac{a_\ce{AB}}{a_\ce{A}\cdot a_\ce{B}}} $$
  Since all components are pure solid substances, all activities equal 1 and therefore, $\Delta G = \Delta G^⦵$.

A comment on this: the fallacy here is to assume that $a_i=1$ at equilibrium. Only for the pure reagents and products in their standard states it is strictly required by definition that $a_i=1$.

Note also that since $\Delta G^\circ = G_{mAB}^\circ-G_{mA}^\circ-G_{mB}^\circ$ and  $G_{mi} = G_{mi}^\circ+RT\ln{\left(\frac{a_\ce{i}}{a_\ce{i}^\circ}\right)}$ we can write

$$\Delta G = G_{mAB}-G_{mA}-G_{mB}$$

In words (and repeating myself), $\Delta G$ in this case is for the process of converting 1 total mole of $A$ and 1 mole of $B$ into 1 mole of $AB$, all at constant T, P and composition (or, equivalently, in a sufficiently large mixture, such that a 1 mole change in the amount of the substances does not change the properties).

Schematically, this is how I interpret the two processes, with equation 2 for mixing in the top, and the reaction of A and B to form AB in the bottom:


Assigning absolute configuration when determining crystal structure isn't possible When natural products are isolated, their structures are often determined by x-ray crystallography which gives a complete picture of the atom coordinates, the relative and the absolute stereochemical configurations. If however the isolated molecule won't crystallize, how can absolute stereochemistry be obtained?

I know that 2D NMR can tell about connectivity and also possibly relative stereochemistry ($J$-value analysis), but unless one stereocenter is known the others cannot be implied from it. Optical rotations also are useless unless the compound is known. How then can the absolute configuration be found without single-crystal x-ray diffraction?

This question is possibly a little broad- there are entire books dedicated to methods of structure determination and assignment of stereochemistry.

The tl;dr answer is that there is no single method by which absolute stereochemisty can be assigned unambiguously, it depends on the nature of the natural product (terpene, polyketide, polypeptide, etc) and the exact structural features.

Some common methods which are often found in isolation papers:

1. Degradation

Historically, one of the most common ways of identifying new natural products was to attempt to degrade them to known compounds. If you imagine taking a new peptide natural product (this is actually a bad example as most peptides are made up of natural amino acids), you could easily hydrolyse it back to the constituent amino-acids (or smaller peptide chains) for which rotational data is available.

2. Derivatisation to make crystals

If your material doesn't crystallise, make small structural modifications to make it crystalline.

Being able to do this somewhat depends on knowing the 2D structure (by NMR). Once you identify the functional groups present, you can begin to modify the structure in order to obtain something crystalline (to do X-ray analysis on).

There are too many possibilities to go through in detail, but you could, for instance, take a carboxylic acid and form an ester, or some kind of salt.

This method is often carried out in the pharmaceutical industry, where they want their active pharmaceuticals to be in the solid state in a known form (different forms have different properties, such as solubility, so great lengths are taken to obtain uniform material).

3. Derivatisation for NMR studies

If the molecule of interest has an alcohol present, its possible to form esters with a special kind of carboxylic acid, this is the Mosher ester assignment.

The alcohol of interest is reaction with two enantiomeric acids (the Mosher acids), to form a pair of diasteromers. The Mosher esters shield the molecule with the phenyl group, causing shifts to change on the nearby protons around the chiral alcohol. By considering the relative changes, the assignment may be assigned.  This method is often found in structure elucidation papers, however there are some instances where it can give misleading results, so shouldn't really ever be used on its own without any other evidence.

It's worth noting that the Mosher esters themselves are often crystalline, allowing a crystal structure to be obtained in addition to the assignment by the NMR analysis discussed above.

4. Computational work

Computation is one of the newer methods of assigning absolute stereochemistry, so new in-fact that the FDA (the people who approve drugs) don't accept computational assignment on its own, unless accompanied by some other, more traditional method, of assigning configuration.

In theory, there are methods to predict optical rotation, or other properties such as circular dichroism, which can then be compared to the real experimental values. In practice, this rarely works well for complex molecules as there are too many variables involved which effect the computational result.


Determine concentration Using Beer's law How do you determine concentration of a sample using wavelength and absorbency? Say our wavelength for a sample of Red 40 is 508.50 nm and absorbance is 0.283 when we use 6ml water and 4 ml Red 40. What would be our concentration and the slope ε? This is really frustrating me so any help would be appreciated.

I don't know anything about Red 40, but you will need the extinction coefficient.  Perhaps you determined this from a plot of concentration vs absorbance in the lab?

I am going to attempt to answer you question with reference to NADH, which has an accepted extinction coefficient of 6220 M-1.cm-1 at 340 nm.

Say I have a solution of NADH in a test-tube, I take 4 ml of this and dilute with water to 10 ml.  I then take this diluted solution and measure its absorbance at 340 nm in a spectrophotometer using a cuvette with a path-length of 1 cm.  The value I obtain is 0.4.
(Edit.  In the above measurement, I take care to 'blank' the spectrophotometer against water).

What is the concentration of NADH in the original solution?

From the Beer-Lambert law (where c is concentration)

0.4 = 6220 x c x 1

Therefore

c = 0.64 x 10-6 M  = 0.64 µ M

But the solution on which the measurement was made was
diluted 4 in 10 or 1 in 2.5

Therefore the concentration of NADH in the original solution is

c = 2.5 x 0.64 x 10-6 M = 160.77 x 10-6 M = 160.77 µ M

So, the concentration of NADH in the original solution is about 161 µ M (micro-Molar).

I am making a number of assumptions here, the main one being that the spectrophotometer obeys the Beer-Lambert law in the absorbance region of interest.  You should determine that this is so by measuring  the absorbance of series of NADH solutions and make sure that the resulting graph of absorbance vs concentration is linear and through the origin.

