40

The classifications endothermic and exothermic refer to transfer of heat $q$ or changes in enthalpy $\Delta_\mathrm{R} H$. The classifications endergonic and exergonic refer to changes in free energy (usually the Gibbs Free Energy) $\Delta_\mathrm{R} G$. If reactions are characterized and balanced by solely by heat transfer (or change in enthalpy), then you'...


21

Why won't water freeze if you put ice in it? It will, even at room temperature. You just need a big enough, cold enough ice cube. Don't believe it? Add a few drops of water to an ice cube in an ice cube tray (which is the same as adding an ice cube to a few drop of water). Wait a few seconds, turn the tray upside down. No water will fall, presumably ...


21

Due to symmetry constraints ($D_\mathrm{3h}$) in $\ce{SO3}$ there 6 electrons in $\pi$ type orbitals. In a wider sense of the term this molecule is Y-aromatic, but the HOMO actually represents two electrons in in-plane lone pair orbitals of oxygen. The LUMO is an antibonding $\pi$ orbital with respect to all $\ce{S-O}$ bonds. The following valence molecular ...


18

Reproducibility is more important than stability. To obtain black phosphorus, you have to heat your sample under high pressure for quite a while, and even then it may still contain a significant number of random crystal defects, so its properties are not quite the same each time you make it. White phosphorus, on the other hand, is prepared by sublimation and ...


15

Getting a large exotherm per mole of molecules is a bit of a bad cheat if you're allowed to make molecules arbitrarily big! Energy per mass is the only honest way to do it (or energy per mole of atoms would work too I guess). However, there really is no chance to chemically match a nuclear exotherm; nuclear energies are simply in a class of their own. ...


15

The Enthalpy $H$ is defined as $H=U+PV$. Therefore, $$\Delta H=\Delta U + P\Delta V +V\Delta P$$ For an adiabatic process, $q=0$. therefore from the first law of thermodynamics, $$\Delta U = q +w =q-P\Delta V$$ $$\Delta U=w=-P\Delta V$$ Substituting this in the first equation you get, $$\Delta H=V\Delta P$$ If $\Delta P$ is zero during the process (...


12

Average bond enthalpies (note the word "average") are calculated in a different way from formation enthalpies. Formation enthalpies are well-defined and precise, meaning that two different people could measure them the same way and get the same answer. Average bond enthalpies are averages over many different types of bonds. For example, there is only one ...


12

Explanation of notation: $H$ is the enthalpy of the system. $\Delta$ means change of, so $\Delta H$ means change of the enthalpy This symbol means standard condition, standard condition is defined as a pressure of $10\text{kPa}$ and reactants and products are in their standard state, or concentration of solutions are $1\text{M}$. However, since LaTeX ...


11

A strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use $\Delta_{\mathrm{mix}}H = \Delta_{\mathrm{mix}}G + T \Delta_{\mathrm{mix}}S$ later): for isothermal and isobaric conditions it is \begin{equation} G = \sum_{i} n_{...


11

Yes, adding at least 495 kJ/mol of kinetic energy one way or another (thermally, photochemically by irradiation with photons of that energy, sonication, etc.) will cause $\ce{O2}$ to dissociate into monatomic oxygen. $$\ce{O2 ->C[energy]\ 2O}$$


11

You are on the right track - diamond is not the thermodynamically stable carbon phase at STP. Taking two figures from A.T. Dinsdale, 'SGTE Data for Pure Elements', CALPHAD 15(4) 317-425 (1991) one sees: and Since graphite is the thermodynamically stable phase of carbon at STP, it is usually selected as the reference phase so it has $\Delta H^0_f = 0$. In ...


11

As far as I can tell, your reasoning is sound, though the question is actually not written spectacularly well. By "enthalpy of the reaction" it seems to be implicitly assuming that the stoichiometric numbers are referring to moles (making all the answers have units of kJ instead of kJ/mol). The question doesn't specify standard enthalpy change of reaction, ...


11

It is purely a matter of definition. A standard enthalpy of formation describes the change in enthalpy during the formation of 1 mol of a target compound by reacting the (pure) elements it consists of, whereby each element is expected to be in its most stable modification for the given temperature. Now take the case of oxygen as an example: At standard ...


11

Because calculation with bond energies are notoriously imprecise. Those bond energies that you use do not reflect the exact C-C bond energy in ethanol, for example. It is merely an "average" C-C bond energy taken from a wide variety of organic compounds. Obviously if you use it to calculate the enthalpy change of a reaction specifically involving ethanol, ...


