In Thomson's atomic model, why did he not assume that there are positively charged particles embedded in a negatively charged sphere? Why did Thomson assume that electrons are embedded in a positively charged sphere? Why did he not assume that they are positively charged particles embedded in a negatively charged sphere?

J. J. Thomson is recognized with blowing apart a long-held belief that atoms were indivisible through his discovery of electrons while studying the deflection of discharges from cathode ray tubes.  He postulated that the particles formed in this way were small with a very high charge to mass ratio.  At this point he had two pieces of information: (1) an atom has an overall neutral charge and (2) part of the atom consisted of these small negatively charged entities which we now call electrons.  He therefore postulated that the rest of the atom consists of positive charge to balance the negative charge of electrons.  It is believed that the plum pudding analogy stems from his proficiency as an instructor and a desire to relate the atomic model to something his students could understand.


What does it mean if a catalyst support is doped? I have seen literature refer to catalyst supports being doped with another chemical, such as Nb doped $\ce{TiO2}$. What does it mean if a catalyst support is doped?

If a substance is doped, it means that some small amount of another substance has been introduced into it. More specifically, a substance is doped by a dopant, the latter of which is the minority component. The dopant can be an additive to or a replacement within the bulk material.

In your case, some fraction of $\ce{Ti}$ atoms are replaced by $\ce{Nb}$ atoms (the replacement scenario is the usual one for crystalline materials such as yours). Thus, the physical properties of your substate will be slightly different that those of pure $\ce{TiO2}$.

See this Wikipedia entry for a more complete summary of what I have written here.


Can the rate of reaction of a Voltaic/Galvanic cell be determined theoretically? I am conducting a chemistry practical in which I measure the voltage of a Zn-Cu Voltaic cell over time.

Assuming that the surface area of the electrodes are known, concentration of reactants are known, wire resistance is known (Could assume negligible for simplicity).
Is it possible to determine the rate of the reaction for this setup?
Could I then determine respective concentration of reactants? This would enable the use of the nernst equation to determine theoretical voltage.

Many thanks,

Since battery scientists got Nobel this week, this question is worth pondering. It is not a trivial question. However consider this question:

A car battery has 12 V, whereas 9 V cells are also common. If we connect two 1.5 V cells and one 9 V cell on series, one can generate 1.5+1.5+9 =12 V, yet this arrangement cannot start a car despite producing the same voltage.

What is so magical about the lead-acid car battery and this series arrangement? All this shows is that voltage and current produced by battery (which is the rate, hence kinetics and hence the rate of reaction) are two separate things. The main problem in batteries, the rate of electron transfer, is always a limiting factor as to how much current can be drawn from them. So far the lead acid battery is still one of the best because it has a very fast electron transfer rate unparalleled by other batteries. I do remember there was another Nobel Prize winner (Marcus?) who did a lot of work on electron transfer kinetics on the electrodes.

The Nernst potential you see for batteries is the potential when a very very small amount of current is drawn (ideally none) from the battery. This way of measurement of the potential is called null point potentiometry. Under those conditions, the rate of reaction is nearly zero because no current is being drawn from the battery.

In short one cannot measure the rate of the electrochemical reaction just from the Nernst potential because it heavily depends on the amount of current being drawn from the battery. Indeed a non-trivial problem worth someone's PhD.


What does $\mathrm{A}_r(\mathrm{H})$ represent when calculating the average natural weight of an elemental atom? I have a question that explains the weights and abundances of various hydrogen isotopes, and the question concludes by asking to:

  Calculate the atomic mass of hydrogen $\mathrm{A}_r(\mathrm{H})$ (with the $r$ in subscript).

The answer simply summed the mass of each isotope multiplied by its natural abundance. However, the text never mentioned the meaning of $\mathrm{A}_r(\mathrm{H})$ and I couldn't find anything online.

The relative atomic mass $A_r$ is 'the ratio of the average mass of atoms of an element ... to 1/12 of the mass of an atom of carbon-12' (from http://en.wikipedia.org/wiki/Relative_atomic_mass).

The atomic mass $m_a$ is the mass of a particular isotope of an atom, and if it is in 'u' (unified atomic mass units) it is relative to 1/12 of the mass of an atom of carbon-12.

So, to get $A_r$ you take a weighted average of $m_a$ in 'u' for each isotope.

Note also that the standard atomic weight is the relative atomic mass of an element using natural abundance on Earth (http://en.wikipedia.org/wiki/Relative_atomic_mass).


Why are the mass ratios of O to KClO3 and O2 to KClO3 the same? I just did a lab that involved heating KClO3 (with MnO2 as a catalyst) to decompose it into KCl and oxygen gas. That part is easy. What's throwing me is the % yield calculations.

The theoretical yield of O should just be the mass percentage of O in KClO3 (48.00/122.55 = KClO3 is 39.17% O by mass).

The actual yield of O2 (measured by subtraction) was 0.840 g out of a initial mass of 2.062 g KClO3, or 40.7%.

My question is this: Given that the theoretical yield was 39.17% based on moles of O (not O2) , why is the actual yield 40.7% (well with margins of error outlined by the professor) when it's based on O2? Shouldn't it be half as much? Or did I mess my calculations up?

Please feel free to edit tags, I'm out of my depth on this particular SE site in that area.

Your calculations are correct. The key is that you are measuring percentage by mass: the mass of oxygen in 2.062 g of $\mathrm{K}\mathrm{Cl}\mathrm{O}_3$ is the same whether it is considered the mass of individual oxygen atoms or of $\mathrm{O}_2$ (or of any other allotrope of oxygen, for that matter). You have half as many moles of $\mathrm{O}_2$, but the mass is the same (since the mass of 1 mol $\mathrm{O}_2$ is twice the mass of 1 mol O).


What's the difference between isoionic point and isoelectric point? What's the difference between isoionic point and isoelectric point? My teacher uses them interchangeably, but Wikipedia states that isoionic point is when charges are balanced, while isoelectric point is when there is a net charge of 0. Is this not the same thing?

The difference, according the Journal of Research of the [US] National Bureau of Standards (1940) is:

  The isoionic point is defined as the point at which dissociable groups of the substance combine equally and only, with hydrogen and hydroxyl ions [6]. This is identical with the isoelectric point only when the substance does not combine with ions other than hydrogen or hydroxyl. It follows then that although it may be possible to determine the isoelectric point by several methods, as by titration, provided that the substance combines with only hydrogen or hydroxyl ions, only an only an electrokinetic method can determine the isoelectric point in every case. The isoionic and isoelectric points of some soluble proteins have been found to be approximately the same. However, in the case of the insoluble materials, especially those which are more or less crystalline in nature, the isoelectric and isoionic points may be far apart. This is true of crystals of even relatively simple substances, such as amino acids [8]. It has been suggested that in addition to the amphoteric properties at the surface, negative ions are also adsorbed at the surface, which results in a shift of the isoelectric point to lower pH values.

In other words, "isoionic" only applies to a zwitterionic substance in pure water, whereas "isoelectric" can be used when there are ions such as Na+ or Cl- present in the solution along with the zwitterion.


How is NaOH able to clear a solution of Paraformaldehyde? While making a solution of $4~\%$ para-formaldehyde for immunohistochemistry (to fix the tissue), we add para-formaldehyde into water and stir. It looks like a chalky supersaturated solution even after stirring with a magnetic stirrer. And as per protocol, you are supposed to add 2 drops of $2~\mathrm{N}\ \ce{NaOH}$ to clear the solution. It surprisingly turned crystal clear!

I was amazed at how just 2 drops could make such a huge difference. Does it react with para-formaldehyde? Does changing pH change solubility? But para-formaldehyde is a polymer, how could it suddenly become soluble? Does adding $\ce{NaOH}$ make it a monomer?

para-formaldehyde consists of long chains of the following type:

$$\ce{HO-CH2-O-[CH2-O]_{$n$}-CH2-OH}$$

When being dissoluted in water, this needs to be broken down into formaldehyde monomers (and then further into formalin):

$$\ce{(CH2O)_{$n$} + H2O <=> $n$ CH2O + H2O <=>> CH2(OH)2}$$

Acids and bases are both able to speed up the first step. Bases by deprotonating one end and causing a domino effect of bonds being broken and formed with a hydroxide released on the other end. Acids by protonating one end and then the domino occuring in the other direction.

$$\ce{HO-[CH2-O]_{$n$}-CH2-O-CH2-O-CH2-OH ->[\ce{OH-}] \\HO-[CH2-O]_{$n$}-CH2-O-CH2-O-CH2-O- ->\\ HO-[CH2-O]_{$n$}-CH2-O-CH2-O- + H2C=O -> \\HO-[CH2-O]_{$n$}-CH2-O- + 2 H2C=O ->}\\ \dots\\\ce{HO- + ($n$+3) H2C=O}$$


Effect of vibration on wine quality Most people that have ever visited a viticulteur (winemaker) will know that it is common wisdom that vibrations are bad for the wine quality. When I was just a kid the winemaker always told me that we should be silent because the wines needed to sleep, but later a different one explained that this must have been to stop me from running around and causing vibrations.

Even scientific research (e.g. Chung et al. 2008 Journal of Food Composition and Analysis (21)8) has shown that vibrations will indeed change the wine and, among others, accelerate the aging process.

What I am wondering is why these vibrations matter? Is it a chemical effect, in which the vibrations accelerate certain chemical reactions? Or is it more physical in nature, in which the vibrations increase mixing rates and therefore somehow change the evolution of the wine?

If it helps to get your minds working: Vibrating wine in a glass

One of the effects could be that vibrations increase mixing on phase boundary between wine itself and air above it.

Other thing is that vibration are generally not good for crystal growing process (instead big and nice monocrystals one gets polycrystalline solid).  Usually you can find nice crystals of tartaric acid in the bottom of bottles and barrels. I do not know chemistry behind it but certainly it will change taste of the wine if the tartaric acid stays longer in wine (slow crystallisation) instead of falling rapidly out of solution when you shake it.

Of course if there is still a sediment earlier during fermentation, certainly that will be not good if this mixes with clear wine.


When does phenolphthalein containing solution change colour? We were estimating the amount of dissolved $\ce{CO2}$ in water using the American Public Health Association method. It was a Titrimetric Method using phenolphthalein indicator.

Titrant used was $\ce{NaOH}$ and analyte was sample water.

The following questions came to my mind:

Why does the phenolphthalein $\ce{[HIn]}$ solution change colour at pH 8.2-10 when addition of a little titrant $\ce{NaOH}$ could result into production of some $\ce{In−}$ ions, the ions that give the colour to the solution. Why does it show colour change in only  when there's $10^{-5.8}$ to $10^{-4}$ free $\ce{OH-}$ in the water?
Is the source of this resulting basic pH, the titrant $\ce{NaOH}$? I think so because all the $\ce{OH-}$ formed from the reaction of titrant $\ce{NaOH}$ and analyte $\ce{CO2}$ reacts with the $\ce{H+}$ ions of phenolpthalien. So its only possible that the shift from pH 7 to pH 8.2 is due to the $\ce{OH-}$.

I am not absolutely sure I understood the question completely, but I will try to shed some light on the colour change of an indicator.

