# Tag Info

6

The problem begins with your first statement: $$\left(\frac{\partial S}{\partial V}\right)_U=0$$ Because in fact: $$\left(\frac{\partial S}{\partial V}\right)_U=\frac{p}{T},$$ which is clearly non-zero (except in limit of zero pressure or infinite temperature). More generally, being at equilibrium does not equate to a partial derivative being zero! ...

5

[Comment by Poutnik] Important is also en.wikipedia.org/wiki/Grotthuss_mechanism for the proton interchange. Mobility of H3O+ and OH- gives a hint it must be fast. If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable ...

5

The reason why statement #2 doesn't make sense to you is because it is not generally true! For a closed system with P-V work only, $$dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV = C_v dT + \left(\frac{\partial U}{\partial V}\right)_T dV$$ The difference between that statement and your statement is ...

4

Given you know and understand Charles' and Gay-Lussac's laws, it's not about chemistry, rather, simple ratios: $$\begin{cases} T \propto V\\ T \propto p \end{cases} \implies p \propto \frac 1 V$$ which, as Zenix commented, is a math form of Boyle's law.

3

The Ellingham diagram doesn't actually use molar Gibbs energies of formation $\Delta G_\mathrm{f}^\circ$ per se; it is more accurate to say that it uses molar Gibbs energies of reaction $\Delta G_\mathrm{r}^\circ$. The difference is that the formation energy is only relevant to one specific chemical equation, for example: $$\ce{Ca + 1/2O2 -> CaO} \qquad \... 3 Short answer is "not necessarily". If we isolate V in the ideal gas law, we have$$V=\frac{nRT}{P}.$$If n doubles and nothing else changes, then V does also double. But the doubling of n could also be offset by a doubling of P or a halving of T, either of which would result in no change in the volume. And yes, if the volume increases, pressure-... 3 \require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}} I would start from$$ dH = \pd HTp dT + \pd HpT dp$$This gives rise to$$ \pd HTV = \pd HTp + \pd HpT \pd pTV$$in which you can recognize some components of the solution. Since$$dH = dU +PdV + VdP $$so that$$ \pd HTV = \pd UTV + V \pd ...

3

From a statistical standpoint, the mean energy of a system is given by: $$\langle E \rangle =E \cdot P(x)=\frac{\int_{-\infty}^{+\infty} E {e^{- \beta E}}}{\int_{-\infty}^{+\infty} {e^{- \beta E}}}$$ Where $\beta =\frac{1}{k_B T}$ and $P(x)$ is the probabily of the system being at a particular energy, $E$. Now, if your energy dependance is quadric in some ...

3

Entropy is indeed a state function, and thus depends only on the state of the system. Hence it doesn't matter how you get from state A to state B, the entropy change will be the same. The analogy would be that it doesn't matter which path you use to get from the base of a mountain to the summit, your elevation change will be the same. This is because ...

2

Firstly, I would replace the word "feasible" with "favorable". I'm also going to replace $E_{cell}$ with $E_{OCV}$, where the open circuit voltage (OCV) is the potential of the cell without any applied electric potential. So you're right in that if the $E_{OCV}$ is positive, the net reaction is thermodynamically favorable: it will occur spontaneously if ...

1

Yes, you are right, the reason primarily looking at it from a theoretical point of view is the high bond dissociation energy of $\ce{N#N}$ Let's look at this aspect from every point of view Theoretical- High bond dissociation energy of $\ce{N#N}$ Kinetics- We have the Arrhenius equation given below, which states the relationship between the rate constant ...

1

We may interpret the energy variable as free energy of formation per mole of $\ce{O2}$. Thus, for instance, silicon would be favored to react with a limited amount of $\ce{O2}$ versus iron, because silica has a more negative free energy of formation per mole of $\ce{O2}$ than iron oxides; even though $\ce{Fe3O4}$ might be more negative per mole of compound ...

