New answers tagged thermodynamics
1
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-...
-1
In the dissociation of $H_2$ into $2$ $H$, one mole of something yields $
2$ moles of something else. The number of moles increases. As $V$ is proportional to $n$, at constant pressure, the volume $V$ of the gas must increase.
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 ...
0
For the Ellingham diagram:
(1) We write down the reaction/reactions that we are interested in. The $\Delta G_f$ (formation) per mole of product is calculated using the individual free energies of any intermediate reaction, and using Hess' law to sum them up.
(2) This energy is then normalised per mole of $O_2$, using stoichiometry.
(3) The normalised ...
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! ...
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 ...
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:
$$\ce{Ca + 1/2O2 -> CaO} \qquad \...
1
$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$
If you write
$$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 \...
3
Proving a thermodynamic relation between $(\partial H/\partial T)_p$ and $(\partial U/\partial T)_V$
$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$
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 ...
-2
Hopefully I can provide you with some physical intuition on this phenomenon through a simple example.
Imagine a gas heated by a flame in a closed cylinder with a piston on the top balanced by the same pressure as the gas inside. As you heat the gas, the particles move faster (gain kinetic energy) and collide with the walls of the container harder, ...
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 ...
0
At the triple point the water system is assumed to be adiabatic. The system has 3 phases, ice from pure water, liquid water, and pure water vapor. The various equilibria are shown below.
Note that the ice floats on the water. So the ice is in contact with both the liquid phase and the gas phase. Furthermore The ice can't cover the whole surface like a ...
-1
The triple point of water is an example of Thermodynamic Equilibrium, which is explained by Wikipedia as:
simultaneously in mutual thermal, mechanical, chemical, and radiative equilibria
Your equation only covers the chemical facet of this equilibrium and hence is not the correct way of representing total thermodynamic equilibrium. I am also not sure of ...
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 ...
0
The enthalpy change is always Products minus Reactants.
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 ...
0
The Gibbs energy has not been derived this way. It came from a thought process. About 100 years ago, Gibbs was puzzled by the very existence of spontaneous endothermic reactions. In mechanics, objects always fall down. They never "fall up" spontaneously. They never gets more energy spontaneously. In exothermic reaction, the matter looses spontaneously energy,...
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 ...
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 ...
0
The heat of formation of $CH_4$ is $-75$ kJ/mol. This is not much : y is small. As the dissociation energies of $C-H$ bonds are higher than $400 kJ/mol$, the heat of reaction of the first equation is more than $1600 kJ/mol$. This first reaction is about $20$ times more exothermic than the second reaction. x >> y !!
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 ...
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.
...
0
To understand internal energy*, I prefer to start with this:
Consider a thermodynamic system. There are only three ways we can increase its internal energy (which is the sum of its internal potential energy and internal kinetic energy): flow heat into it; do work on it; or add matter to it (and the converse for decreasing its internal energy). If the ...
0
The internal energy is defined in (classical) thermodynamics for a finite change as $\Delta U=q+w$ where $U$ is the energy contained in the 'system' called the internal energy, and $q$ is the heat absorbed by the system and $w$ the work done on the system. Often the word 'system' means some ideal gas (often imagined to be in a cylinder with a frictionless ...
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.
-2
The internal energy U is the sum of all energies stored in the chemical bonds of the examined sample of matter. It is a potential energy.
The enthalpy H is the same, but with a correction due to the presence of the atmosphere. The difference between H and U is similar to the difference between weight and apparent weight. Let me go back to this measurement.
...
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 ...
0
I'll address part of your confusion here:
The following statement is not generally true:
Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.
For instance, in a closed system $\Delta U = q + w$ so what is independent of path isn'...
0
This has to do with the simplifications we make regarding the energy of molecules in an ideal gas (they've been largely discussed here if you want to have a look).
But essentially, in an ideal monoatomic gas, we can state that all the energy is translational kinetic energy, and so we can write that the mean energy is: $$U=n \frac{3}{2}RT=n c_vT$$
So, the ...
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 ...
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 ...
1
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 ...
0
I am a 12th grader, so I'll explain in the most colloquial way possible.
Internal energy is a state function.
A state function is a property whose value does not depend on the path taken to reach that specific function or value. ... Path functions are functions that depend on the path taken to reach that specific value. (1)
This implies a derivation for ...
0
You may have calculate dU and/or dH in any transformation, at constant volume or constant pressure, because U and H are state functions. These variations of U and H may not be easy to obtain, but they can and must always be calculated.
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 ...
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.
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