As a rule of thumb the human eye can make out a change in colour when there is a 10:1 ratio of the components. Since phenolphtalein, $\ce{HIn}$, is without colour, and its deprotonated form $\ce{In-}$ is pink, it means that a solution with

$\displaystyle \frac{n(\ce{HIn})}{n(\ce{In-})} = \frac{10}{1}$ will appear clear, while
$\displaystyle \frac{n(\ce{HIn})}{n(\ce{In-})} = \frac{1}{10}$ will appear pink.

Since phenolphtalein is a weak acid, you will see, that the colour change will theoretically happen around equilibrium constant, i.e. $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$. (Which is also a simplification, because phenolphtalein has multiple acidic protons.) At that point you will probably not notice significant changes. Though phenolphtalein is a bit of an exception here, because it has only one colour. The human eye can probably see a smaller surplus of the colourful component. (It is much more difficult for an indicator like methyl red.)

If you substitute the above fractions into the Henderson-Hasselbalch equation, you can approximate the range in which the colour change will happen:
\begin{align}
\mathrm{pH}
  &= \mathrm{p}K_\mathrm{a} - \lg\left(\frac{n(\ce{HIn})}{n(\ce{In-})}\right)\\
  &=%
  \begin{cases}
     \mathrm{p}K_\mathrm{a} - \lg\left(\frac{10}{1}\right); & \text{clear}\\
     \mathrm{p}K_\mathrm{a} - \lg\left(\frac{1}{10}\right); & \text{pink}  \end{cases}\\
\mathrm{pH}
  &= \mathrm{p}K_\mathrm{a} \pm 1
\end{align}

When you are estimating the amount of carbon dioxide dissolved in water, then the following equilibrium will be present:
$$\ce{CO2 + H2O <=> HCO3- + H+ <=> H2CO3}$$

Adding hydroxide ions you are shifting this equilibrium to the right, essentially deprotonating the acid. Once you have fully (in first approximation) converted it to carbonate, you start deprotonating the indicator until you notice the colour change.


How can they "wash away" the caffeine from Nescafe Gold to produce that "de-caf" version? I buy and drink Nescafe Gold all the time. I also buy the smaller "caffeine-free" version of the same product. It tastes very differently, which makes me assume that the caffeine does have a lot of taste.

However, maybe it's not simply due to the lack of caffeine? Maybe the process with which they "de-caf" the original product (something about "washing it away", if I recall correctly from what I read/heard somewhere at some point) also removes other ingredients?

How can they possibly "wash away" just the caffeine from instant coffee like that?

Unless the truth is that the entire products are "built" from the ground up, separately for the caffeine-including and caffeine-less products, and the latter is not "based on" the former at all?

Even if so, they still have to "wash away" the caffeine from the raw coffee, which puts us back on square one...

In general they use different types of solvents and methods to wash away the caffeine (or at least a great percentage of it) from the coffee beans , you can find more information here : https://www.todayifoundout.com/index.php/2011/04/how-decaffeinated-coffee-is-made/


Why does thallium form TlF and TlF3 but aluminium only forms AlF3? Does this have something to do with the inert pair effect?

Ahh, Thallium is a very nice element. Because it's weird.

The +3 oxidation states are favorable except for the heavier elements, such as Tl, which prefer the +1 oxidation state due to its stability; this is known as the inert pair effect. Chemwiki.UCDavis
As an example in group 13 the +1 oxidation state of Tl is the most stable and Tl(III) compounds are comparatively rare. Inert pair effect, Wikipedia

So, why?
Inert pair effect was introduced to describe why the facts above are correct. However, the original explanation, provided by Sidgwick, relied on the fact that the $s$ subshell has an "inert" pair of electrons. This can't be correct, and Wikipedia uses the trend in ionization potentials of "$s$" electrons to prove it wrong:
$$\ce{In < Al < Tl < Ga}$$
A better explanation lies within the lower bond enthalpies of $\ce{MX}$ where M is one of the p-block heavy elements.
To reach a high oxidation number, elements will need to have a strong bond. They wanna bond in order to lose energy (i.e. become more stable) after all! If the bond is weak, this energy isn't provided. Thus, a heavy element is better off with being less oxidised.
See this and this. (Sorry, they might be behind a paywall)


Why does an acid have a low pH value? When study biology in class, I don't know how my teacher get to the topics on pH value. So, she says that pH stands for percentage hydrogen. Then, I thought there is something strange about it cause acid has a higher percentage of hydrogen then base. But why does an acid have a low pH value? Is there any mistake on the pH definition or my understanding?

Actually in pH the p is for potenz or power of hydrogen. The concentration of proton $(\rm H^+\equiv H_3O^+)$ in water is often $10^{-3},10^{-5},10^{-14}$ which is very tedious to write, so pH is defined as $3,5,14$ for these solutions. Or:
$$\rm [H^+]=10^{-pH}\iff \log_{10}[H^+]=-pH\iff pH=-\log_{10}[H^+]$$

Note that pH scale and proton concentration works in reverse direction, one increases so the other decreases.

The pH Scale at normal room temperature is from 0 to 14 (debatable) and 7 is neutral (no excess proton or no deficiency of proton) and 0-7 is acidic (excess proton) wheras 7-14 is basic (deficiency of proton). This is defined because purest form of water at this temperature has pH 7 and pH more than this (and hence concentration of proton less than this) is caused by bases which take up proton and and pH less than this (and hence concentration of proton more than this) is caused by acids which release proton.


Diesel additive spill. Odor elimination I am not chemist, however I am interested in science and know some basics.
So I have real life problem to solve and I hope there will be some people interested to help me or at least understand more how chemistry works. Anyway I would appreciate any answer.

So I have unfortunate spill of diesel addtive VIF (czech republic product). Composition is like this from the site(some translations maybe wrong):

Name - % in mixture

2-ethylhexyl nitrate - 50-100

solvent naphtha light - <5

solvent naphtha hard - <5

2,6-Di-tert-butylphenol - <5

1,2,4-Trimethylbenzene - <2.5

1,3,5-trimethylbenzen - <2.5

So I spilled this stuff in my car's trunk. Don't ask how it was, this is a fact and I cannot revert it back. It was a canister with it, so I spilled around 2 litres of it. I noticed it only after a day.

I took out the rug from the trunk and it was full of this liquid. The rug is textile. I tried to take it out with water and actually it was kind of eager to go out with water. It was lot of it, and I didn't want to do it manually. I cleaned my hands with soap and no smell was left on my hands after washing them, so I assumed I can clean the rug in the washing machine. Which was probably the worst mistake.

So instead of it being cleaned, I have

Rug that smell like the worse mechanical/technical smell in the world, however with no stains.
Washing machine which not only smells worse, but it gives this smell to anything that was washed in the machine and the smell is kind of concentrated.

So I have several questions on what can give this smell? Which component exactly? I wrote to the manufacturer with this questions, but they only told me I can try some solvent like technical petrol. But it was not an answer to the question "how to get rid of the smell"

So after reading some threads, everybody advising activated charcoal as best deodorizer.
I bought a kilo of it, and put around 200 grams to the washing machine for like 20 hours, and after that I didn't notice any change in the smell.

So the questions is: what is the formula for absorption with activated carbon? How to calculate how much of it I need to deodorize well? Will it deodorize at all? Maybe activate carbon pores are too small for the molecules or etc?

To get the smell out of your washing machine and rug you have to do two things

1.) Get the source of the smell into the water and
2.) Keep it in the water for flushing.

To get it into the water I suggest using dish soap. Dish soap is far more design to handle the greasy materials you listed and will dissolve them better than regular laundry detergent.  The draw back to this is that dish soap will form foam quite readily which laundry detergent is formulated not to do. If you have a top loading washer this can cause suds to over flow your washer.  You may obfuscate this issue by monitoring and adding rubbing alcohol as needed to suppress the suds. If front loading washer you do not need to monitor but it will take a long time, as these machines can sense suds and will wait for the suds to die down before restarting. I also suggest using LA's Awesome as it is a very effective degreaser and uses limonene (orange oil) as an emulsifier.

To keep the smell in the water I suggest activated charcoal  which can be bought for use in aquarium filters.  This will absorb the petro-chemicals and carry them away.  I suggest you use activated charcoal after a first attempt with the prior suggest method of soap and degreaser as charcoal can coat your drier affecting the next load or two, but washing a rag can help clean it out.

One of the comments mentions scaling as a possible issue.  LA's awesome has chealating agents and this should be of no issue.  If you choose to etch some scale I would suggest vinegar for this over citric acid.


Addition of combined gas law constant There are two containers connected to each other but separated by a cork. If container(1) has a gas $G_1$ at a volume $V_1$,  pressure $P_1$ and temperature $T_1$ and similarly in container(2) there is a gas $G_2$ at volume $V_2$, pressure $P_2$ and temperature $T_2$. If the cork between the containers is removed and the temperature changed to $T_3$ what is the pressure of the mixture inside the system of containers.

My textbook gives it as:

$\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}=\frac{P_3V_3}{T_3}$

where $V_3$ = $V_1$+$V_2$

$P_3=\frac{T_3}{V_3}\left(\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}\right)$

I don't understand how that comes out. Can we just add the constants(of combined gas law) like that?

Caution: This answer is exclusively from my perspective.

Moments after uploading this question I realized the answer myself. For those who strike upon this question. I 'think' the answer goes like this:

The combined gas law states that :

$\frac{PV}{T}=nR$ ----------  [or $PV=nRT$ ideal gas equation] (where 'n' is the number of moles of the gas in consideration)

Now therefore:

$\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}$ = $n_1R+n_2R$ = $R\left(n_1+n_2\right)$

Now

$n_1+n_2=n_3$ (NOTE: $n_3$ is not the number of moles of a single gas but the gases $G_1$ and $G_2$ combined)

This can be said as the volume gets added up. Therefore further referring to Avogadro's Law (of equal volume) we the latter equation yields the equation in question.

PLEASE DO CORRECT THIS ANSWER IF IT IS FAULTY.


Why is the boiling point of sulfuric acid much higher than that of phosphoric acid? Why is boiling point of sulfuric acid much higher than that of phosphoric acid?

According to the data book,

• The boiling point of sulfuric acid is 337 °C (639 °F; 610
   K) When sulfuric acid is above 300 °C (572 °F), it will decompose
   slowly.

• The boiling point of phosphoric acid is 158 °C (316 °F; 431 K) When phosphoric acid is above 213 °C (415 °F; 486 K), it will decompose slowly.

However, sulfuric acid can only form 2 hydrogen bonds per molecule but phosphoric acid can form 3 hydrogen bonds per molecule. While both of them are strong acids which can undergoes complete ionization, what other factors that I've ignored may lead to the fact that boiling point of sulfuric acid is much higher than that of phosphoric acid?

In addition to the factors Howsikan mentioned, my intuition leads me to believe there's a parallel with the boiling points of $\ce{NH3}$ and $\ce{H2O}$. I think the oxos ($\ce{O=P}$ and $\ce{O=S}$) are better acceptors than the hydroxyls, so the acids would prefer $\ce{OH...O=}$ over $\ce{OH...OH}$. $\ce{H3PO3}$ has three oxos to one hydroxyl whereas $\ce{H2SO4}$ has two of each, so $\ce{H2SO4}$  can create a favourable network more easily.


Concept of Bomb caloriemeters Considering the example where $0.3212\ \mathrm g$ of glucose (molar mass = 180) being burnt in bomb calorimeter of heat capacity $641\ \mathrm{J\ K^{-1}}$: the temperature rises by $7.793\ \mathrm K$. Is the heat (say '$E$') obtained taking $E=641 \times 7.793 \times 180/0.3212$ the internal energy or heat of combustion? Why? Answer explained in detail will be highly appreciated!