At the very least, doubling the concentration of NADH should exactly double the absorbance (0.8) and halving it should give exactly half the absorbance (0.2).  In addition, measurement of water (or whatever you are 'blanking against) should give zero - that is the spectrophotometer should be properly 'blanked'.  If not you will need to subtract this value.

So what happens if we do not dilute our original sample, but just measure the absorbance at 340nm directly?

From Beer-Lambert

A = 6220 x 160.77 x 10-6 = 1

The original solution is predicted to have an absorbance of 1, but IMO even with modern spectrophotometers you are 'pushing your luck' to assume that Beer-Lambert is obeyed at such a high absorbance.  A 'rule-of-thumb' is to keep the absorbance below 0.8.

One final comment.  It is very difficult to accurately determine an extinction coefficient from absorbance versus concentration plots, unless you accurately know the concentration of the original material.  If this figure relies on a 'dry-weight' measurement, you will need to be sure that the compound is pure, for a start.

Reading your question again, I think we need a bit more experimental detail, and if you could supply it, I'll have another go, maybe.

NADH is widely used in the biochemical sciences where (among many other things) the relatively high extinction coefficient of 6220 M-1.cm-1 is exploited in continuous enzyme assays.  If an enzyme produces or consumes NADH (or NADPH) then it is (usually) very easy to design an assay.

EDIT

It would appear that the accepted extintion coefficient (Molar Absorption Coefficient) for Allura Red AC (Red 40) is about 25 000 M-1.cm-1 (or 25 000 mol-1 L cm-1) at about 505 nm. The best reference I can get it here. Azo dyes tend to have very high extinction coeffficients, so this 'looks about right'

The diluted solution has an absorbance of 0.283

Taking the above as the extinction coefficient, and in addition assuming the measurement was made in a cuvette with a path length of 1 cm (l =1), one can calculate from Beer-Lambert that the concentration of Allura Red in the diluted solution is about 11.32 x 10-6 M.

The dilution factor is 4 in 10, or 1 in 2.5.  Therefore, the concentration of the original solution is about 28.3 x 10-6 M.


How to get rid of bubbles? Certain solutions, such as aqueous SDS or soapy waters, after being shaken evolve a lot of bubbles. These bubbles disappear over time, but it can take minutes or sometimes hours (eg. for 20% SDS).

Is there anything that can be done to accelerate the popping of these bubbles? For instance, would heating the solution affect the rate of bubble collapse?

Note: I am interested in methods that do not alter the chemical properties of the solution. For instance, I know that adding certain substances can affect bubbles, but if I added those then the solution would be something different from what I wanted in the first place. It is assumed that procedures such as heating are not sufficiently energetic to cause a noteworthy reaction.

Without addition of chemicals? Try out ultrasound (the same physical principle opticians use to clean your pair of glasses).  If you whish, this may be handled even contact less (some background and here), to mention examples.


Explaining redox reactions to a 6 year old? In a similar vein and for pretty much the same reason as I post this question on the Physics.SE site, what would be a an effective means to promote understanding about redox reactions to a 6 year old?  In particular, what (if any) simulations are available?

I might suggest you take a look at What's Chemistry All About by Frith, Gillespie and Larkum.  (ISBN: 1409547078)  It is a very approachable book that in my mind pays close attention to the science jargon introduced to its audience.  Redox reactions are described in the section "Let's make things happen" under the types of reactions.  A bit of history about where the name oxidation comes from and the safety symbol that you will see on an oxidizing agent are both included in the brief discussion.

The example reactions given are combustion and firecrackers, which I think are appropriate and accessible to a child.

As for experiments to do, I'll suggest another book, the Illustrated Guide to Home Chemistry Experiments by Thompson. (ISBN: 0596514921) There is a chapter on electrochemistry with two experiments that might be of interest to a young learner: the electrolysis of water to make oxygen and hydrogen gas, and the oxidation of iron.  Of course, there's also the "make your battery" type experiments which are always fun.

As for simulations, I would suggest heading on over to the PHet website and have a stroll through all the activities there.  I don't recall off hand what they have for electrochemistry, but there's plenty to keep a budding scientist interested.


Defining the position of equilibrium I've read quite a few other answers on this site such as this one, but can't quite seem to understand fully yet.

Say we have a container with gases reacting such that the number of moles on both sides of the equation are not equal. If we then, for instance, reduce the volume of the container, the total pressure increases and the value of the quotient Q will also change. The net reaction then favours a particular direction so that the mole fractions change sufficiently for Q to return to the value of K, which remains constant.

This is commonly described by saying equilibrium shifts to either the left or the right. Whilst it is true that after the system has again reached equilibrium the mole fractions have changed (so in a sense, the new equilibrium has shifted to one side in terms of number of moles), the value of Q is equal to its initial value so in this sense the position of equilibrium as defined by K has not changed.

The same applies to changing the concentration of a species; the actual concentrations of every species after the system has re-equilibrated will be different, however the value of Q will have returned to its previous value.

The only situation that I can think of where the position of equilibrium actually shifts is when a change in temperature alters the value of K.

My question is, essentially, is the term 'position of equilibrium' more of a qualitative descriptor of the composition of reactants and products? I initially thought it meant the value of Q at equilibrium (or K), but evidently in gaseous reactions the equilibrium yield does change with changes in pressure though K remains constant, so such a definition would not work.

Thank you!

"Shifting" an equilibrium

Unfortunately for learners of chemistry, statements like "the equilibrium shifts to the right" are quite common jargon.

What is meant by this, speaking in more accurate technical terms, is the following: A reaction had reached equilibrium (net reaction is zero), and then some change occurred so that it was no longer in equilibrium. Now, there is a net forward reaction in the direction of the products ("to the right" refers to reactants turning into products).

"Shifting to the left" describes a similar sequence of events, just in the reverse direction.