1

$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$ If you write $$dS=\pd SVT dV + \pd STV dT$$ then $$\pd SVU=\pd SVT + \pd STV \pd TVU \tag{1}$$ Next use $$dT = \pd TUV dU + \pd TVU dV$$ to obtain $$\pd TUS = \pd TUV + \pd TVU \pd VUS$$ or $$\pd TVU = \pd UVS \left[ \pd TUS - \pd TUV \... 1 I feel Koushal has not seen the difference between \Delta G and \Delta G°, because G of all reactants and products change during the reaction. G and G° are very different concepts. G° is the Gibbs energy of 1 mole of any reactant and of any product in the pure state at 25°C and 1 atm. G° is a constant of the substance, independent of the ... 1 The Gibbs energy of reaction \Delta_r G determines in which direction equilibrium lies, i.e. in which direction there has to be a net reaction (with a change in concentrations) to reach equilibrium. When equilibrium has been reached already, there is no net reaction (i.e. concentrations are constant). Nevertheless, at the molecular level, reactions in ... 1 Not quite an answer, but too long for a comment: Here's a link to the official CODATA site showing the values: codata.info (via the Internet Archive). It references the following source: Cox, J. D., Wagman, D. D., and Medvedev, V. A., CODATA Key Values for Thermodynamics, Hemisphere Publishing Corp., New York, 1989. Unfortunately, that source isn't readily ... 1 There's a general principle that applies anytime you calculate a change in a property: The change in the value of a property in going from an initial state to a final state is always value(final state) - value(initial state). For instance, what is the change in elevation (which we'll call height, h; not to be confused with enthalpy, H!) when we go from 500 ... 1 It really is that simple. Stating that a thermodynamic system observes Euler's Theorem can be considered axiomatic if the geometry of the system is Cartesian: it reflects how extensive variables of the system scale with size. In a homogeneous system with a Cartesian geometry all extensive properties scale identically and so any extensive property can be ... 1 Maurice has stated why equation 1 is much more exothermic than equation 2. A visualization of the processes may help you understand the details. Equation 1 involves the reaction of a carbon atom with four atoms of hydrogen to form methane in the gas phase. The bond dissociation energy (BDE) of the hydrogen molecule is +104 kcal/mol. Accordingly, the heat of ... 1 The cracking reactions transforming heavy fuel, tar, domestic or heavy oils into lighter molecules like gasoline, kerosene or white spirits, they are all carried out at temperatures higher than 200 °C. The electrolysis of alumina to produce metallic aluminium is done at about 960 °C. The production of iron from ore and charcoal is done in a blast furnace at ... 1 The key word is "mixing". If you take a glass cylinder and fill it halfway with D2O (the heavier water, so put it on the bottom) and fill the cylinder ever so gently with H2O, the mixing will be determined by diffusion (NO mechanical mixing). But the experimental techniques required to analyze and determine the rates would be expensive and time-consuming. ... 1 To solve this problem, you suppose that \Delta$$H°$and$\Delta$$S° do not change with temperature. You calculate \Delta$$G°$by two approaches :$1)$-$RT$ln$K_p$, and$2)\Delta$$H° - T\Delta$$S°$. Then$T\$ can be easily obtained.

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This is related to the difference between how Cv is measured experimentally and how it is applied in solving practical problems. To measure Cv experimentally, we hold the volume constant and determine the change in internal energy U by measuring the amount of heat added. However, for practical purposes, we know that, for an ideal gas, U and Cv are ...

1

Adding impurities(which mostly have low heat capacity, like salt) decreases the enthalpy of vaporization. This is because the impurities lower the heat capacity of the solution as a whole, making the enthalpy of vaporization lesser than the original solvent. This is why even though the boiling point of a solution increases when impurities are added it is ...

1

Adding heat is risky and could lead to overpressure or damage. I'd also expect it to boil more of the water off with the ammonia, which is probably not ideal. Not sure about efficiency per se, but you can definitely improve the effectiveness with fans. It's a heat pump that maintains a given temperature delta not between outside and inside exactly but ...

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