What is the fundamental difference when we completely burn fumaric acid in a bomb calorimeter? Why do we take the heat released as the internal energy? IS it because volume is constant?

Reference: Elements of Physical Chemistry (Paula, Atkins) Q.no:3.21 and 3.22

Let me list some equations first: $$ w = -p\Delta V,\qquad\Delta U = q + w,\qquad \Delta H = \Delta U + \Delta (pV)$$
where $w$ is work, $U$ is internal energy, $q$ is heat transfer and $H$ is enthalpy

Now, when you are trying to calculate the internal energy, enthalpy or anything else, you should first note what values are constant and what values change through out the reaction. For example, if I burnt a sample of glucose in an open beaker, the volume would obviously change, but the pressure would remain constant. This is because the beaker is open, which means that pressure exerted on the products and reactants is the external atmospheric pressure which is always remains a constant value. Noting the fact that pressure is constant is important as it allows us to simply the equations that I have listed above. $$q = \Delta U - w = \Delta U + p\Delta V$$ $$\Delta H = \Delta U + p\Delta V = q$$
Therefore, by realising that pressure is constant, we are able to state the $\Delta H = q$

Now lets consider the case of burning glucose in a bomb calorimeter. Here is a diagram of one below:

So lets consider what values remain constant and what values change throughout the combustion. As you should be able to see, the bomb calorimeter has a fixed volume (it can't expand or contract), hence $\Delta V = 0$. Since the bomb calorimeter is a closed object, the pressure on the system is the internal pressure, not the atmospheric pressure which is a constant. This internal pressure will change as the reaction proceeds (in this case it will increase as gases are produced from the combustion). Therefore $\Delta p \neq 0$.

Now, using the data from above, we can obtain the following expressions: $$w = 0,\qquad\Delta U = q,\qquad\Delta H = q + V\Delta p$$

So when you do the calculation: $641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K$ you are finding the total heat transfer ($q$) from the system to the surroundings (the bomb calorimeter). Now form the above information we know that the total heat transfer is also equal to change in internal energy, therefore: $$641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K = \Delta U$$

I think you were trying to find the molar internal energy for the reaction, which the calculation will be then: $$\Delta U_\mathrm m = \frac{641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K\times180\ \mathrm{g\ mol^{-1}}}{0.3212\ \mathrm g}$$


Product of reaction between aluminium chloride and trimethylamine Can anyone tell the product of the reaction between $\ce{AlCl3}$ (aluminium chloride) and $\ce{N(CH3)3}$ (trimethylamine)?

Here are a few tips to get you started:

Lewis acids are electron pair acceptors; Lewis bases are electron pair donors.
This looks like a Lewis acid-base reaction. You mention that this is an acid-base reaction but there are no easily accessible protons in either reactant. Aluminum trichloride has no protons; trimethylamine has does have protons but these are $\ce{sp^3~C-H}$ type protons - which are extremely weak. Also, there is no path to stability; deprotonation of one of the methyl groups leaves a localized negative charge on a carbon.

There are other heuristics one can use. For example, note that aluminum trichloride has an electron-deficient aluminum atom (the Al doesn't have an octet). Acceptance of an electron pair will provide it with an octet. This heuristic works fairly well; consider boron trichloride for example. Boron is similarly electron-deficient. And boron trichloride is also a Lewis acid.

Also note that the amine has a lone pair. Keeping in mind that Lewis acid-base reactions are about the movement of electrons (rather than protons) then we might reasonably suspect this to be a Lewis acid-base reaction. Lone pairs generally indicate Lewis basicity, as these are generally available for donation to a Lewis acid. Delocalization of lone pairs can weaken the lone pair's Lewis basicity. Thus, the presence of a lone pair only serves as an indicator of Lewis basicity rather than the strength of Lewis basicity. And of course you can also donate bonding electrons rather than lone pairs (halogenation of ethene would be an example; the $\ce{C=C}$ double bond is broken and the electrons previously involved in that double bond are put into the antibonding orbital of some diatomic halogen)


Mechanism of hydroboration oxidation reaction I know that the hydroboration reaction is used to add $\ce{H2O}$ to the double bond in an anti-Markovnikov fashion.

I went to the Wikipedia page and found the following mechanism:

But I could not understand as to why the addition is syn.

The addition is concerted, i.e., everything happens at the same time; there aren't multiple steps to the addition. You can expect the boron to go on one face of the double bond while the hydrogen goes on the other face. There's just no way for the hydrogen to reach over to the other side for the anti-attack. It has to be syn.


What is pure oxygen (gas) density as a function of T or P? I'm looking for a set of data (ideally graphed) of $\ce{O2}$ density as a function of $T$ or $p$ (with constant $p$ or $T$, respectively).

The best I've found is this 1948 paper which provides an empirical model but (1) it's not as recent as I'd like and (2) I want solid experimental data, not empirical fits. One can calculate ideal-gas densities by $$\frac{pM}{RT}$$ but I'm looking for real values.

A recent empirical model would also be very useful!

Something like this, which is $\rho(T)$ for air:

A colleague showed me a website that can render almost any thermodynamic quantity as a function of another for various substances: http://webbook.nist.gov/chemistry/fluid/.

Using NIST's software allowed me to get an isobar graph of $\rho (T)$ from 55K to 1000K (at 1.0atm).

And $\rho (P)$ from 0 - 800atm (at 273.15K):

It even has an HTML data-point option (and downloadable tab-delimited form). Perfect! In the case of $\ce{O2}$ it uses data from papers in the '80s and '90s (experimental and (from the software) empirical)


Why does K⁺ have 0 valence electrons? If an electron is removed from an atom of $\ce{K}$, why does it then have 0 valence electrons as it states in my General Chemistry textbook? I would think that if this happened, it would have the same exact electron configuration as $\ce{Ar}$, which has 8 valence electrons, correct? What's going on here?

What you say is correct.  The  [Ar] configuration we are left with does have 8 valence electrons.  But I think it is just semantics.  Elemental potassium has an [Ar] 4s1 electron configuration.  One would say it has one valence electron.  If we take that one valence electron away, it makes sense to say that it now has zero valence electrons because "1 - 1 = 0".


Why is absorbance spectrum continuous and not quantized? 1) In text books, the absorption event is usually described by a figure shown below1: light excites electrons to higher states in atom or molecule. The difference between the ground state and the excited one matches the energy of the absorbed light. But electron states in atoms and molecules should be quantized, so why do we observe absorbance spectra that are smooth and continuous but not spiked and quantized?

2) Where does the energy go when the excited electron returns to its normal state? In fluorescence, the electrons return to lower states and excite light at higher wavelength than the absorbed one. But if I had a molecule that did not exhibit fluorescence, just plain absorption, does the molecule emit light at the same wavelength it absorbed or does the energy turn to heat? I mean, the electron cannot be excited forever.

3) What is the physical phenomenon behind absorption coefficient? I have been thinking that it has something to do with the lifetime of the excited state – shorter lifetimes increase the probability of a photon to be absorbed, resulting in larger coefficients. Am I completely wrong?

1Image taken from http://www.photobiology.info/Visser-Rolinski.html

The image shows the absorption and emission spektra of molecules. For isolated atoms, you would indeed observe sharp lines. Think in the Fraunhofer lines.

UV/VIS absorption of molecules

In the case of molecules, absorption occurs from the vibrationally relaxed electronic ground state ($S_0$) to various vibration levels of the electronically excited $S_1$ state. This results in a broad absorption band instead of a sharp line.

The manifold of $S_1$ with different vibrational levels does not live forever. Within the shortest time, relaxation to the vibrational ground state of $S_1$ occurs. The electron distribution of the excited state is typically significantly different from the $S_0$ state. This results in a completely changed reactivity.

Radiationless deactivation

If you don't observe emission from an $S_1$ or $T_1$ state, the excited state is deactivated by internal conversion, i.e. by vibrational exchange with the bulk (solvent).


Concentration and Half Life of reactants A and B $3.12$ A reaction $\ce{A + B ->C}$, follows first order kinetics with respect to each reactant, with an overall rate constant $k=\pu{2.0 x 10^{-3} M-1s-1}$, the initial concentrations of A and B are $\pu{0.10 M}$ and $\pu{6.93 M}$. respectively. Given $\ln 2 = 0.693$, choose the correct statement(s).
(A) The concentration of A remaining after $100$ seconds is approximately $\pu{2.5 x 10^{-2} M}$.
(B) The concentration of B remaining after $100$ seconds is approximately $\pu{1.73 M}$.
(C) The half life for A is about $50$ seconds
(D) The half life for B is about $50$ seconds
Since reaction is first order with respect to both reactant A and B, I can write,

$$ k=\frac{1}{t} \ln{\frac{a}{a-x}}
 $$
where a is initial concentration of reactant and a-x is remaining concentration.
On solving the above for A and B after 100 seconds, I get $A_r = \frac{0.1}{e^{0.2}} $ and $B_r = 0.693/e^{0.2}$
The answer does not match with any options for both concentration as well as half life.
Can you please help me with this?

The whole reaction isn't first order and that can be arrived at using one simple concept. The unit of rate constant $k$ is $\pu{M^{-1} s^{-1}}$ and not $\mathrm s^{-1}$.

Since reaction is first order with respect to both reactant A and B, I can write,
$$ k=\frac{1}{t} \ln{\frac{a}{a-x}} $$

This would have been correct if and only if the whole reaction was a first order reaction. Not a first order reaction with respect to each reactant individually. For example:
$$\ce{A->B}\quad \text{Rate}=k[A]$$
That is not the case here. Here the reaction would be better phrased as:
$$\ce{A + B->C}\quad \text{Rate} = k[A][B]$$
Which means that this is a second order reaction and not a first order reaction.
Now, the other issue is that this second order reaction isn't of the form $\ce{2A-> B}$ where the rate would have been relatively easy to find using a generic law and hence we move to a more basic proof for a reaction $\ce{A + B -> C}$. As given above, the rate law for this expression is "Rate $= k[A][B]$" Now we draw a RICE table for this generic reaction to get:
\begin{array}{|l|c|} \hline
\mathrm{R} & \mathrm A & \mathrm B &\mathrm C \\ \hline
\mathrm I & a & b & 0 \\
\mathrm C & -x & -x & x \\
\mathrm E & a-x & b-x & x \\ \hline
\end{array}
Hence, we can see that for a time t, the rate of reaction can be defined as:
$$ \text{Rate} = -\frac{\mathrm d[A]}{\mathrm dt}=k(a-x)(b-x)$$
Differentiating $[A]$ with respect to $t$, we get:
$$-\frac{\mathrm d[A]}{\mathrm dt} = \frac{\mathrm dx}{\mathrm dt}$$
Substituting this in the previous equation and integrating with respect to x, we get:
$$kt = \frac{1}{b-a}\ln \frac{a(b-x)}{b(a-x)}$$
Now, in the given question, we are asked to find the value of $[A]$ and $[B]$ after $\pu{100 s}$. Therefore, $t=\pu{100 s}$, $a = \pu{0.1 M}\,,b=\pu{6.93 M}$ and $k = \pu{2 \times 10^-3 M^{-1} s^{-1}}$.
Solve for x in this equation and find the value of $a-x$ and $b-x$.
This isn't a complete solution, but the rest is just manipulating the expression using mathematics and has no actual relation to kinetics.
A hint to make calculations simpler would arise from noticing that $a<<b$ and so $b-a$, $b-x$ can both be approximated to $b$.