New equilibrium or reestablish equilibrium?

The OP is correct in saying that unless the temperature changed, the equilibrium constant does not change, and the reaction quotient Q of the first equilibrium state is equal to that of the second. So you could say you are back to the same equilibrium.

Typically, however, the mole fractions, concentrations or partial pressures will be distinct for the first and the second equilibrium. So you could also say the reaction reaches a new equilibrium.

If you say it reaches equilibrium again, you are sufficiently vague to be correct.

Position of equilibrium

The term "position of equilibrium" is ill-defined. To avoid any confusion, it is better to distinguish between reaction quotient (which is equal to the equilibrium constant for both the first and second equilibrium) and the set of concentrations (which is typically different).


Reacting sodium metal with aqueous sodium hydroxide, what would happen? I know sodium metal, when reacted with water, produces flammable hydrogen gas but I was wondering if that would also be the case if I had aqueous 1M sodium hydroxide?

Would this reaction be safe or will it also produce a harmful compound like hydrogen gas?

Consider first the solvation of sodium metal in water to give a sodium ion and a hydrated electron.
$$\ce{Na(s) \xrightarrow{\ce{H_2O}} Na^+(aq) + e^-(aq) }$$
The hydrated electron will then act as the reactive species to reduce water to hydrogen atoms and hydroxide anions.
$$\ce{e^-(aq) + H_2O(l) \rightarrow H(aq) + OH^-(aq)}$$
Note that two of these hydrogens come together to form the diatomic hydrogen gas.
$$\ce{2H(aq) \rightarrow H_2(g)}$$
When we multiply the coefficients in the first two equations by two, we are able to consider the net ionic equation through the elimination of the reactive intermediates.
$$\ce{2Na(s) + 2H_2O(l) \rightarrow 2Na^+(aq) + 2OH^-(aq) + H_2(g)}$$
We see that the dissolution of sodium metal in aqueous sodium hydroxide would be affected by Le Chatelier's principle due to the presence of the sodium hydroxide. We would expect there to be a reduction of solubility and rate of dissolution. The greater the initial concentration of sodium hydroxide, the more appreciable the effect.
Given that a solution of sodium hydroxide at $\ce{25 ^{\circ}C}$ is saturated at $\ce{25 M}$, a $\ce{1 M}$ solution would not be expected to have an appreciable effect. Hydrogen gas would remain a product, evidenced by the net equation.
As far as the reaction of molten sodium and molten sodium hydroxide, I expect that is what they were referring to on the Wiki page for sodium oxide when they gave this equation.
$$\ce{2NaOH + 2Na \rightarrow 2Na_2O + H_2}$$
The source given is Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. ISBN 978-0-08-037941-8.
The solubility of sodium hydroxide was resourced from the Wiki page for sodium hydroxide. The source given appears to be Haynes, CRC Handbook of Chemistry and Physics.
The first set of equations were sourced from Walker, David C. Production of Hydrated Electrons. Canadian Journal of Chemistry. Volume 44. 1966.


IUPAC naming of double salt My book gives the IUPAC name of $\ce{KAl(SO4)2.12H2O}$ which is nothing but emperical formula of potash alum as

"Potassium aluminium (III) sulphate -12 water"

But Wikipedia gives it as "Aluminium potassium sulfate dodecahydrate" , though there could be change in name because of number of water molecules but there is significant difference between the two names

So, my question is how to give IUPAC name to a double salt (potash alum is a double salt), and why is there difference between two iupac names I.e the one which is given in my book and the one in wikipidea.

According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), alphabetical order is used.

Alphabetical order is used in formulae as follows.
(a) Within the group of cations and within the group of anions, respectively, in formulae of salts and double salts. Deviations from this rule may be acceptable if it is desired to convey specific structural information (…)

and

Alphabetical order is used in names as follows.
(d) In compositional names, the names of the formally electropositive components and the names of the formally electronegative components are each arranged alphabetically with the former group of components preceding the latter. (…)

Note that the order of components in the name may deviate from the order of the corresponding components in the formula. For example:
$\ce{KMgF3}$ (since K < M)
magnesium potassium fluoride (since m < p)
$\ce{ZnI(OH)}$ (since I < O)
zinc hydroxide iodide (since h < i)


My Misconception of Entropy It was recently brought to my attention that my understanding of entropy is wonky at best.
In my experience, entropy was introduced (superficially at best) during general chemistry/foundations of inorganic chem and described in terms of order/disorder of a system, usually followed by some messy room analogy. In one of my textbooks, the author, Gary Wulfsberg, was discussing how a reaction tends to favor the products "if it increases the dispersal of ions/molecules over a larger volume of space..."(Wulfsberg, G., 2018). Wulfsberg goes on to say that

The measure of this dispersal or disorder is known as the entropy (S) of the substance. A
positive entropy change for a reaction indicates increasing dispersal or disorder
(Wulfsberg, G., Foundations of Inorganic Chemistry; ch. 4, pg. 200)

which I interpreted it as him saying that all systems that increase in entropy have a corresponding increase in disorder.
Later in my undergraduate career, Boltzmann's entropy was introduced during thermodynamics and described by my professor as the amount of micro-states available for the particles within the system and that an increase in entropy corresponds to an increase in the states that the system can exist in.
All was fine and peachy until recently, when I came across an article that discusses the Shannon Measure of Information (SMI) and entropy. In an article by Arieh Ben-Naim, he talks about how the idea of order/disorder in relation to entropy is not necessarily correct, in fact it is a fallacy that does not hold true nor can order/disorder be measured definitely for all systems. He states,

It is true that many spontaneous processes may be viewed as proceeding from an ordered to a disordered state. However, there are two difficulties with this interpretation. First, the concept of order is not well-defined, and in many processes it is difficult, if not impossible, to decide which of the two states of the system is more or less ordered.
J. Chem. Educ. 2011, 88 (5), 594–596

Additionally, he talks about how some systems do have "order parameters", however, it is not in relation to entropy and that not every process where an increase in entropy is observed has a corresponding increase in disorder. He later goes on to describe the SMI treatment of entropy "as the number of binary questions one needs to ask to find the location of the particle." Hence, if the number of yes/no questions one needs to ask to find the location of particle increases, so has the thermodynamic entropy.
So here is my question
Is entropy described by Boltzmann/statistical mechanics the same as the entropy described by Shannon information theory?
Additionally, is there any validity in relating order/disorder in describing the change in entropy of chemical systems?

