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2

This is in answer to (primarily) the third question. The association with disorder is naturally anthropocentric, "disorder" is an idea based on our experience of the world. The association with "disorder" is straightforward once "order" is properly defined, which, given the number of meanings attributed to the word, is not ...


1

Such books forget to mention, or you may have not noticed such notes, that $\mathrm{d}T=0 \implies \mathrm{d}U=0$ does not apply to phase transitions because there is heat exchange at constant temperature, related to different energy of different phases at the same temperature. E.g. by water boiling, water gains about 5.5 times more energy than by heating it ...


1

The mathematical answer is based on Clapeyron's equation (written in what follows in form inverted relative to the more usual): $$\frac{dT}{dp}= \frac{\Delta V}{\Delta S}$$ which must hold on any point on the phase coexistence line. If $\frac{dT}{dp}=0$ then it follows that $\Delta V=0$.


3

In general for a perfect or ideal gas, $$C_p=C_V + R'$$ (using your notation) where the heat capacities are molar quantities. It follows that for a perfect gas mixture $(C_p)_\text{mix}=(C_V)_\text{mix} + R'$.


7

The issue seems to be not the reaction between $\ce{P4O6}$ and water, but whether $\ce{P(III)}$ can form at all. The OP's first reference says no whereas the second reference says yes. Basically the two references used different thermodynamic assumptions. The first one assumes an atmosphere in equilibrium while the second assumes only a partial ...


0

A non-spontaneous reaction does not proceed without the continuous supply of energy: A spontaneous reaction is one that, once started, continues without assistance. Under a particular set of conditions (e.g. room temperature), the reverse of a spontaneous reaction will always be non-spontaneous. For example, if A to B is spontaneous at 300 K, B to A is non-...


0

Well, essentially, the Gibb's Free Energy Change of a reaction is the property that determines how much a reaction wants to progress in any particular direction. It's a property denoted by ΔG. Basically, the more negative the value of the Gibb's Free Energy change, the more the reaction "wants" to happen. Essentially, the reaction tends to achieve ...


2

To carry out the B transformation, you have to heat a lot. If you don't heat, the volume will of course increase, but the pressure will decrease. B is a process difficult to carry out, because it is not easy to heat enough a gas who is inflated to as to maintain the inner pressure. Then, at the end of B, the temperature is very high. You block the position ...


0

Start by remembering that when we say that a process is spontaneous we mean under certain conditions e.g. at room temperature. The reverse of spontaneous processes are non-spontaneous processes. Non-spontaneous processes do not occur naturally according to classical thermodynamics and it is statistically extremely improbable that they will occur naturally ...


1

There are at least two fundamental issues you have to address. First, you have to distinguish between the Gibbs energy $G$ and the Gibbs energy of reaction $\Delta_r G$. In you diagram, one is the value on the y-axis (without defined zero point) and the other is the slope of the line, as labeled in your sketch. The second issue is that in the expressions on ...


1

I can't see what you did in your derivation, but, for what it's worth, here's my derivation. My starting equations are $$G=\sum{n_i\mu_i}$$and, along your contour, at constant temperature and pressure, $$\mathrm dG=\sum{\mu_i\mathrm dn_i}$$The changes in the number of moles of the various species are given by $$n_i=n_{i0}+c_in$$where $c_i$ is the ...


0

You heat your calorimeter with Al and chlorine (electrically, recording the energy you put in) from 298K until the reaction starts. You record the final temperature after the reaction. After cooling the calorimeter down to 298K, you heat it (with the AlCl3 in it) until it is at the same temperature again, and record the energy needed. Now you subtract the ...


1

The molar enthalpy of formation at room temperature is not dependent on the reaction effectively happening at the room temperature. It follows the general principle of state variables, being related to the Hess's law, which refers to the fundamental law of energy conservation. The principle says, that the change of enthalpy ( or any other state variable ) ...


1

Think about Coulomb's inverse-square law: $$F = k\frac{q_1q_2}{r^2},$$ where $q$ is the magnitude of the charge, $r$ is the distance, and $k$ is the Coulomb’s constant. Mathematically speaking, as $r$ increases, the magnitude of $F$ becomes smaller. As far as the charge density, as it increases, the magnitude of $F$ increases. So both explanations are saying ...


0

What am I doing wrong here? You are missing the possibility of non-PV work. According to the second law of thermodynamics: $T\,\Delta S\ge\Delta Q$ and to my understanding, $\Delta H$ is just the same as $\Delta Q$. $\Delta H$ is just the same as $\Delta Q$ under the following conditions: You are working at constant pressure, and the only work is ...