Structure of nitric oxide These are the two resonating strctures of NO given in my textbook. In the first resonating structure, oxygen is shown to have 9 electrons. How can that be possible since oxygen does not have the third shell so it has no orbital available to accommodate any more electron?

Also, why don't elements donate only one electron in such cases. I know unpaired electrons cause repulsions but in cases like NO where its not possible to avoid an unpaired electron, cannot oxygen donate only one electron to nitrogen. If it does that, we could have this structure. Is this structure possible?

Also, just out of curiosity, between oxygen and nitrogen, on whom will be one electron pair more stable? I am not able to decide this on the basis of electronegativity.

There is actually an erroneous explanation in your textbook. There is how the nitrogen monoxide molecular structure is explained in school: as there are no extended methods known by pupils or moreover the teacher, it's ok children to think that third unpaired electron on nitrogen atom is resonating between two atomic kernels as it's presented on picture upstaris. But it's wrong. Of course, that third electron cannot entirely belong to an oxygen atom (look to the right structure of first picure), because it (ox. atom) simply doesn't have enough free "quantum cells" to take it up to.

Things became much easier ever since people could describe the situation using molecular orbitals. Consider the molecular orbital diagram below:

It's to say, that

As the modern theory of molecular orbitals considers, there is some inter-atomic space between two atoms, where is the bonding electron allocation is the most likely situated in by the moment. There is exacly where our "third" elecron situated towards to that molecule. It doesn't ever belong to oxygen atom, but, technically, we can consider that structer as a resonant. It would me more correct, if to represent it's structure like that:

As that molecule has one antibonding electron π*2px it is a molecular diradical and it has an unstable structure. That also follows because of it's real physical-chemical properties (semi-stable only under low temperatures, and is very chemically active molecule).


Is it correct to say H2SO4 behaves like an acid in this reaction? I came across this equation :
$$\ce{ZnCO3 + H2SO4 → ZnSO4 + CO2 + H2O}$$
I wanted to know if $\ce{H2SO4}$ behaves like an acid in this reaction?

How do we generally determine if a compound acts as an acid?

In general, this depends on the definition of acid that you are using.

For this specific case, the Broensted definition would suffice.
A Broensted acid is a reagent that donates a proton ($\ce{H^+}$) to its reaction partner.

The protons of the sulfuric acid are transfered (donated) to the carbonate anion, which gives carbonic acid. Carbonic acid decomposes into water and carbon dioxide, as you will experience when drinking sparkling water.

The key thing now is the affinity to donate said proton, which leads to the concept of acid strength and can be expressed in terms of $K_\mathrm{a}$ values.

If the anions were reversed and $\ce{ZnSO_4}$ would react with $\ce{H_2CO_3}$, one would find no reaction occuring.

This is a very easy explanation of the things going on, and I'd suggest you read a bit into

Acid definitions: Broensted and Lewis acids
Acid strength: Equilibrium of dissociation reactions
Conjugated Acid / Base pairs


Why is ammonia still considered polar despite undergoing rapid inversion? Here it is stated that a molecule with a rotation axis cannot have dipole moment perpendicular to the axis, as any dipole moment would be changed upon rotation about this axis (which preserves the molecular geometry and hence the dipole).

The ammonia molecule (and other amines) are known to undergo inversion rapidly. When the molecule inverts, its dipole moment would be reversed. So why is ammonia still considered a polar molecule?

The ammonia molecule inverts itself roughly 25 billion times per second (https://en.wikipedia.org/wiki/Nitrogen_inversion , http://courses.washington.edu/phys432/NH3/ammonia_inversion.pdf). Thus averaged over a long time the dipole moment of an individual ammonia molecule will be zero, as for every moment there is a dipole pointing in one direction, this will be cancelled at a different time by a dipole in the opposite direction (as the system is charge neutral the dipole takes the same value independent of where you place the origin, so molecular motion won't affect this)

However note the word "long" in the above. Any process that happens on a short time scale will not experience this long time averaging of the property, but rather the instantaneous, non-zero, value of the dipole. So what is short and what is long in this context? Well 25 billion inversions per second means that 1 inversion takes 0.04ns. How far does a molecule move in that time? Well taking an order of magnitude value from https://en.wikipedia.org/wiki/Thermal_velocity of 400m/s a molecule moves 160 Angstroms in the time it takes one inversion of ammonia to take place. This is a HUGE distance for a molecule, thus the motion is dictated by the instantaneous, non-zero, value of the dipole. Thus we talk of Ammonia as a polar species.


Chemical bond-covalent-valence electrons OH- I would like to ask how many valence electrons took part in creating chemical bond in $\ce{OH-}$? Is the minus related to $\ce{O}$ or to $\ce{H}$? Are there any spare valence electrons which didnt take part in creating chemical bond in $\ce{OH-}$?

In the photo the orange electrons are non bond electrons.
And the black electrons are bond electrons.

There are 2 electrons took part in creating chemical bond in $\ce{OH}$.
The first one is from oxygen.
The second is from hydrogen.

This kind of bond is called "Covalent bond".
For more about Covalent bond :
https://en.wikipedia.org/wiki/Covalent_bond

The minus is  related to  the black and white electrons that are free (in Oxygen).

The white electron is come from other element.


Why are tertiary alcohols more reactive towards Lucas reagent but show low rates of esterification? Generally the order of reactivity of alcohols follows the order tertiary > secondary > primary. But why is it reverse in case of esterification.

Do you know the mechanism of esterification? The first step - which is also the rate determining step - involves the nucleophilic attack of the lone pair on the alcohol's oxygen atom on the electrophilic carbon atom of the carboxylic acid. Hence, the more nucleophilic the oxygen atom, more will the equilibrium shift to the right, favoring esterification.

Now, can you tell which degree alcohol is the most nucleophilic? And can you tell now which degree alcohol will favor esterification the most?

For Lucas reagent, the reaction involves the formation of a carbocation (at the position where the hydroxyl group originally was) by elimination of the hydroxyl group as an oxonium ion. Thus, the more stable the carbocation, the more the Lucas reagent favor the white precipitate.


Aluminum and aluminum oxide I have a new grain mill that I am hesitant to use because the rollers are anodized aluminum.  As I understand it the anodized aluminum is actually aluminum oxide and supposedly non-toxic.  I realize that the aluminum/alzheimer connection is in doubt (but I have seen it fluctuate as so many "certainties" do).  I do not wish to take a chance playing with aluminum in my food, so can you answer the following questions:

Given that eventually the aluminum oxide would wear off, exposing me to the aluminum.  Would the aluminum form a new protective oxide and if so, how fast would the aluminum re-oxidize and reform a protective barrier? (instantaneously, days, weeks?)

I am not an expert in food safety, so understand this is for information only.

Aluminum, when exposed to the atmosphere, will immediately oxidize.  Aluminum oxide is not very soluble an will likely pass straight out of the body when ingested.  Aluminum oxide is in certain toothpastes as an abrasive.

If aluminum is exposed to fluoride, chloride, or sulfate there could be an issue if it converts to a soluble form.  As long as grinders are cleaned according to the manufacturer's recommendations this should not be an issue.

The amount of aluminum oxide which wears off a grinder is related to the abrasiveness of the material being ground.  Stainless steel wears about six times less than aluminum.  A titanium grinder is even better.


Is propanone responsible for photochemical smog? Many volatile organic compounds (VOC) are responsible for photochemical smog. Propanone is a VOC. It's produced in a great amount for industrial purposes. So, propanone can be a potential contributor to photochemical smog.

I have 2 questions:

Is propanone really responsible for photochemical smog?
If it is, then what kind of photochemical reactions happen with propanone? (Please describe the mechanism also)

Most of the information is taken from Wikipedia or is mentioned therein

First some definations.

What is a VOC?

Volatile organic compounds (VOCs) are organic chemicals that have a high vapor pressure at ordinary room temperature. Their high vapor pressure results from a low boiling point, which causes large numbers of molecules to evaporate or sublimate from the liquid or solid form of the compound and enter the surrounding air, a trait known as volatility. For example, formaldehyde, which evaporates from paint and releases from materials has a boiling point of only –19 °C (–2 °F).

What is a photochemical smog?

Noxious haze of minute chemical droplets and ozone, resulting from sunlight induced chemical reaction between industrial and automobile pollutants (mainly nitrogen dioxide, hydrocarbons and aldehydes).

Indeed, propanone(acetone) is a VOC. Majority of acetone is released from industries origins. It indeed evaporates rapidly, from water and soil and has a 22-day half-life and degraded by UV light via photolysis primarily into methane and ethane or other alkyl, alkoxy or acyl radical.

As said, the main components of photochemical smog is nitrogen dioxide, hydrocarbons and aldehydes. The reaction include breakdown of hydrocarbons and aldehydes into alkyl and alkoxy radicals. These radicals react with nitrogen dioxide to form PAN(peroxyacetyl nitrate) which is the main cause of photochemical smog. The reaction mechanism is given here:

Another mechanism is given here:

What is observed is that the active component is acyl radical which is necessary in the PAN formation. Both aldehyde and acetone necessitate in acyl radical through their decomposition. From here:

  The primary photochemical decomposition of aldehydes produces organic
  free radicals. For ex: acetaldehyde decomposes into methyl and formyl
  radicals.

  $$\ce{CH3CHO + hv → •CH3 + HCO•}$$

  Similarly, acetone decomposes into methyl and acetyl radicals.

  $$\ce{(CH3)2CO + hv → •CH3 + CH3CO•}$$

But acetone contribution to photochemical smog is negligible as acetone is quite stable to irradiation and photolysis. From here:

  [...]Acetone has shown little reactivity to irradiation. Its
  contribution to produce significant amount of PAN or ozone in ambient
  air is negligible.


Which of the following octahedral complexes absorbs the photon with the longest wavelength, λ? I wonder which of the following absorbs photons with the longest wavelength, $\ce{[Cr(CN)6]^3-}$, $\ce{[Cr(SCN)6]^3-}$ and $\ce{[Cr(H2O)6]^3+}$.
I know the one to absorb the highest wavelength should also have the highest spin. Since all of these are coordination isomers, I think the strong-field ligands should have higher energies (making the answer $\ce{[Cr(CN)6]^3-}$).
I hope someone can clarify the relation between ligand field strength and energy. Any help is much appreciated!

Since CN- is a strong field /pi-acceptor ligand, therefore, it will create larger splitting. So for transition higher energy will be needed. It means that in case of Complexes with CN- higher energied photons i.e. photons with shorter wave length will be absorbed.


What were the expectations of Rutherford's gold foil experiment? At the time when Rutherford's gold foil experiment was performed, Thomson's plum pudding model was believed to be true (at least by Rutherford himself and his students).
With this model in mind Rutherford predicted that most of the alpha particles will be deflected by at most a fraction of a degree (sourced by this Wikipedia page), but why?
In my opinion, since according to the plum pudding model the mass of an atom was assumed to be uniformly distributed and the atomic mass of gold is nearly 50 times larger than the mass of an alpha particle, and gold is solid, therefore much less intermolecular space will be present, so, most of the alpha particles should rebound or get deflected by a large angle.
Help me improve my opinion.