The association of entropy with disorder is anthropocentric. "Disorder" is a concept derived from our experience of the world. The association with "disorder" is clearer once we explain what we mean by "order". The following definition among those provided by Merriam-Webster most closely fits the intended meaning:

a regular or harmonious arrangement

Since regularity always implies lower entropy, all else being equal, it is therefore fair to associate "order" with lower entropy.
The association with "order" or regularity also syncs with the concept of entropy according to Boltzmann's statistical mechanical definition ($S= k_\mathrm B \log \Omega$). $\Omega$, the number of microstates available to the system, can be quantified using the entropy. More possible unique microstates implies a higher entropy. Greater regularity implies more constraints regarding arrangement of the system and therefore less possible microstates. Solids are usually more regular and therefore have lower entropy than fluid states at the same T. Same when comparing gases and liquids. The possible microstates increases from solid to liquid to gas.
This also jives with the informational-content definition. You can use less information (use a more compact description) to describe an orderly (regular) system. You need more information to describe all possible arrangements of molecules in a gas or liquid than in a solid. Think of entropy as measuring the length of the recipe required to build all possible arrangements of the system.


What does a number in brackets after another number mean? I.e. 21(1)cm-1 What does the number in brackets mean in these two examples?

$$21(1)\ \mathrm{cm^{-1}}$$

and

$$1.0(3)\times10^{-7}$$

The results of measurements and other numerical values of quantities are often given with an associated standard uncertainty. A numerical value and the associated uncertainty may be expressed as shown in the question:

$$\begin{align}
y&=21(1)\ \mathrm{cm^{-1}}\\[6pt]
&=a(b)\ \mathrm{cm^{-1}}
\end{align}$$

where
$y$ is the estimate of the measurand (e.g. the result of a measurement) expressed in the unit $\mathrm{cm^{-1}}$,
$a$ is the numerical value, and
$b$ denotes a standard uncertainty expressed in terms of the least significant digit(s) in $a$.

It is important to note that the given uncertainty refers to the least significant digits of the given numerical value. For example, in the expression

$$l=23.4782(32)\ \mathrm m$$

the $(32)$ represents a standard uncertainty equal to

$$u(l)=0.0032\ \mathrm m$$

Many physical constants are also given in this form. For example, the previously recommended value for the molar gas constant $R$ as given by NIST from 25 June 2015 until the new value became available on 20 May 2019:

$$R=8.3144598(48)\ \mathrm{J\ mol^{-1}\ K^{-1}}$$

where the $(48)$ represents a standard uncertainty of

$$u(R)=0.0000048\ \mathrm{J\ mol^{-1}\ K^{-1}}$$

This form is in accordance with various current standards, in particular

JCGM 100:2008 Evaluation of measurement data – Guide to the expression of uncertainty in measurement
ISO/IEC Guide 98-3:2008 Uncertainty of measurement   Part 3: Guide to the expression of uncertainty in measurement (GUM:1995)
ISO 80000-1:2009 Quantities and units – Part 1: General

The Guide to the Expression of Uncertainty in Measurement (GUM) also shows other permissible forms:

  7.2.2 When the measure of uncertainty is $u_\mathrm c(y)$, it is preferable to state the numerical result of the measurement in one of the following four ways in order to prevent misunderstanding. (The quantity whose value is being reported is assumed to be a nominally 100 g standard of mass $m_\mathrm S$; the words in parentheses may be omitted for brevity if $u_\mathrm c$ is defined elsewhere in the document reporting the result.)

  1) “$m_\mathrm S=100{,}021\,47\ \mathrm g$ with (a combined standard uncertainty) $u_\mathrm c = 0{,}35\ \mathrm{mg}$.”

  2) “$m_\mathrm S=100{,}021\,47(35)\ \mathrm g$, where the number in parentheses is the numerical value of (the combined standard uncertainty) $u_\mathrm c$ referred to the corresponding last digits of the quoted result.”

  3) “$m_\mathrm S=100{,}021\,47(0{,}000\,35)\ \mathrm g$, where the number in parentheses is the numerical value of (the combined standard uncertainty) $u_\mathrm c$ expressed in the unit of the quoted result.”

  4) “$m_\mathrm S=(100{,}021\,47\pm0{,}000\,35)\ \mathrm g$, where the number following the symbol $\pm$ is the numerical value of (the combined standard uncertainty) $u_\mathrm c$ and not a confidence interval.”

Note that item 2) corresponds to the form given in the question.

Concerning item 4), however, the GUM notes

  The ± format should be avoided whenever possible because it has traditionally been used to indicate an interval corresponding to a high level of confidence and thus may be confused with expanded uncertainty (…).

Furthermore, ISO 80000 notes

  Uncertainties are often expressed in the following manner: $(23{,}478\,2\pm0{,}003\,2)\ \mathrm m$. This is, however, wrong from a mathematical point of view. $23{,}478\,2\pm0{,}003\,2$ means $23{,}481\,4$ or $23{,}475\,0$, but not all values between these two values. (…)


Why is finding [H+] when given [OH-]= 3.57e-5 M wrong when I use -log[OH-]? Question/Problem

I know that [H+][OH-]=$10^{-14}$

And the problem asks to find out the concentration of H+ when the concentration of OH- is $3.57*10^{-5}$M

By using the equation above i find that the concentration of H+ is 2.67E-10

But if I use a different method as follows:

-log[OH]=pOH so by substitution, -log[$3.57*10^{-5}$]

then that gives me 4.425.......

then i subtract 14 from that number to get 9.574......

which that is pH

then i set 9.574=-log[H+]
so then you "e" everything which the answer comes to be  6.95E-5

why is this wrong?