11

No, absolutely not. Precipitation reactions can be either endothermic and exothermic. Table 1. Thermodynamic data of precipitation for some salts \begin{array}{cccccc} \hline \text{Salt} & \Delta G_\text{ppt}^\circ & \Delta H_\mathrm{ppt}^\circ & -T\Delta S_\mathrm{ppt}^\circ(\pu{25 °C}) \\ \hline \ce{Be(OH)2} & -121 & -31 & -90 \\ \...


11

The issue here is in your definition of entropy. According to your definition: $$\mathrm dS = \frac{\text{đ}q}{T}.\tag{1}$$ However, there is a small difference. As you may know, the entropy of a system is a state function and so it only depends on the initial and final states. Therefore, if the entropy of the system were to be calculated for a specific path ...


10

I tried to use the curated dataset from Mathematica to answer this question. You can ignore the following if you have no interest in Mathematica and just jump down to the end for my ideas. The following code gets all of chemicals currently in the curated dataset and grabs the melting points and NFPA health rating. The second line finds the chemicals with ...


10

At constant volume, the pressure is not necessarily constant. The pressure will usually increase when the system is heated. This makes the $PV$ term not constant. For this reason, you get: $$\Delta H = \Delta (E + PV) $$ $$\Delta H = \Delta E + \Delta (PV) $$ $$\Delta H = \Delta E + V\Delta P $$ Now heat is just $\Delta E$, because there is no work. So, $$...


10

There is a lot of uncertainty regarding the far future of our Universe, but it seems that chemistry as we know it will be gone long before the end. Both free and bound protons (and neutrons) are expected predicted to decay through at least one of several mechanisms, with a half-life somewhere between the range of $10^{35}$-$10^{200}$ yr (far shorter than the ...


10

For example, suppose you have a solid block of TNT. It explodes and releases much energy. $\Delta H$ is negative. Gaseous products like nitrogen, carbon dioxide and water vapor are formed. The system has become more disordered, so entropy has increased.


10

This seeming contradiction can be reconciled by examining the thermodynamic quantities involved. First, per Wikipedia, the enthalpy of vaporization is "the enthalpy change required to transform a given quantity of a substance from a liquid into a gas at a given pressure." Written symbolically: $$ \Delta H_{\mathrm{vap}} = H_{\mathrm{gas}} - H_{\mathrm{liq}...


10

According to the 97th edition of the CRC Handbook of Chemistry and Physics, table 1 Bond Dissociation Energies in Diatomic Molecules [1, p. 9-76]: $$D^\circ_{298}(\ce{H-O}) = \pu{(429.74\pm 0.03) kJ mol-1}.$$ The value has been taken from the article by Ruscic et al. [2]. Note that this value is strictly for the diatomic molecule $\ce{OH}$ in gas phase and ...


9

The question to the Newton - Ask a Scientist program by Danna C Griffiths (to be found via the Internet Archive) has an answer for it. What I understood from the explanation on that site: Solubility is defined as the concentration of the solute in a saturated solution. So when considering the increase in solubility with temperature, you have to check the ...


9

For free radical reactions, the most important parameter in assessing bond strength is bond dissociation enthalpy (BDE). Typical values for $\ce {C-F}$ bonds are around $\mathrm {100\ kcal/mol}$, while for $\ce {C-Cl}$ bonds are around $\mathrm {80\ kcal/mol}$; $\ce {C-Cl}$ bonds are therefore weaker. This can be rationalized by considering the poorer ...


9

NIST webbook does have a lot of data, though they are not in any kind of an API form as far as I know. http://webbook.nist.gov/chemistry/


9

To preface, the reason that this is not a combustion reaction is because this is a redox reaction that occurs at room temperature. In fact, chloroform is noncombustible. Though at first glance, this may simply appear to be an unbalanced equation that could be rewritten: $$\ce{CHCl3 + 1/2O2 -> COCl2 + HCl}$$ It is better to take it for what the question ...


9

It is possible, in a significantly different way that you envision. Freezing point depression via the cryoscopic constant is an example of a colligative property, which holds only for relatively dilute solutions. Once the solutions get very concentrated, intermolecular interactions become more complicated and do not generalize, meaning different compounds ...


9

With enough effort, Born–Haber cycle can be extended to polyatomic ionic solids, however it's practically never done in practice due to the lack of experimental data or because it's impossible to obtain any. From [1, p. 117–118] (emphasis mine): Lattice energies cannot be measured experimentally since they represent hypothetical processes: $$\ce{M^n+...


9

The textbook is referring to the entropy change of the system. While the textbook is correct that absolute zero can never be attained, its statement that the entropy change is infinite is wrong. The authors' rationale for thinking it is infinite likely stems from a misinterpretation of the definition of entropy change: $dS = \frac{\text{đ}q_{rev}}{T}$ where ...


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