The nucleus has a radius roughly 10⁴ times smaller than the size of the atom itself (imagine a sports ball in a stadium). That would mean that its volume were 10¹² smaller than the volume of an atom. Sure the gold nucleus is ~30 times as charged and is ~50 times heavier. But dilute that charge and mass by a factor of a trillion, and suddenly those don't seem so significant anymore.
That's the worst thing about those schematic diagrams that you see for the Rutherford experiment. They blow up the size of the nucleus to prove a point, but those diagrams are definitely "Not to Scale™".


Calculating dissociation degree from pH and Ka without given initial acid concentration Is it possible to calculate the dissociation degree (DD) from pH and Ka without directly been given the initial acid concentration?

I found the following formula:

$$\text{DD} = \frac{K_\mathrm{a}}{10^\mathrm{-pH}}$$

Is this true?

Even though the degree of dissociation $α$ depends both on the nature of the dissolved electrolyte (e.g. acid) and the concentration, for the homogeneous medium it is possible to determine $α$ by $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ only, without any auxiliary information such as initial concentration since $\mathrm{pH}$ is a function of concentration.

I would suggest you not to remember certain formulas, since they have been derived only for particular systems and cannot be universally applied. Instead, use the law of mass action, especially when it comes to multi-component systems and multi-step dissociating electrolytes.

The following is heavily inspired by the German Wikipedia article for the degree of dissociation. Consider all $n$ steps of the dissociation of the $n$-protic acid $\ce{H_nA}$:

$$
\begin{align}
\ce{H_nA + H2O &<=> H_{n-1}A- + H3O+}&\quad &K_{\mathrm{a}1} \tag{1}\\
\ce{H_{n-1}A- + H2O &<=> H_{n-2}A^2- + H3O+}&\quad &K_{\mathrm{a}2} \tag{2}\\
&\ldots &&\ldots\\
\ce{H_{n-(m-1)}A + H2O &<=> H_{n-m}A^m- + H3O+}&\quad &K_{\mathrm{a}m} \tag{$m$}\\
&\ldots &&\ldots\\
\ce{H2A^{(n-2)-} + H2O &<=> HA^{(n-1)-} + H3O+}&\quad &K_{\mathrm{a}(n-1)} \tag{$n-1$}\\
\ce{HA^{(n-1)-} + H2O &<=> A^n- + H3O+}&\quad &K_{\mathrm{a}n} \tag{$n$}
\end{align}
$$

where $n\in\mathbb{N}$, $m\in\mathbb{N}$, $m < n$; $K_{\mathrm{a}m}$ – $m$-th dissociation constant. Dissociation degree $α$ for any component of the composition $\ce{H_{n-m}A^m-}$ formed at the $m$-th step is by definition

$$α_m = \frac{[\ce{H_{n-m}A^m-}]}{c_0}$$

where $c_0$ – initial concentration:

$$
\begin{align}
c_0 &= [\ce{H_nA}] + [\ce{H_{n-1}A-}] + \ldots + [\ce{H_{n-m}A^m-}] + \ldots + [\ce{HA^{(n-1)-}}] + [\ce{A^n-}]\\
  &= \sum_{m=0}^{n}{[\ce{H_{n-m}A^m-}]}
\end{align}
$$

Each concentration in this set can be found recursively from the dissociation constant:

$$K_{\mathrm{a}m} = \frac{[\ce{H_{n-m}A^m-}][\ce{H3O+}]}{[\ce{H_{n-(m-1)}A^{(m-1)-}}]}$$

$$\implies$$

$$
\begin{align}
[\ce{H_{n-m}A^m-}] &= \frac{K_{\mathrm{a}m}[\ce{H_{n-(m-1)}A^{(m-1)-}}]}{[\ce{H3O+}]}\\
  &= \frac{K_{\mathrm{a}m}K_{\mathrm{a}(m-1)}[\ce{H_{n-(m-2)}A^{(m-2)-}}]}{[\ce{H3O+}]^2}\\
  &\ldots\\
  &= \frac{\prod_{i=0}^m{K_{\mathrm{a}i}}[\ce{H_{n-m}A^{m-}}]}{[\ce{H3O+}]^{n-m}}
\end{align}
$$

so that for $α_m$ after rearranging sums of the products the following universal equation  of two arguments ($K_\mathrm{a}$ and $\mathrm{pH}$) emerges:

$$α_m = \frac{[\ce{H3O+}]^{n-m}}{[\ce{H3O+}]^n + K_{\mathrm{a}1}[\ce{H3O+}]^{n-1} + K_{\mathrm{a}1}K_{\mathrm{a}2}[\ce{H3O+}]^{n-2} + K_{\mathrm{a}1}K_{\mathrm{a}2}\cdot\ldots\cdot K_{\mathrm{a}n}}\cdot\prod_{i=0}^{m}{K_{\mathrm{a}i}}$$

For the monoprotic acid $\ce{HA}$ ($n = 1$, $m=0$) the formula becomes much simpler:

$$α_0 = \frac{K_{\mathrm{a}}}{K_{\mathrm{a}} + [\ce{H3O+}]} = \frac{K_{\mathrm{a}}}{K_{\mathrm{a}} + 10^{-\mathrm{pH}}},$$

and if the acid is very weak ($K_{\mathrm{a}} \ll 10^{-\mathrm{pH}}$, e.g. $K_{\mathrm{a}} \ll \text{ca.}~10^{-7}$), only then you can use your formula:

$$α_0 \approx \frac{K_{\mathrm{a}}}{10^{-\mathrm{pH}}}$$


Solubility of a tribasic weak acid I have a question that reads:

  $\ce{H3A}$ is a tribasic acid. The three deprotonations can be written as

  $\ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$

  What is the predominant species in a 0.10M solution of $\ce{H3A}$ at pH 5?

  a) $\ce{H3A}$

  b) $\ce{H2A-}$

  c) $\ce{HA^{2-}}$

  d) $\ce{A^{3-}}$

I worked this out as

$\ce{Ka1 = \frac{[H+][H2A-]}{[H3A]}}, \ce{Ka2 = \frac{[H+][HA^{2-}]}{[H2A-]}}, \ce{Ka3 = \frac{[H+][A^{3-}]}{[HA^{2-}]}}$

At pH 5, $\ce{[H+] = 10^{-5}M}$

So,

$\ce{[H+] = 10^{-5} = Ka1\frac{[H2A-]}{[H3A]} = 10^{-2.9} \frac{[H2A-]}{[H3A]}}\ce{\implies \frac{[H2A-]}{[H3A]} = 10^{2.1} }$

and similarly,

$\ce{\frac{[HA2-]}{[H2A-]} = 10^{0.8}}$

$\ce{\frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$

Since $\ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $\ce{HA^{2-}}$, so c).

This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).

You can also use the Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{\ce{[A-]}}{\ce{[HA]}}$$

rearranges to:

$$\mathrm{pH} - \mathrm{p}K_{\mathrm{a}} = \log \frac{\ce{[A-]}}{\ce{[HA]}}$$

$\log \frac{\ce{[A-]}}{\ce{[HA]}} > 0$ for conjugate base dominating.

$\log \frac{\ce{[A-]}}{\ce{[HA]}} < 0$ for acid dominating.

$\log \frac{\ce{[A-]}}{\ce{[HA]}} = 0$ if they're present in the same concentrations.

The equilibrium of interest is the third dissociation.

Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($\ce{HA^{2-}}$) dominates over the conjugate base ($\ce{A^{3-}}$). You should be able to extend this for each equilibrium.


Why is ionic strength (mu) calculated differently for these 2 examples? In the following examples; I don't understand why $\ce{Na_3PO_4}$ is a much more complicated process when calculating $\mu$ compared to $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$. Specifically, I don't understand how to determine which species to include in the $\mu$ equation. All of the examples I've come across just behave just using water and the original species at most. What is more confusing also, is why doesn't the $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$ include the concentrations and charges of their conjugates as the last example with $\ce{Na_3PO_4}$ does.Thanks for any help!!! Here is the work:

$0.1000M$ of each reagent to start.

$\ce{NaH_2PO_4}$

$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] << 0.1M$$
$$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-1)^2)$$

$$\mu = \frac{1}{2}\sum([0.1M](+1)^2 + [0.1M](-1)^2)$$

$$\mu = 0.1000M$$

$\ce{NaH_2PO_4}$

$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] << 0.1M$$
$$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-2)^2)$$

$$\mu = \frac{1}{2}\sum([0.2M](+1)^2 + [0.1M](-1)^2)$$

$$\mu = 0.3000M$$

$\ce{Na_3PO_4}$

Without using Quadratic (which is just to keep things simple):
$$[OH^-]=0.044M$$
$$[PO_4^{3-}]=0.056M$$
$$[HPO_4^{2-}]=0.044M$$
$$[H^+]=\frac{K_w}{0.044M}=2.27*10^{-13}M$$
$$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [PO_4^{3-}](-3)^2+[HPO_4^{2-}](-2)^2+[H^+](+1^2))$$
$$\mu = \frac{1}{2}\sum([0.3M](+1)^2 + [0.056M](-3)^2+[0.044M](-2)^2+[2.27*10^{-13}M](+1^2))$$
$$\mu=0.5120M$$

The third pKa of phosphoric acid (going from disodium to trisodium phosphate) is 12.32.

When trisodium phosphate is added to water it is a relatively strong base, beginning to titrate H20 to OH-, to a significant degree.  In other words, to be above the pKa of 12.32, the concentration of OH- needs to be at least 0.02M.

In the other examples, since the other two pKas of phosphoric acid are 2.15 and 7.2, the pH in those examples will be about midway between 2.15 and 7.2 for the monosodium salt and about midway between 7.2 and 12.32 for the disodium salt, so the concentrations of H+ and OH- are insignificant compare to the other ions.


Does sodium borohydride reduce C=N group? Q.15 the major product of the following reaction is:

I have read that $\ce{NaBH4}$ reduces only aldehyde, ketone and acid chlorides. In some cases it may reduce double bonds conjugated with carbonyl groups. So shouldn't the answer here be option 1?

Waylander's comments answer the title question as well as I could.  Sodium borohydride, and even modified versions where some of the hydride ligands are replaced by others like $\ce{NaBH3CN}$, can reduce imines.

But, if we have only a limited amount of borohydride we will likely see the carboxyl group primarily reduced, as it has a more polar pi bond and thus it is a more reactive electrophile.  So you may in principle get mostly 1 with limited sodium borohydride, or 4 if you have enough sodium borohydride to attack both the carboxyl and less reactive imine functions.

The carbon-carbon bond here, not conjugated to either electronegatove function, is interactive to borohydride and thus products 2 and 3 are not reached.


Non-carbon based life forms on hot planets I am writing a story with people made of elements other than carbon. A couple of planets in this series are really hot, so I need compounds with high melting points.

For example, there is a lone girl who lives on Venus, which is hot enough to melt lead. I decided to use tungsten carbide for her house.

But on another planet, there are humanoids made of silicon who survive at higher temperatures than humans on earth.

Is this scientifically possible?

If you're interested in my book and don't mind a bit of self-promotion, watch for the next few years and keep an eye out for:

  The New Worlds Series #1: Into the Flames
The New Worlds Series #2: Call of the Cybans
Milky Fight #1: The Authority Gains Strength.