Well You have to note that you cannot simply "e" everything because the base of log in chemistry is base 10 while you assumed that it was all base e. Ie ln(x). in order to get x, we use e, while if we say log(x) we have to assume that it is base 10 which is the general norm


Relationship between Gibbs free energy, enthalpy, entropy and rate of reactions I'm having a little trouble constructing the bigger picture of chemical reactions.

From the collision theory we know that molecules react if: 1) they collide 2) they collide bearing a sufficient amount of energy to surmount the activation energy and collide at a specific geometric orientation.

From the relationship between Gibbs free energy, enthalpy, entropy and temperature we know: $\Delta G = \Delta H - T\,\Delta S$,
and that a reaction is only thermodynamically favored when $\Delta G < 0$. And even when $\Delta G < 0$ the reaction may still not occur at a measurable rate because it has too high an activation energy.

From the Arrhenius equation we know: $k = A ~\mathrm e^{-E_\mathrm a / (RT)}$

Linking all these together:
1. Reactions only occur when particles can collide with each other.
2. Colliding particles will only react if they bear sufficient energy (can be increased through temperature and/or adding other external energy sources) and have correct orientation (chances can't be increased?).
3. Only particles in thermodynamically favored reactions have enough energy to overcome $E_\mathrm a$ and so will spontaneously occur without external energy applied. But they may occur extremely slowly if they have a very high $E_\mathrm a$ (according to the Arrhenius equation).

Is this reasoning correct?

I'm having a little trouble constructing the bigger picture of chemical reactions.

Let's go step by step.

  From the collision theory we know that molecules react if: 1) they collide 2) they collide bearing a sufficient amount of energy to surmount the activation energy and collide at a specific geometric orientation.

Yes, in order to react, molecules taking part in a (bimolecular) reaction need to:

be in contact;
be correctly aligned to each other (a random alignment may not react);
have enough energy to overcome the activation barrier ($\Delta G^\neq$).

  From the relationship between Gibbs free energy, enthalpy, entropy and temperature we know: $\Delta G = \Delta H - T\,\Delta S$,
  and that a reaction is only thermodynamically favored when $\Delta G < 0$. And even when $\Delta G < 0$ the reaction may still not occur at a measurable rate because it has too high an activation energy.

What you're describing is the Gibbs reaction energy: the Gibbs energy between products and reactants.
This indeed tells us how thermodynamically favourably the reaction is ($\Delta G < 0$).

But the phrase in bold above is wrong.
The reason is that the Gibbs activation energy ($\Delta G^\neq$) is the Gibbs energy difference between transition state and reactants.

As Mithoron mentioned, Gibbs reaction energy would give you the equilibrium constant for the reaction:

$$K = e^{-\tfrac{\Delta G}{R T}}$$

  From the Arrhenius equation we know: $k = A ~\mathrm e^{-E_\mathrm a / (RT)}$

Generally, from Eyring theory,

$$k = \frac{k_B T}{h c_0} e^{-\tfrac{\Delta G^\neq}{R T}}$$

Comparing to Arrhenius equation, see:

$$k = \frac{k_B T}{h c_0} e^{-\tfrac{\Delta H^\neq - T \Delta S^\neq}{R T}}
= \frac{k_B T}{h c_0} e^{\tfrac{\Delta S^\neq}{R}} e^{-\tfrac{\Delta H^\neq}{R T}}$$

Thus,

$$A = \frac{k_B T}{h c_0} e^{\tfrac{\Delta S^\neq}{R}} \\
E_a = \Delta H^\neq$$

That's what you would find from an Arrhenius's plot.

  Linking all these together:
  1. Reactions only occur when particles can collide with each other.
  2. Colliding particles will only react if they bear sufficient energy (can be increased through temperature and/or adding other external energy sources) and have correct orientation (chances can't be increased?).
  3. Only particles in thermodynamically favored reactions have enough energy to overcome $E_\mathrm a$ and so will spontaneously occur without external energy applied. But they may occur extremely slowly if they have a very high $E_\mathrm a$ (according to the Arrhenius equation).

  Is this reasoning correct?

The reasoning is correct, except for point three above (in bold), for which the same error is found as before: reactions may have a big $K$ (small $\Delta G$) but small $k$ (big $\Delta G^\neq$) or vice-versa.


Separation of a solution of n-butanol and nitroethane Theoretically speaking, how could one separate a solution of n-butanol and nitroethane? They have a similar boiling point, so fractional distillation would be not possible. Wikipedia also states that nitroethane is more miscible in water than butanol, but that would probably give a water nitroethane azeotrope (?). Can anyone tell how the separation could be achieved? Also, would it be a difference between a 50:50 solution and a 5:95 (vice versa)?

Use counter-current extraction.
I'm not sure about solvents, but let's say that I can separate nitromethane from water by distillation. Also that butanol can be separated from ether by distillation. (as I remember ether and water aren't miscible...)
So I line up 7 sepatory funnels.
Into sepatory funnel #4 I put my mixture and 50 ml of water and 50 ml of ether. Equilibrate then water phase goes to sepatory funnel #5 and ether phase goes to sepatory funnel #3. Sepatory funnel #4 is now empty.

In sepatory funnel #5 I add 50 ml ether, Equilibrate then water phase goes to sepatory funnel #6 and ether phase goes to sepatory funnel #4
In sepatory funnel #3 I add 50 ml water, Equilibrate, then water phase goes to sepatory funnel #4 and ether phase goes to sepatory funnel #2

So basically I have the water extracts going one way and the ether extracts the other way.