0

This is a very difficult question you have asked, and it has taken me a few days to figure out the answer. This question is closely related to a Challenge Problem that I issued in PhysicsForums.com 3 years ago. Even though I knew the answer to the challenge problem, it still took me significant time to figure out the connection to your question. Here is ...


0

In the absence of non-PV work (at constant pressure), the change in enthalpy of the system is equal to the heat exchanged between system and surrounding. This follows from the first law of thermodynamics and the definition of enthalpy. The equation you are showing is based on the definition of the Gibbs energy (the chemical potential is defined via the Gibbs ...


2

The electrode redox reactions of the Daniell cell: $$\ce{Zn(s) <=> Zn^2+ + 2 e- }$$ $$\ce{Cu(s) <=> Cu^2+ + 2 e- }$$ maintain at the respective electrode a particular potential where both opposite reactions have the same rate, implying there are no external galvanic causes that affect this potential. Rates of oxidation electrode reactions ...


2

BTW, electrons don't actually actually "flow" very fast. The drift velocity of electrons is incredibly slow, on the order of a millimeter every 5 or ten seconds in a room-temperature conductor. The propagation of current is like the particles in Newton's Cradle, where the electric field bumps each electron down the line, a much faster process. ...


1

Here I'm assuming you are interested in comparing the magnitude of $\Delta S^{\circ}_{fus}$ at $T_{fus}$ with that of $\Delta S^{\circ}_{vap}$ at $T_{vap}$, for the elements in their standard states (so, for instance, hydrogen would be $\ce{H_2}$ rather than $\ce{H}$). Like you, I was unable to find a comparative tabulation of these values on the internet. ...


1

As already said in the comments, you cannot use any unit you want in the reaction quotient; it has to be formulated in terms of activity (for solutes) or fugacity (for gases). The "standard state" of a chemical species is a reference state in which the properties of the chemical species are defined. This standard state is a theoretical state, it ...


5

The reason might be that while drawing the reaction energy profile, we forget to mention what energy we are mentioning in the Y-axis. The following conventions are generally used: If reaction conditions are constant NVT, energy in Y-axis should represent internal energy. If reaction conditions are constant NPT, energy in Y-axis should represent enthalpy. If ...


2

An isolated system is one that can exchange no mass, heat, or work with the surroundings. So, from the first law, the change in internal energy of an isolated system in any process within the system is $\Delta U=0$. Now for my analysis of this problem in greater detail to determine the enthalpy change. I'm taking as the initial state of the system one mole ...


8

All flammable organic liquids could, in principle, create large fuel-air explosions but the conditions are very hard to achieve accidentally The first important point to note is that benzene should not be a particular worry. There are plenty of other, widely used, chemicals and mixtures that could, in principle, cause large explosions. Standard automotive ...


0

Could all benzene explode at once? No, this is highly unlikely. While liquid benzene is flammable, it is not explosive. In the worst case, there will be a large fire, but chances of spontaneous combustion as compared to Trinitrotoluene are very, very low. The explosion in Jilin in China took 1 hour so that seems to be the case. According to the same ...


5

$\Delta G^0$ is change in free energy change in going reversibly from molar stoichiometric quantities of pure reactants, each at temperature T and pressure 1 bar, to corresponding stoichiometric quantities of pure products, each at temperature T and pressure 1 bar. So $\Delta G^0$, $\Delta H^0$, and $\Delta S^0$ are all functions of T, but not pressure (...


0

A wonderful explanation by All difference between follows: 1. $\Delta$ G relate to these two quantities ΔH° and ΔS° while $\Delta$ G° does not relate to the quantities: $\Delta$ G definitely does relate to those two quantities. ΔH° and ΔS° represent the change in enthalpy and entropy between product and reactant but they do not mean a “100% complete ...


2

If I've understood your question properly, you are asking - in the condition $E^\circ_{\ce{M^3+/M^2+}} =\pu{4 V}$, $E^\circ_{\ce{M^3+/M}} =\pu{1 V}$, $E^\circ_{\ce{Y+/Y}} =\pu{3 V}$ - whether $\ce{M^3+}$ is a stronger oxidising agent than $\ce{Y+}$ irrespective of it going to $\ce{M \text{or} M^2+}$. This would not be true since $\ce{Y+}$ has a higher ...


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