Given that a $\ce{C-C}$ bond dissociation energy is $83\ \mathrm{kcal\ mol^{-1}}$, and $\ce{Si-Si}$ bond dissociation energy is $52\ \mathrm{kcal\ mol^{-1}}$, a silicon based life form is much worse off on a high temperature planet. It's actually for this very reason that there aren't nearly as many silicon based compounds as there are carbon based compounds. Thats not to say they don't exist, however. There are silicon hydrides that in the form of silicon chains, analagous to alkanes, existing up to about 8 silicon long. They are all readily hydrolyzed and are thermally unstable, especially the longer chains.

Silicon is also not quite as versatile as carbon in terms of the types of bonds it can form. Carbon is capable of forming single, double, and triple bonds due to its ability to hybridize its $\mathrm{s}$ and $\mathrm{p}$ orbitals. Silicon is a larger atom than carbon, and as a result it is much more difficult for its $\mathrm{p}$-orbitals to overlap with the $\mathrm{p}$-oribatls of adjacent atoms. Again, this is of course not to say that this doesn't happen, as it does in the case of diselenes and disilynes, but again, these are (very) unstable.

Most often silicon is found combined with oxygen, as in the very stable $\ce{SiO2}$, or as you might know it: glass, sand, or quartz; not exactly ideal for a moving, living creature. Just an idea, but if you're trying to make scientifically feasible life forms on a hot planet, you might just be better off having carbon based life forms in an inert atmosphere. In other words, an atmosphere that doesn't lend to them bursting into flames.

I'm not sure how set you were on making these different races out of other elements, but no element by itself really matches the chemical variability that carbon offers.

EDIT: After doing a little more research, I thought it might be beneficial to include polysiloxanes, or silicone, which consist of $\ce{C-(Si-O)_{n}-Si-C}$ bonds. These actually are actually very interesting because they are extremely elastic, functional over a wide range of temperatures ($-100^\circ \mathrm{C}-300^\circ \mathrm{C}$), resistant to ozone and UV radiation, do not promote bacterial growth (but who knows about extraterrestrial bacteria), have a low chemical reactivity, high electrical resistivity, and most interestingly, have self-healing properties.

I would now change my tune and say that maybe a race of silicon-carbon based life forms isn't too far fetched at all! At least for sci-fi writing purposes.


Is s-p mixing referring to hybridization or is it the mixing of one atoms s orbital with the other's p orbital? According to molecular orbital theory s and p orbitals can mix if they are close enough in energy to each other. For period 2 diatomics, this occurs for $\ce{Li}$, $\ce{Be}$, $\ce{B}$, $\ce{C}$ and $\ce{N}$ - in the case of oxygen and fluorine the s and p orbitals are two far apart in energy. The effect can be seen by realizing that s-p mixing causes the $2\sigma_\mathrm{g}$ orbital to be higher in energy than the $1\pi_\mathrm{u}$ orbital.

My question is: Does the mixing of the s and p orbitals occur between the orbitals of one atom (basically hybridization right?) or does the s orbital of one atom mix with the p orbital of the other (and vice versa)?

This has confused me because I realize that concept of hybridization is widely used and functions well so I initially believed that the mixing should occur in one atom - forming hybrid atomic orbitals - and  then bonding. However, after I thought about it some more I began to doubt that this is the case. Firstly, the s and p orbitals on one atom do not have the correct symmetry to mix effectively (the s orbital would simultaneously constructively interference with one lobe of the p orbital and destructively interfere with the other). Also, the molecular orbital diagram of carbon monoxide reveals that s-p mixing must be occurring since the  $3\sigma$ orbital is higher in energy than the $1\pi$ orbital.

This also seems to contradict the idea that the s and p orbitals mix on the same atom because in $\ce{O_2}$ there is no s-p mixing so why would oxygen mix its s and p orbitals when it's bonding with carbon.

According to molecular orbital theory s and p orbitals can mix if they
  are close enough in energy to each other.

We can make that statement a bit more general so that we can apply it to all orbitals, atomic and molecular, by saying,

According to molecular orbital theory orbitals can mix if they are

close enough in energy to each other, and
have the same symmetry

  Does the mixing of the s and p orbitals occur between the orbitals of
  one atom (basically hybridization right?) or does the s orbital of one
  atom mix with the p orbital of the other (and vice versa)?

Good question, you are really thinking about this!  First off, yes, mixing of the s and p orbitals occurring between the orbitals of one atom is called hybridization.

When I think of these problems I

first, hybridize the atomic orbitals (s and p) on an atom to generate hybrid atomic orbitals
second, mix these hybrid atomic orbitals that exist on separate atoms to form a bond, a molecular orbital

But that's not how it really happens.  It is important to realize that hybridization only occurs in response to some type of interaction between two atoms.  In other words, only when atoms begin to approach and interact with one another does hybridization start to occur.  When hybridization starts to occur, bonding between the atoms starts to occur at the same time.  To first hybridize and then bond (like I think of it in my mind) is an artificial construct that allows me to sort things out.  Hybridization on an atom and bonding between two atoms both occur simultaneously.

  the s and p orbitals on one atom do not have the correct symmetry to
  mix effectively (the s orbital would simultaneously constructively
  interference with one lobe of the p orbital and destructively
  interfere with the other).

Yes they do.  As this picture shows we can add and subtract an s and a p atomic orbital to form two $\ce{sp}$ hybridized atomic orbitals.

picture source

  in $\ce{O2}$ there is no s-p mixing so why would oxygen mix its s and
  p orbitals when it's bonding with Carbon.

I think they are leaving oxygen unhybridized.  So, for example, they are mixing (adding and subtracting) a carbon $\ce{sp}$ orbital with an oxygen s orbital to form the $\ce{1\sigma}$ and $\ce{2\sigma}$ molecular orbitals in your MO diagram.

Edit: Response to OP's comment

  Do you know why s-p mixing raises the energy of the $\ce{2σ_{g}}$ orbital?
  (it lowers the $\ce{1σ_{g}}$)

Nice diagrams and a helpful explanation can be found here.  Basically, It has to do with the symmetry of the orbitals, like I mentioned above.  Look at your $\ce{1σ}$ orbital, when you add or subtract it with the $\ce{3σ}$ orbital, the symmetry is such that there is no constructive or destructive overlap. Therefore there is no interaction and there energies will not change.  On the other hand, the symmetries of the $\ce{2σ}$ and $\ce{3σ}$ orbitals are such that they can constructively and destructively overlap.  As a result they interact and their energies change, one is pushed up and one is pushed down.


Regioselectivity in Kharasch addition of bromotrichloromethane When hex-1-ene is treated with bromotrichloromethane in the presence of a peroxide initiator (e.g. dibenzoyl peroxide), what is the regioselectivity of the addition?

I know that the radical addition of HBr to an alkene is considered anti-Markovnikov, but is this reaction considered a Markovnikov or anti-Markovnikov addition?

This looks like a typical Kharasch (chain) reaction. The reaction is initiated by the formation of peroxide radicals (by homolytic cleavage of the weak O–O bond). These peroxide radicals then go on to abstract Br• from the starting material:

Note that the alternative generation of Br• is unfeasible, as radicals tend to attack at less hindered sites: hence the phenyl radical attacks the bromine which only has one attached atom, and not the carbon which has four flanking halogens. The weaker C–Br bond is also preferentially broken over the other three C–Cl bonds, so there is no chance of generating a CCl2Br• radical.

The trichloromethyl radical then adds to C-1 of the alkene, resulting in the more stable secondary radical. This can now abstract a bromine atom from unreacted bromotrichloromethane to give the product and a new Cl3C• radical.

Unfortunately, with additions of X–Y across a C=C double bond (as opposed to additions of H–X), the designations Markovnikov and anti-Markovnikov lose their meaning. A Markovnikov addition of H–X would be one where the hydrogen adds to the less substituted carbon. However, here it is not clear whether to define "Markovnikov" as the regioselectivity where X adds to the less substituted carbon or where Y adds to the less substituted carbon.


Reviving Li-ion battery in freezer? I have read that it is possible to revive a dead Li-ion battery by putting it in the freezer for three to seven days, then letting it get back to room temperature.

Can this process work, and if so, how does it work?

Notes:

Somehow the opposite, i.e. heating (cooking) the battery, is more intuitive to me. After all, in the dead battery, there is "something stuck" (crystallized?), and heat could break that. But I don't know if it's a good idea to try that at home. ;-)
I also found that hitting a Li-ion battery, against a table, for example, has some effect. At the moment, I have a battery in an IBM ThinkPad T41 laptop that, although still working, is on it's way to death. When it runs flat and I connect the laptop to mains, then the battery is not detected properly (battery status LED continues blinking orange). So, I take it out, hit it, put it back in, and the problem is solved (LED constantly orange). Also, I have the impression that hitting gives some extra charge. It could be all just a coincidence, of course.
I am asking in part for a friend whose laptop battery has been dead, I believe, already for some weeks or months. He said that the battery is not that old, but he has let it run flat in cold weather (up in the mountains). Anyhow, whether the process works for this particular battery should not influence my decision on accepting an answer. I am primarily interested in understanding a little bit more about battery chemistry, just to broaden my general knowledge.

There are a number of rumors on the internet about different ways to treat your battery. Some of these involve putting batteries in the refrigerator or freezer for varying lengths of time to improve performance in some way. Snopes has concluded that there is no basis for this. However, batteries and battery aging mechanisms are very complex and so I think it's difficult to completely rule out the possibility that under some situations there could be some beneficial effect from freezing.

Nevertheless, I would recommend that you avoid using your battery in any way that it was not designed for. The reason is that internal damage to the battery (such as from extreme temperatures, moisture, or mechanical stress) could create a safety hazard. Even if cooking, freezing, or bashing does have some positive effect, it is probably not worth putting yourself at risk. The best thing to do is to recycle your old battery and buy a replacement.


How was the potassium battery invented? It is said that mobile phones using potassium battery can be charged within a few minutes. Was it an accidental discovery or the companies made a new design for faster charging?

The search for a suitable replacement for lithium ion batteries has been ongoing pretty much since the invention of lithium ion batteries. However, the motivation for this research has primarily been cost, and shorter charge times (along with other advantages) is simply a fortuitous property of the cheaper potassium ion batteries.

Part of the reason that potassium ion batteries are much cheaper that their lithium counterparts is the relative prevalence of potassium in the Earth's crust, potassium being present at levels about 900 times that of lithium.

According to this Wikipedia article on the development of the potassium ion battery:

  The prototype device used a potassium compound Prussian blue as the cathode material for its high electrochemical stability. The prototype was successfully used for more than 500 cycles. A recent review showed currently that several pragmatic materials have been successfully used as the anode and cathode for the new generations of potassium-ion batteries.
  [...]
  The potassium battery has certain advantages over similar lithium batteries (e.g., lithium-ion batteries): the cell design is simple and both the material and the fabrication procedure are cheaper. The key advantage is the abundance and low cost of potassium in comparison with lithium, which makes potassium batteries a promising candidate for large scale batteries such as household energy storage and electric vehicles. Another advantage of potassium-ion battery over lithium-ion battery is the possibility for charging faster. This means that the next generation of mobile phones based on potassium-ion batteries can be charged within a few minutes only.

So, while faster charging times are certainly a desirable characteristic, this was not the driving force for the development of potassium ion batteries.