Why is o-nitroaniline not more basic than aniline? We know from SIR effect (Steric Inhibition of Resonance) that in o-nitroaniline, the nitrogen's p-orbitals will change their plane, causing the conjugation/resonance of its lone pairs to stop. So, the only effect acting on it now is −I effect from the $\ce{NO2}$ group.

Comparing with just aniline, the $\ce{NH2}$ lone pairs are in constant resonance with the benzene ring making them unavailable for donation. Thus, aniline should be less basic than o-nitroaniline. But it isn't as per experimental values. So what's going on? What's the logic behind this?

Yes it is true that SIR is observed in o-nitro aniline so it will not show -M effect.

But I think you're confused in the resonance of the lone pair of NH2 group with the benzene ring which in fact occurs in both cases. The thing that makes o-nitro aniline less basic is the -I effect which makes the lone pair on - NH2 group less available for donation .

In fact the complete order of basic strength of nitro substituted anilines is:
[Aniline>m-nitro aniline>p-nitro aniline>o-nitro aniline ]


Anomalous electronic configuration of platinum Why is the electronic configuration of platinum
$$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$

and not $\mathrm{[Xe] 4f^{14} 5d^{10} 6s^0}$ or
$\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$?

The Madelung energy ordering rule gives the energy of the orbitals approximately:
$ \mathrm{1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d <
5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p}$
That would speak for $\mathrm{[Xe] 4f^{14} 5d^{8} 6s^{2}}$ if you follow the Aufbau principle (from German Aufbau = setup). But $\ce{Pt}$ is an exception (as there are some). A rule of thumb is that half-filled shells are stabilized. So that means in the case of $\ce{Pt}$, $\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$.
The real answer is much more complicated. It comes from relativistic effects, electron correlation and shielding effects. There is an interplay between the attraction of nucleus and the electrons, and the electron repulsion between all electrons. The heavier atoms become, the more important relativistic effects become, since the inner electrons are moving much faster as they are in a stronger electric field from the higher charge of the nucleus. Often it is said that the outer orbitals are less compact than the inner ones, which is true if one calculates and analyses them, but a stronger effect is that outer electrons are shielded and therefore the nucleus attraction is weaker. So there is no simple way, learn the exceptions or solve the Schrödinger/Dirac equations.
Also be careful when speaking about orbitals. An orbital is a one-electron wavefunction and they are more a chemical concept than reality in multi-electron atoms.


How are Organic Compounds named? I have heard of of a special system which specially deals with naming of organic compounds.

How do we get names such as 2-methylpropan-1-ol.
I have tried googling it and found few links but their explanation was way above my head. I also asked my teacher about the same question and he said this is not required in your standard, but I really want to know about naming.

Few Links that I found on Google were

http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry
Another is wikihow very difficult, I need a simple explanation.

I myself while searching and talking to my dad found the answer.Sounds weird, sorry for posting.

THE NAMES OF ORGANIC COMPOUNDS

Cracking the code

A modern organic name is simply a code. Each part of the name gives you some useful information about the compound.

For example, to understand the name 2-methylpropan-1-ol you need to take the name to pieces.

The prop in the middle tells you how many carbon atoms there are in the longest chain (in this case, 3). The an which follows the "prop" tells you that there aren't any carbon-carbon double bonds.

The other two parts of the name tell you about interesting things which are happening on the first and second carbon atom in the chain. Any name you are likely to come across can be broken up in this same way.

Counting the carbon atoms

You will need to remember the codes for the number of carbon atoms in a chain up to 6 carbons. There is no easy way around this - you have got to learn them. If you don't do this properly, you won't be able to name anything!

code  no of carbons
meth    1
eth 2
prop    3
but 4
pent    5
hex 6
Types of carbon-carbon bonds
Whether or not the compound contains a carbon-carbon double bond is shown by the two letters immediately after the code for the chain length.
code    means
an  only carbon-carbon single bonds
en  contains a carbon-carbon double bond
For example, butane means four carbons in a chain with no double bond.

Propene means three carbons in a chain with a double bond between two of the carbons.

Alkyl groups

Compounds like methane, $\ce{CH4}$, and ethane, $\ce{CH3CH3}$, are members of a family of compounds called alkanes. If you remove a hydrogen atom from one of these you get an alkyl group.

For example:

A methyl group is $\ce{CH3}$.

An ethyl group is $\ce{CH3CH2}$.

These groups must, of course, always be attached to something else.

Types of compounds

The alkanes

Example 1:  Write the structural formula for 2-methylpentane.

Start decoding the name from the bit that counts the number of carbon atoms in the longest chain - pent counts 5 carbons.

Are there any carbon-carbon double bonds? No - an tells you there aren't any.

Now draw this carbon skeleton:

Put a methyl group on the number 2 carbon atom:

Does it matter which end you start counting from? No - if you counted from the other end, you would draw the next structure. That's exactly the same as the first one, except that it has been flipped over.

Finally, all you have to do is to put in the correct number of hydrogen atoms on each carbon so that each carbon is forming four bonds.

If you had to name this yourself:

Count the longest chain of carbons that you can find. Don't assume that you have necessarily drawn that chain horizontally. 5 carbons means pent.

Are there any carbon-carbon double bonds? No - therefore pentane.

