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38

Well, a lot of things happen to the reactants. Some bonds stretch (and maybe eventually break), the others shrink, and your molecules morph into different molecules, which are the products. (source) As for staying at the very peak, that would be kinda unnatural, but luckily, not every peak looks like this; sometimes there is a tiny dent near the top, and ...


28

The combustion of alkanes like butane is fearsomely complicated involving dozens of transient compounds and hundreds of different reaction. If you have a few spare hours there is a dissertation that presents a nice summary of the process here (this is a 1MB PDF). A butane molecule is pretty stable and doesn't react with oxygen on contact so you need some ...


20

It is all about minimizing the energy of a molecule. In the case of carbon, the only molecule that adopts a perfect hexagonal geometry in its ground state is benzene (and its derivatives that possess a 6-fold rotational axis). In this case the hexagonal geometry is adopted because all of the carbons are $\ce{sp^2}$ hybridized. The ideal geometry (lowest ...


20

Unfortunately, the question as stated is thermodynamically impossible. Let's look at the proposed reaction: $$\ce{CO2(g) -> CO(g) + O(g)}$$ This reaction is simply a bond dissociation (specifically, a carbon-oxygen covalent double bond is broken). We can look up the enthalpy change associated with it. From a table of values on Wikipedia, we find in the ...


20

Better is to say "Energy released by forming bonds of combustion products is bigger than energy needed to break bonds of combustion reactants.". Particularly breaking $\ce{C-C}$, $\ce{C-H}$ and $\ce{O=O}$ bonds needs less energy than is released by forming $\ce{O-H}$ and $\ce{C=O}$ bonds. The nature of released energy of chemical bonds is ...


19

Absolutely yes. Lighting a torch in such an environment would simply be the reverse physical process (and same chemical process) of what is done in our oxygen-containing atmosphere. In the chamber or alien world of hydrogen gas, providing an ignition source to a stream of oxygen would give a flame. The chemical reaction would actually be the same as if ...


18

There is a simple (some might say simplistic) way to get an intuition about this and it involves thinking at the molecular or atomic level rather than about the bulk properties of the reaction (the thermodynamic view). For a simple reaction where molecule A has to bang into molecule B to create molecule C, the reaction will only happen if the amount of ...


17

Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of ...


17

Now that's a mildly non-trivial observation. Why would they be equal, really? Let's say a particle with mass $m$, charge $q$, and initial velocity $v$ enters an area of length $L$ where an electric field $E$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $x=vt,\;y={at^2\over2}$, where the ...


15

Bond formation is alway strictly exothermic in the sense of the change of enthalpy. exothermic reaction A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative. A bond can only exist, if it needs energy to break it, i.e. the bond dissociation energy is always positive. bond-dissociation energy, $D$ The enthalpy (per mole)...


15

Authors may be sloppy about notation in this matter. I recommend considering $R_\ce{H} \approx \pu{10973 cm-1}$ and $Ry \approx \pu{2.18e-18 J}$, noting $Ry = hc \cdot R_\ce{H}$. Units of wavenumbers $(\pu{cm-1})$ and energy are often considered interchangeable in practice because they are proportional to each other by the constant value $hc$. In my notes, ...


15

This is a very fundamental question and for really understanding the "why" some advanced physics is involved. I will describe the process rather superficially. As you might know, the level energies of atoms and molecules can be calculated (in principle) using quantum mechanics. The simplest system is the hydrogen atom as it consists of a single ...


14

Internal energy, $U$, is the total energy contained in a thermodynamic system. However, absolute internal energy is hard to determine, and even relative internal energy and changes to internal energy are hard to determine. Here's why: Changes to internal energy usually occur through heat transfer $q$ or work done $w$. There are other ways to transfer energy ...


14

$\ce {MgO}$ ($\approx 3800$ $\pu{kJ mol^{-1}}$) has higher lattice energy than $\ce {LiF}$ ($\approx 1045$ $\pu{kJ mol^{-1}}$) mainly because of the greater charge on $\ce{Mg^2^{+}}$ ion and $\ce {O^{2-}}$ as lattice energy is directly proportional to the charges of the combing atoms. Your reasoning isn't incorrect but remember that $\ce{Li}$ and $\ce{Mg}$...


13

A bond is formed between the oxygen of water and the phosphorus of the gamma-phosphate. Here is a good link. Bonds are both broken and made in chemical reactions but many biology teachers and textbooks state that "Breaking ATP bonds releases energy." In reactions bonds are broken and made. If the strength of the bonds formed exceeds the ...