Is there a visible change to baking soda when it reacts with odour? For example, would a box of baking soda in a fridge lose weight as it absorbs odor? How long would it take to drop 100 g?

Let us assume that baking soda works by not simply absorbing, but reacting with acidic/odiferous substances via an acid-base reaction. So, for example if you had rancid butter in the fridge and it gave off the nasty-smelling butyric acid, that acid would react with the baking soda to give sodium butyrate plus carbonic acid (which breaks down to carbon dioxide and water). Let us assume that the carbon dioxide and water would evaporate. That would be a net gain of weight since the butyric acid molecule has a bigger mass that carbon dioxide and water.

Most odiferous substances will likely have a higher molecular weight than carbon dioxide plus water, so I believe over time the baking soda will increase in weight. This will also be true for those cases where there is no reaction - but simply absorption.


Do I have to systematically show the protons in my reaction equation to find the Pourbaix diagram's line equations? I was trying to get the line equation for the Pourbaix diagram for the couple $\ce{Ni(OH)2}/\ce{Ni}.$ I can write two chemical equations for basic and acidic conditions, respectively:
$$
\begin{align}
\ce{Ni(OH)2 + 2 e- &-> Ni + 2 OH-}\tag{R1}\\
\ce{Ni(OH)2 + 2 H+ + 2e- &-> Ni + 2 H2O}\tag{R2}
\end{align}
$$
Now I can determine the line equation for both chemical equations using Nernst equation:
$$
\begin{align}
E^\circ + \frac{0.06}{2}\log\left([\ce{H+}]^2\right) &= E^\circ - 0.06\,\mathrm{pH};\tag{1}\\
E^\circ - \frac{0.06}{2}\log\left([\ce{OH-}]^2\right) &= E^\circ + 0.06\,\mathrm{pOH}.\tag{2}
\end{align}
$$
I need to make the $\mathrm{pH}$ appear in the previous equation. To do so, I use the relation
$$\mathrm{pH} + \mathrm{pOH} = 14 \quad\Leftrightarrow\quad \mathrm{pOH} = 14 - \mathrm{pH},\tag{3}$$
which gives
$$E^\circ + 0.06(14 - \mathrm{pH}) = E^\circ + 0.84 - 0.06\,\mathrm{pH}.\tag{4}$$
I've found two slightly different expressions depending on whether I choose basic or acidic conditions. I've spend quite a few hours trying to figured out what is wrong here, and I've found a PDF Diagrammes E-pH = Diagrammes de Pourbaix (courtesy of Académie de Clermont-Ferrand) stating on page 4 that even if the species only exists in basic conditions I need to make the $\ce{H+}$ appear in the half-reaction.
Is there a convention when writing chemical equation prior to the determination of the line's equation for Pourbaix diagram? Or did I make a mistake somewhere? If yes, could help me please?

the standard potentials are given in the literature generally in an acidic medium at pH = 0. In this case, we must therefore write the equations in an acidic medium with $ H ^ + $ and not with $ HO ^ - $


Incomplete combustion I cannot find a definitive equation that describes the incomplete combustion of a hydrocarbon. Equations $1$, $2$ and $3$ are some of the variations I have found.

Is it,

$(1)$ $\ce{C_xH_y + O_2 \rightarrow CO + H_2O}$,

$(2)$ $\ce{C_xH_y + O_2 \rightarrow CO2 + CO + H_2O}$ or,

$(2)$ $\ce{C_xH_y + O_2 \rightarrow CO2 + CO + C + H_2O}$?

There is no definitive equation for an incomplete combustion. Sometimes incomplete combustion of hydrocarbons may even form species such as polyaromatic compounds and other hydrocarbons. The result may differ by variation of factors such as temperature, oxygen concentration, and the hydrocarbon undergoing combustion. Basically what you're asking is like asking "is there a definitive number between 0 and 1?" You may already know that numbers between 0 and 1 is an infinite set. Incomplete combustion may not be an infinite set, because nobody proved that it is, but it's still long ways off from a definitive one equation solve it all thing.

There is only a definitive equation for complete combustion; like there is only two numbers within a set of {0, 1}.

It's easier to recognize that any combustion that is not a complete combustion as a incomplete combustion.


Can a reaction of the form A + B → AB ever be endothermic? I understand that forming a bond between $\ce{A}$ and $\ce{B}$ is an exothermic process, with negative enthalpy, but won't there be energy required to get $\ce{A}$ and $\ce{B}$ close enough together for the bond to form?  So we would get an energy hill with activation energy on one side and heat of formation on the other side.

Is it ever possible in a chemical reaction to have more energy required to get the reactants $\ce{A}$ and $\ce{B}$ close enough together than the energy released in forming the bond?  In terms of atoms I am thinking that the repulsion between two atoms may reach a peak at some distance greater than the bond length, so we might get an energy hill diagram where:

What about a situation like breaking and forming bonds on elemental crystals like graphite and diamond?  If graphite pieces $\ce{(A)}$ are sloughed off of a large piece of graphite $\ce{(B)}$ this obviously requires energy to break the bonds, but the graphite flakes don't spontaneously re-bond to the large piece upon light contact.
Likewise, a fractured diamond does not re-bond on contact. Also, in formation of a diamond, why is there a need for a continual input of heat?  If the energy of formation is greater than the energy needed to activate the reaction, then wouldn't the released heat of formation cause other diamond bonds to form without more input of energy.  This indicates to me that some energy is required to begin the process of combining $\ce{A + B}$ to form $\ce{AB}$, even if the actual bond formation is exothermic - that there is a two sided energy hill in reactions of $\ce{A + B -> AB}$, though the activation portion may be smaller than the formation portion.
What about an analogous $\ce{A + B -> AB}$ nuclear fusion reaction with heavy species above iron?  Clearly in this case there is a large input of energy needed to fuse elements above iron, and a small release as the nucleons reach a distance that allows the strong nuclear force to dominate.  In this case the energy hill has a greater required activation energy than an energy of formation.  How is this fundamentally different from chemical bonds?
I understand that a critical component is that nuclear attraction becomes stronger than electrostatic repulsion at a critical distance, but might there be some instances of chemical reactions where $\ce{A}$ and $\ce{B}$ have to pass through a less stable conformation to get to the more stable bonded state?

So there are three possible answers:

$\ce{A + B -> AB}$ reactions can be described energetically solely by the heat of formation of the bonds between $\ce{AB}$.
$\ce{A + B -> AB}$ can be energetically described by an energy hill, but the heat of formation will always exceed the activation energy, or the energy required to move $\ce{A}$ and $\ce{B}$ through any intermediate states that are less stable than $\ce{A + B}$ (far apart).
$\ce{A + B -> AB}$ can sometimes be exothermic where the energy required to move $\ce{A}$ and $\ce{B}$ into an arrangement that will allow bonding can exceed the heat of formation of the bond between them.

A bond does not simply form because two atoms are close enough. You also have to bring them into the right electronic state to do so. Which takes various amounts of energy. So yes, an A+B -> AB reaction can be endothermic.

Also beware of mistaking the abscissa in those energy diagrams as a distance: It's usually just a symbolic "reaction progress", or it is a line on a multidimensional hypersurface describing the energy potential with respect to all angles, distances, electronic parameters etc. in the reaction.


Why are metal oxides basic and non-metals' acidic or neutral Recently I came across the fact: Metal oxides basic and non-metals' acidic or neutral.

Searching for it, partly my query is solved, as I found:

Metal oxides or $\ce{O2-}$ forms $\ce{OH-}$ after hydrolysis in the following reaction so metal oxides are basic in aqeous solutions.
$$\ce{O2- + 2H2O->4OH-}$$

I think this might happen as metal form ionic compunds and non-metals form covalent and some of them may not even react tus neutral.

I've got only a little $5\%$ of the exact reason, therefore I wish if someone could explain the whole process/fact. I considered Arhenius concept.

Note My question is partly answered here, but is not satisfactory.

We have learnt that metallic oxides are basic nature. This is because the oxides of metals like $\ce{MgO}$ or $\ce{CaO}$ or $\ce{Na2O}$ form hydroxides in their aqueous solutions.
For example:
$$\ce{K2O + H2O -> 2KOH}$$
$$\ce{MgO + H2O -> Mg(OH)2}$$

But when non-metallic oxides dissociate into their constituent ions in water, they give $\ce{H+}$ ions in their aqueous solutions forming acids
For example:
$$\ce{CO2 + H2O -> H2CO3}$$
$$\ce{SO3 + H2O -> H2SO4}$$


On the process of heating a gas at constant pressure Calculate the final temperature and the change in internal energy when $\pu{500 J}$ of energy is transferred as heat to $\pu{0.900 mol}$ $\ce{O2(g)}$ at $\pu{298 K}$ and $\pu{1.00 atm}$ at constant pressure. Treat the gas as ideal.

I am trying to understand how to approach this question without simply plugging the numbers into an equation. Can someone please explain their thought process while solving this? For example, does having constant pressure automatically mean there will be a temperature change and if so, why? What about volume? Why do we not need to know the volume for this equation? Basically, I want to understand what's going on conceptually rather than memorize a few rules. Any help would be much appreciated.

OK, let's go one question at a time.

  does having constant pressure automatically mean there will be a temperature change and if so, why?

Not quite.  Adding energy means there will likely be a temperature change. What
constant pressure means is that it will be easy to calculate this change, because you can use the value of the constant-pressure molar heat capacity, $C_p$, which is the same for all diatomic ideal gases.  It is also independent of temperature, volume and pressure.  You will find it in your book.

The molar heat capacity tells you how much energy you need to transfer to one mole of the gas to raise the temperature by one degree.  You now have everything you need to know by how much the temperature changes, and therefore the final temperature.

  What about volume? Why do we not need to know the volume for this equation?

Actually you do know the volume, or can easily figure it out.  And though the volume is not important for finding the temperature change, for the reason explained above, it's helpful in finding the change in internal energy. Why?
The change in internal energy is the energy you put in (500 J) minus whatever energy the gas spends expanding.  Since the pressure is constant the energy spent expanding is simply the product of the pressure by the increase in volume.  You can calculate the increase in volume because you know the pressure, number of moles and initial/final temperature.

Now a bit of meta advice.

  I am trying to understand how to approach this question without simply plugging the numbers into an equation. Can someone please explain their thought process while solving this?

You don't make clear what, if anything, you know (either conceptually or equation-wise) and what, if anything, you've tried.
You'll get help faster if you show you've done your part.

  Basically, I want to understand what's going on conceptually rather than memorize a few rules.

Understanding what's going on conceptually goes hand-in-hand with learning the rules.  You can't really do one without the other.  (Well, you can memorize the rules without understanding the concepts, but you won't know which ones to apply.)


Why does ozone have higher entropy than oxygen? $$\ce{3O2(g) -> 2O3(g)} \qquad \Delta H^\circ (\pu{298 K}) = \pu{+142 kJ mol^-1}$$

Since the RHS of the equation contains less number of gaseous molecules, naturally I expected the $\Delta S$ to be negative. In fact, while solving many thermodynamics questions relating to sign of $\Delta S$, I used this concept to get the correct answers (we were even taught this, that's why).

However, this appears to be an exception as this reaction does happen (thermodynamically possible due to negative $\Delta G$). It's given in my textbook (page 186, NCERT Chemistry) and the textbook accounts for it stating that the reaction has $\Delta S > 0$ without mentioning the reason for it.