There's a methyl group on the number 2 carbon - therefore 2-methylpentane. Why the number 2 as opposed to the number 4 carbon? In other words, why do we choose to number from this particular end? The convention is that you number from the end which produces the lowest numbers in the name - hence 2- rather than 4-.
If you want more detail goto: http://www.chemguide.co.uk/basicorg/conventions/names.html


Supporting electrolytes and increased conductivity? Hi a question about supporting electrolytes (SE) I hope you can help with.  I found some related questions here through searching but looking for a simple answer (if there is one!):
If SE are electroinactive (in the potential range of interest) and so do not pass or receive electrons at the electrodes to complete the circuit then how to they "increase conductivity"?  I am referring to typical DC experiments, e.g. chronoamperometry.  It's not like electrons hop through the solution from one electrode to the other using SE ions as a shuttle.
I understand that one role is that they migrate to the opposite electrodes to neutralize ion charge imbalances that would build up due to reagent consumption or product generation from the electroactive species you are measuring (e.g oxidizing an enzyme product), thereby facilitating these redox reactions that do occur to continue. Is this the answer?  So in just water + SE, for example, it is more conductive because they facilitate the electrolysis of water (H2 and O2 generation (e.g for SE without Cl-) by this method?
I also understand that movement of ions is considered current, but with DC wouldn't this migration slow down and exhaust as time went on? Is it this transient conductivity due to ion movement that they refer to?
Thanks!

Thanks for your answer Maurice.  What you describe sounds like this (maybe not clearly written) part of my own possible answer to my question: "I understand that one role is that they migrate to the opposite electrodes to neutralize ion charge imbalances that would build up due to reagent consumption or product generation from the electroactive species you are measuring,  thereby facilitating these redox reactions that do occur to continue".
I think my confusion arises because the IUPAC definition is:  "A supporting electrolyte, in electrochemistry is an electrolyte containing chemical species that are not electroactive (within the range of potentials used) and which has an ionic strength and conductivity much larger than those due to the electroactive species added to the electrolyte", which makes it sound like the the SE ions themselves are conductive.  Plus the most common explanation offered is along the lines of "because they are ions and ions conduct".
If I understand correctly now as discussed, they actually facilitate the reaction of the true electroactive species (which produce faradaic currents through redox - the current being a measure of conductivity), by maintaining neutrality around the electrode, rather than being "conductive" themselves (along with other roles such as making results more purely diffusion controlled, etc).
Put simply, the SE ions do not increase conductivity directly by contributing to the current (like an electroactive ion would), but increase conductivity by increasing the rate of faradaic processes of the actual electroactive species that are present (by using the SE ions to maintain electrical neutrality around the electrode).
Is this it?
Again, thanks for your time!

Remember that in electrochemistry whenever a direct current is passing through the solution there must be electrolysis occurring at the electrode. Now, all you want to see in electrochemical experiment, is current due to the oxidation or the reduction of the desired solute which is diffusing to the electrode in an unstirred cell.
The gist of your excellent and thought provoking question is:

My question is how electroinactive ions of the SE "increase the conductivity" of the solution, as is commonly stated, (when if they are electroinactive they do not produce any (Faradaic) current themselves).

After giving it a little thought and some literature search, it is clear that supporting electrolytes increase conductivity is an erroneous conclusion which is rampant everywhere. In the electrochemical experiment, the flux of the supporting electrolyte must be zero, it should not be carrying any current, because by definition, it is non-reactive in that potential window. The mass transport of the electro-inactive species must be zero.
*Yes, separately, if we were to measure the conductivity of a solution to which an supporting electrolyte, the conductivity would be higher than the pure solvent..but that is a separate measurement. Remember that conductivity is measured by very high frequency AC current.
The book, Electrochemical Engineering by Thomas F. Fuller, John N. Harb, explains it better, pg 69:


Help with missing output from cubegen utility after density calculation in Gaussian16 I am busy trying to do density calculations in Gaussian16 as I need the .cube output for input to the PIXEL software. But my Gaussian output seems to be missing an entire chunk of data and I don't know why.

The output is supposed to look like:

However, my output is missing the entire chunk of electron density data, formatted as 6E13.5 (the block of floating point values indicating electron density at each point in the block).

My input in the route section is

#MP2/6-31G** guess=core NoSymm Pop=MK Density=MP2

And I use the following to generate the .cube file

Formchk NAME.chk NAME.fchk

Cubegen 0 NAME.fchk NAME.cube 0 h

Am I missing a keyword/command to include the electron density data?

You aren't getting the density data in your cube file because you haven't specified the kind keyword in your cubegen command. Gaussian allows a variety of different types of cube files to be generated, including electronic densities/spin-densities/gradients, MOs, and electrostatic potential, along with magnetic current and shielding densities. Without specifying the type of cube file you want, the program doesn't know what to generate.

The reason it can't determine what you want by parsing the chk file is that the file in general will contain information that could be used for multiple types of cubes. The program could hypothetically use the first valid piece of information or some specific type of cube data by default, but this would often produce an undesired output. It could also make all valid cubes files, but these take up a lot of memory so this would be wasteful. They presumably decided the default case should be to do nothing if a type of cube file isn't specified so as not to produce unnecessary or incorrect output.


Basicity of singlet and triplet carbene Which form of n-heterocyclic carbene is more basic and stable? ( Singlet or triplet)

Singlet carbene is sp2 hybridised whereas triplet carbene is sp Hybridised. More s character, more nuclear force, more Electro negativity so more acidic hydrogen and hence less basicity.

So singlet carbene is more basic.


Why do the unsaturated ketones have C-O bond weaker than saturated ketones? How it is related to resonance structures? I am studying Infrared Spectroscopy and the book says that unsaturated ketones have a carbonyl group which has weaker bond because it's more single-bond character than single bond because it has these three resonance structures.

On the other hand, carbonyl group in saturated ketones have more double-bond character because it only has two resonance structures.

I can't really understand - why do the unsaturated ketones have $\ce{C-O}$ bond weaker than saturated ketones? How it is related to resonance structures?