13

In order to answer this question, one needs to define what the absolute energy of a system is. Energy can be trapped in a system in ways not yet discovered or fully understood. Think of the energy associated with mass ($E=mc^2$), which is a result that was not known to the early founders of thermodynamics. We need to define reference points, which we can ...


13

Of course it would break, just like you said; also, a high-energy $\beta$ particle would kill quite a lot of bystander molecules. Also, if not for other reason, the resulting molecule would no longer be DNA , since the decayed atom would no longer be $\rm P$. Also, the product would no longer be radioactive, so we wouldn't be able to detect it anyway. The ...


12

The butane and oxygen in the room are in a metastable state. An example energy diagram of a metastable state is shown in the image below. Before throwing in the cigarette, the butane and oxygen molecules are in the state labeled as 1. This state is metastable because there is a lower energy state (which makes it statistically more probable) available if ...


12

In short, no, the standard Gibbs free energy change is not constant; it is a function of temperature. The same is true for practically all other standard-state quantities. This gets a little confusing because of how standard-state properties are often explained in lower-level (high-school and college-freshman level) textbooks. The standard Gibbs free energy ...


12

The initial and final thermodynamic equilibrium states of your system are as follows: State 1: $\pu{1kg}$ liquid water at $\pu{0 ^\circ C}$ and $\pu{1 atm}$ State 2: $\pu{1 kg}$ water ice at $\pu{0 ^\circ C}$ and $\pu{1 atm}$ You want to find the change in enthalpy, entropy, and Gibbs free energy between these two states. Now, we know that, to get the ...


11

Single point energy arises in the framework of the Born–Oppenheimer approximation and corresponds to just one point on the potential energy surface. Physically it is the total energy of the molecular system with its nuclei beeing fixed (or clamped) at some particular locations in space. In other words, it is total energy of the molecular system within the so-...


11

No, nickel-62 is the most stable on a binding energy per nucleon basis. Fe-58 is second and Fe-56 is third. See Fewell, M. P., "The Atomic Nuclide with the Highest Mean Binding Energy", Am. J. Phys., vol. 63, pages 653-658.


11

Yes, potential energy increases with increasing temperature for at least the following three reasons: At a higher temperature, more atoms/molecules are in excited electronic states. Higher electronic states correspond to greater potential energy. Potential Energy is -2 times Kinetic Energy. So actually, at higher temperature, when more atoms are in higher ...


11

Dioxygen, $\ce{O2}$ is a very special molecule. A good majority of organisms on earth use dioxygen to breathe and survive. Oxygen is also the second most abundant element in the sky, as well as the most abundant elements in the earth's crust. What makes it so special? Its because of its abundance, which is obvious, and also because of it's strong oxidizing ...


11

Rydberg constant $R_∞$ is usually given in reciprocal length units historically and because it's determined from hydrogen and deuterium transition frequencies [1]. Current value (in $\pu{m-1}$) is listed at NIST [2] website (accessed 2019-06-05): $$R_∞ = \pu{10973731.568160(21) m-1}$$ Since it's an energy unit, one can convert it to SI rather trivially via ...


10

the whole reaction, i.e. for 2 moles of the product? This is your mistake. The balanced equation $\ce{2H2 + O2 -> 2H2O}$ does not mean that 2 moles of the product are being formed. The numbers, called stoichiometric coefficients, only tell you about the ratio in which the reactants react and form products. That is to say, if I had a system in which I ...


10

Your question is quite broad, so I will tackle it in parts. Element abundance there is nothing as familiar to mankind (in terms of the frequency in their occurrence) as the reactions involving this compound This is because frequency depends on abundance. If you have a look to the element abundance on the earth oxygen is the first: Element Amount ...


10

As a chemist, I would agree that textbooks are not clear about the term stability and energy. It is not your fault and undergraduate organic chemistry books make the situation worse. I cannot recall the text, it was an old book, however it clearly said that stability of a compound does not mean anything. We should always ask, stability with respect to what? ...


10

Notice that when $n=1$, we have, $$ E=-E_0=-13.6~\mathrm{eV} $$ which is the negative of the energy required to remove an electron from the ground state of a hydrogen atom. If we increase $n$ to say $n=2$, then we have, $$ E=-E_0/4=-3.4~\mathrm{eV} $$ which is a larger number than for $n=1$. Don't let the minus sign confuse you. This is a very common ...


10

Here's your confusion: You need to consider two different things: The momentum transfer per particle per collision. There, since we assume an instantaneous collision, it doesn't make sense to try to figure out force from acceleration. [I suppose you could do this using limits, and maybe there are applications in which that does make sense, but adding that ...


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