Why is the entropy of ozone more that that of oxygen? Is there any simple explanation to it?

You are mixing up two different quantities (entropy of reaction, and molar entropies of the two gases involved in the reaction) and either your text book is in error, or you have misread it somehow. For that reaction, as written, the formation of ozone from oxygen,
$\Delta G$ is positive and $\Delta S$ is negative (the opposite of what you wrote in your question). It is unfavourable on both enthalpic and entropic grounds. Ozone is thermodynamically unstable with respect to oxygen under standard conditions.

Your experience, regarding the sign of $\Delta S$ and its relation to the change in the number of molecules in a gas-phase reaction, is by-and-large a good guide, because the translational contribution to the entropy typically dominates over the rotational and vibrational contributions in a gas. This reaction is consistent with this rule of thumb.

The relevant thermodynamic quantities are the standard Gibbs free energy, enthalpy, and entropy, of formation of ozone, since the oxygen is in its standard state. Taking the figures from this copy of CRC we have
$$
\Delta_{\text{f}} G^\circ (\text{O}_3) = 163.2 \, \text{kJ}\,\text{mol}^{-1}
\quad
\Delta_{\text{f}} H^\circ (\text{O}_3) = 142.7 \, \text{kJ}\,\text{mol}^{-1}
$$
and for the standard entropies
$$
S^\circ (\text{O}_3) = 238.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
\quad
S^\circ (\text{O}_2) = 205.2 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
$$
So indeed the standard molar entropy of ozone is a little higher than that of oxygen (as the comments say, if you want to explain this precisely you need to take into account internal degrees of freedom, as well as properly calculating the translational contributions), but this does not lead to a positive entropy of reaction. To calculate this, i.e. the entropy of formation of ozone, we need
$$
\Delta_{\text{f}} S^\circ (\text{O}_3)
= S^\circ (\text{O}_3) - \tfrac{3}{2} S^\circ (\text{O}_2)
= -68.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
$$
remembering that all these standard quantities of formation are per mole of ozone. Equivalently you can calculate this number from $[\Delta_{\text{f}} H^\circ (\text{O}_3)-\Delta_{\text{f}} G^\circ (\text{O}_3)]/T$ giving the same answer.

So the reaction has a negative $\Delta S$ and a positive $\Delta G$.

[EDIT following OP comment]

This means that, in the oxygen-ozone equilibrium, both forward and backward reactions are happening, at the same rates,
but the rate constants are such that the equilibrium concentration (or partial pressure) of O$_3$ is much lower than that of O$_2$.

This is not how things happen in the stratosphere, nor in most methods for preparing ozone. Generally, these involve a multi-step process, in which
the first step is the dissociation of O$_2$ into two oxygen atoms,
by absorbing a photon of UV light, or through an electrical discharge (for example). Wikipedia provides a reasonable summary of the Ozone-oxygen cycle, or Chapman cycle, with links to further resources, while various production methods are described on the Ozone page. The above equilibrium thermodynamics relations cannot be applied to processes like this, because of the injection of energy.

Instead, in the case of the Chapman cycle, a steady-state kinetic scheme can be set up for creation and destruction of O$_3$, and O atoms, where some of the rate constants depend on the intensity of UV light. The net result is a steady (small) population of O$_3$, and O atoms, accompanied by the conversion of energy from UV light into heat which is dissipated in the atmosphere. In real life, the situation is more complicated still, because of the roles of other species and reactions: books such as RP Wayne's Chemistry of Atmospheres give a lot more detail.

[2nd EDIT following further OP comments]

For the equilibrium between oxygen and ozone, the standard Gibbs free energy change for the reaction $\Delta_r G^\circ$ may be related to the equilibrium constant $K$ (see e.g. here)
$$
\Delta_r G^\circ = -RT \ln K
$$
where $K$ is expressed (for ideal gases) in terms of the partial pressures
relative to the standard pressure $p^\circ$ (1 bar)
$$
K = \frac{(p_{\text{O}_3}/p^\circ)^2}{(p_{\text{O}_2}/p^\circ)^3}
$$
For large positive $\Delta_r G^\circ$, $K$ is extremely small,
so the equilibrium partial pressure of ozone is extremely small,
if we are given that the partial pressure of oxygen has a reasonable value.
Indeed, if we take (for the reaction as written, producing two moles of ozone)
$$
\Delta_r G^\circ=2\times\Delta_{\text{f}} G^\circ (\text{O}_3) = 326.4 \, \text{kJ}\,\text{mol}^{-1}
$$
then $K=\exp(-\Delta_r G^\circ/RT)\approx 6\times10^{-58}$;
if the partial pressure of oxygen is $\approx 0.21$ bar
(as in the atmosphere around us),
then the equilibrium partial pressure of ozone is $p_{\text{O}_3}\approx 2.4\times 10^{-30}$ bar.

It is wrong to conclude that the reaction producing ozone "cannot happen"
because the free energy change is positive.
$\Delta_r G^\circ$ is simply the free energy change associated with
the complete conversion of reactants into products.
In reality, an equilibrium is always established somewhere in between.
In this case, it is very far over to one side, but not 100%.
If $\Delta_r G^\circ$ were not so large (but still positive) the
equilibrium would not be so far in favour of reactants.
Thermodynamically, you can think of the position of equilibrium
as being determined by both $\Delta_r G^\circ$ and the entropy of mixing of the reactants with the products:
this will always result in an equilibrium position that is neither 100% reactants
nor 100% products.

It is also possible to give a dynamical interpretation of chemical equilibria
in terms of forward and reverse reaction rates,
but I emphasize that this argument only applies, at least in this simplified form, to one-step chemical reactions. The oxygen-ozone equilibrium certainly involves multiple steps
(nobody is pretending that three molecules of oxygen miraculously collide
together at once). To see why this more complicated mechanism
does not affect the argument above about the equilibrium constant, see
How is it that the equilibrium constant does not depend on the mechanism? and you might also find Transition state and free energy useful. So consider the dynamical equilibrium
$$
\text{A} + \text{B} \leftrightharpoons \text{C} + \text{D}
$$
and assume that the forward and reverse reactions are simple one-step processes
obeying the rate equations
$$
\text{Rate}(\rightarrow) = k_\rightarrow [\text{A}][\text{B}],
\qquad\text{and}\qquad
\text{Rate}(\leftarrow) = k_\leftarrow [\text{C}][\text{D}].
$$
At equilibrium both forward and reverse reactions are happening at the same rate because otherwise the concentrations would be changing.
We can write
$$
\text{Rate}(\rightarrow) = \text{Rate}(\leftarrow)
\qquad\Rightarrow\qquad
K = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}
= \frac{k_\rightarrow}{k_\leftarrow}
$$
So, if $\Delta_r G^\circ$ is large and positive, $K\ll 1$,
and $k_\leftarrow \gg k_\rightarrow$.
The rate constant for the reverse reaction is much larger than that for the
forward reaction.
However, both reactions are still happening, and the rates of these reactions
are equal.

As I said, you can't apply this argument directly to a multi-step reaction scheme, but the two links I gave above show how the same ideas apply in such a case.
Also, most of the above material is covered in standard physical chemistry texts.


Applying the Le Chatelier principle I am having issues understanding the consequence(s) of a decrease in pressure on a chemical reaction.

I have understood that when we increase the pressure, the side having more number of moles wants its moles to go to the other side where it is more empty(this is how I learnt it). But I fail to understand why if we decrease the pressure by increasing volume the side having less moles goes to the side having more moles?

Also in the case of adding inert gases why does the equilibrium shift in the direction in which larger no of moles of gas are formed?

The system doesn't "want" anything.  This is a major misconception with Le Chatelier's principle.

Think about the following reversible reaction, everything in the gas phase:

$$\ce{A + B <=> C}$$

We'll simplify the rate law of the forward reaction to

$$\text{rate} = K_\mathrm{fwd}[\ce{A}][\ce{B}]$$

The reverse rate law is

$$\text{rate} = K_\mathrm{rev}[\ce{C}]$$

If the volume is doubled, what happens to the rate of the forward reaction?  It is quartered.  What happens to the forward reaction?  It is halved.  So, the reaction "shifts to reactants".

If the volume is halved, what happens to the forward rate?  It is quadrupled.  The reverse rate is doubled.  The reaction "shifts to products".

What happens when an inert gas is added?  Since the inert gas appears in neither the forward nor the reverse rate law, nothing happens.

I'm afraid you have learned some misconceptions.

Le Chatelier's principle was developed before much of what is studied in a general chem course today was known.  It has been used to manipulate the outcome of reactions.  All of the predictions made by Le Chatelier's principle can be made by understanding how reactions work at the particle level.  Le Chatelier's Principle is from the macroscopic perspective and leads many students to generate misconceptions.


Can a change in internal energy always be expressed as the product of the constant volume heat capacity and the change in temperature? It makes sense to me conceptually that $dU=C_VdT$ but is that always the case because I have seen that result derived thus:
$$dU=U(V,T)=(\frac{\partial U}{\partial T})_VdT+(\frac{\partial U}{\partial V})_TdV$$

(Not the typical expression for internal energy $dU=TdS-pdV$

But for a perfect gas $(\frac{\partial U}{\partial V})_T=0$ so for a perfect gas internal energy is only a function of $T$: $U=U(T)$. Thus, it follows that;

$$dU=(\frac{\partial U}{\partial T})_VdT=C_VdT$$

However, is this only true for a perfect gas? I ask this because, the knowledge of this equality is required to show that entropy is a state function and I was wondering whether the proof that uses the above argument is only valid for a perfect gas or whether it's valid more generally?

I will show you the proof now:

$$dU=dq-pdV=C_VdT$$
$$dq=C_VdT+pdV$$
$$\frac{dq}{T}=\frac{C_V}{T}dT+\frac{nR}VdV$$ (using pV=nRT)

Then if you differentiate appropriately you can show it's an exact differential. However, is this only valid for a perfect gas?

The statement in the title question "internal energy [can] always be expressed as the product of the constant volume heat capacity and the change in temperature" would only apply to a calorically perfect gas.

If you instead said "an infinitesimal change in internal energy can be expressed as the product of the constant volume heat capacity and an infinitesimal change in temperature", that is only true for a thermally perfect gas.

For real gas, the heat capacities depend upon both temperature and pressure.


Using electronegativity and atomic size to compare acidity of cysteine with serine In proteins, the alcohol group of serine is generally more difficult to deprotonate than the thiol group of cysteine.

Serine and cysteine respectively:

In the literature, the explanation given is that the sulfur atom is larger than the oxygen atom, and therefore the resultant negative charge is stabilised over a larger surface.

But I am confused. Oxygen is a more electronegative atom than sulfur and therefore can more stably carry a negative charge. How are these two assertions compatible?

oxygen can carry negative charge more easily because it is more electronegative, and
sulfur can carry negative charge more easily because it is larger?

Clearly, they cannot both be true and in the case of these two amino acids, the size of the sulfur prevails over the electronegativity of the oxygen. Why is that? Why in some circumstances is it the other way around?

You are correct that Oxygen is more electronegative than Sulfur.  However, the relative size of the atoms affects more than their ability to carry charge.  In this case, the larger radius of the Sulfur atom actually causes the H-S bond in Cysteine to be longer than the H-O bond in Serine.  The longer bond is weaker and, thus, more easy to break, making Cysteine easier to deprotenate.