The statement can actually be said like this. In Unsaturated aldehydes/ketones the $\ce{C -O }$ bond has significant single bond character associated with double bond characters(i.e. partial double bond character), whereas, saturated aldehydes/ketones the $\ce{C -O}$ bond has almost pure double bond character.
 In unsaturated carbonyl compounds, as the no. of stable resonating structures increase,( the structure where the $\pi$ electron cloud of $C =C$ double bond delocalises is also very stable) the contribution of  that canonical form where the $C$ and $O$ has double bond between them also decreases . So, that's why, in the resonance hybrid of the molecule( which will give the IR spectra) the bond order of $\ce{C -O}$ bond becomes significantly less, and become close to the midway between double and single bond. But in saturated carbonyls, the other resonating form is relatively very less stable ( not complete Lewis Octet and very close  charge seperation)  than the canonical form of pure double bond. So, contribution of that other form becomes significantly less. So, the resonance hybrid becomes almost close to that consisting of pure double bond.So, in the IR spectra, the streching frequency of saturated carbonyls correspond to almost perfect $\ce{C=O}$ double bond.


What is Hydrogen bond Auto-correlation function Can somebody explain the Hydrogen bond Auto-correlation function in Computational Chemistry (Molecular Dynamics and Simulations) and what does Intermittent and Continuous Hydrogen bond means?

Why the graph decays over time? and it's applications?

This is with reference to the paper: Chandra, A. Effects of Ion Atmosphere on Hydrogen-Bond Dynamics in Aqueous Electrolyte Solutions. Phys. Rev. Lett. 2000, 85 (4), 768–771. DOI: 10.1103/PhysRevLett.85.768.

In eqn 1 and 2 the correlation functions are written in the form $<A B >/<A>$ , the denominator  is the average of $A$ and the numerator the correlation function. This is the the integral product $U(t_d)=\int A(t) B(t-t_d)dt$ where $t_d$ is some time delay. When $A$ and $B$ are the same thing then this is the auto-correlation. It measure how similar the signal A is to B at different times. If $A$ and $B$ are random signals (such as white noise, or from a random number generator) then $U$ is 1 at zero time delay (both signals exactly superimposed) and zero at all other delays and this is because the correlation has no 'memory'; any point in a random signal is not related to any other. If $A$ and $B$ are constant then $U$ is one at all times, i.e perfect 'memory'. In your example the H bonds have a certain lifetime because are being broken due to molecular motion (as explained in the paragraph under eqns 1 & 2) and so the correlation function decays away. A plot of the correlation gives the lifetime of the H bonds.


Epimerization in peptide synthesis How can epimerization during peptide synthesis be detected? And how can we change or modify the coupling conditions to stop epimerization occurring?

The technique of Circular dichroism (CD) is used to detect changes and the presence of chirality in peptides and can detect both helical chirality in polypeptide chains that form helices and in individual chiral centers.

Polarimetry. The optical activity of a polypeptide (oligopeptide) has a specific rotation. A change in the optical rotation of such a peptide. Chirality is a property of a molecule as a whole. Such a change will alter the optical rotation of the peptide.


Using table salt as a way to limit humidity in an unoccupied house We own a static caravan in the UK (Snowdonia region, so rainy and wet all year round, moderately cold winters) and we leave it unoccupied during Winter (Oct-Feb).

We were advised of an apparently common practice among caravan owners which is to leave open boxes of table salt in each room, under the premise that they will capture moisture in the air and lock it away chemically in order to reduce the risk of moisture related decay in the building's structure.

I was skeptical that this would work but in the interests of science I did it anyway last winter.  I took 7 plastic ice-cream tubs and put roughly 1 kilogram of domestic table salt into each and put one in each room of the home (two in the largest room - the kitchen/living room space).

I (stupidly) never took photos of this phenomenon but I have attempted to illustrate my findings...

Fig. B is complete speculation on my part as I did not return to the home during Winter to observe it.  However in Spring when I returned, all of the tubs had experienced a change in their appearance.

All the tubs were now dry again, presumably down to evaporation due to increasing weather temperatures.  And therefore releasing all that moisture back into the building again!

Three of the tubs were largely unchanged with some noticable "caking" together of the salt into crumbly, grainy lumps which returned to normal looking salt grains when crushed.

The most profound change from the remaining tubs was as you see in Fig. C of the diagram.

The salt had actually accumulated on the walls of the tub as a fine sediment.  This suggests that water had accumulated in large amounts in the tub and had in fact risen higher than the original depth of the dry salt grains!  I'd estimate that the tub would have had to accumulate about 0.5kg of water in order for the water/salt solution to reach the depth indicated by the dry sediment.

The salt had solidified into a single, large mass.  The volume seemed to have increased noticeably but the density had also decreased accordingly, so the salt had basically expanded in it's container and solidified.  It was crumbly and brittle and some of it had been reduced to a very fine sediment.

The home is a single storey, about 12m x 4.5m x 2.5m in volume.

My questions then:

Is this a valid technique for capturing excess moisture over Winter?
Are my observations and presumptions reasonable...  Is Fig. B what really happened?
What is the chemistry / physics process that caused the salt to be transformed from Fig. A to Fig. C?
How many times did the tubs cycle between states B and C?  Was it a single cycle that lasted all of winter, or a daily cycle following ambient weather temperatures?  I could not tell just by looking at C on the last day of the experiment.

According to Transportation Information Service: Salt:

  At up to 74% relative humidity and 20°C, salt does not absorb any appreciable quantities of water vapor. The critical water content of sodium chloride (NaCl) is 0.5% at 74% relative humidity, which is the flow moisture point at which salt begins readily to absorb water vapor, and increases such that, at 75% relative humidity, the salt dissolves.

See this graph for better understanding.

So at high humidity, salt can absorb lots of moisture, enough to become liquid.

When it is liquid, it will creep up the walls of the container by adhesion.

Then, when humidity decreases, the salt will lose almost all its moist, and contract, but leave residue up to the height of the former meniscus.