New answers tagged thermodynamics
2
As indicated by @Buck Thorn,the work done by the substance on its surroundings is $P_2(V_2-V_1)$. So, from the first law, $$U_2-U_1=Q-P_2(V_2-V_1)$$Solving for Q then gives:
$$Q=(U_2+P_2V_2)-(U_1+P_1V_1)-(P_2-P_1)V_1=H_2-H_1-(P_2-P_1)V_1$$Here, $$H_2=H_1+(H^{sv}_1-H_1)+(H_2-H^{sv}_1)$$where $H_1$ is the enthalpy of saturated liquid in State 1, and $H_2$ is ...
0
But how can we find this range of temperature or pressures? IS there an experimental methods?
The range of temperatures is fairly easy. Mix saturated aqueous NaCl (liquid solution, concentration about 5 mol/L) with ice (pure water, solid state), and measure the temperature. It will be about equal to the temperature of a household freezer (no coincident, ...
1
Basically you want to break up the problem into two steps:
Transition from liquid to gas at temperature T
Expansion of gas to final volume
For the first step you compute the heat involved from the provided value of $\Delta_{vap} H_m$. For the second step you assume the gas behaves ideally, and then, since the process is isothermal, you can assume that $q=-...
9
Summarizing the relevant point (pun intended) of my answer to a related post: a mixture does not exhibit a "triple line" if its composition is constant. It is only a "triple line" insofar as you get to vary the proportion (mole fractions) of the components. Therefore different samples of water with different isotopic composition will differ in the value of ...
3
Triple points are typically defined as points on the phase diagram of a pure substance where three phases coexist. Their existence requires that the phase rule be observed, which is $$f=C+2-P$$ where $$\begin{align}f &: \text{number of degrees of freedom} \\ C &: \text{number of components} \\ P &: \text{number of phases}\end{align}$$ The number ...
1
Yes, the triple point of the system will change. Instead of a single point specified by temperature and pressure at which multiple phases of water coexist, the addition of solutes (not solvents) will cause the conditions for 3 phases to exist to become a range of temperatures and pressures.
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Let $(V_0,P_0)$ represent the coordinates of the center of the ellipse, and let $P_\mathrm{max}$ and $V_\mathrm{max}$ represent the maximum pressure and maximum volume, respectively over the cycle. Then, the work is the area of the ellipse, which is given by: $$W=\pi(P_\mathrm{max}-P_0)(V_\mathrm{max}-V_0)$$This is, $\pi$ times the product of the semi-...
3
In your question, you didn't mention how low freezing temperature solvent system you are looking for instead of telling that the solvent has to remain liquid at the lowest possible temperature. And, Ivan Neretin mentioned that ethanol alone would get you about freezing temperature close enough to get you any ethanol+acetone solvent would get. For example, ...
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its me again, I want to post the answer I found because I think it might be useful for somebody in the future; it appears a 2:3 molar acetone:methanol proportion gives the lowest freezing temperature ~158 K (see attached picture below).
It appears the difficulty finding these curves was because they are usually called "solid-liquid equilibrium" curves ...
-4
People really need to look at the fundamental definition of the natural logarithm! Calculus (and mathematics in general) always works correctly with units.
The argument to a Log most certainly can have units. People aren't just ignoring the argument's units, and it isn't magic.
The result of taking a Log is always unitless. Even if the argument has units ...
1
The second formula you quote follows directly from the first. Starting with $W=N!/(n_1!n_2!\cdots)$ for $N$ distinguishable particles spread over set of $i$ energy levels with energy $\epsilon_i$, then $\ln(W)=\ln(N!)-\sum\ln(n_1!)$. Using the Sterling formula for factorials the entropy becomes, after some fiddling around and simplifying, $\displaystyle S=...
2
The only time I've ever seen that second notation is when referring to the substitution of the Helmholtz energy, as A = U-TS and therefore A = -kTlnQ. If you want an in depth explanation of how the thing is derived I suggest looking at Chapter 10 of Biomolecular Thermodynamics: From Theory to Application by Douglas E. Barrick.
Here are some photos of the ...
1
The key to the answer lies in this comment in the answer to the linked OP:
the dissolution of NaOH is exothermic, yet it's solubility increases with temperature.
So in short it's the opposite of #2: dissolving the first couple of molecules of $\ce{LiCl}$ is exothermic, but then if you dissolve more it become endothermic. The heat of dissolution to form ...
6
The primary flaw in your reasoning is assuming that $K$ is proportional to $-\Delta G^\circ$, so that a reaction with $\Delta S^\circ >0$ and $\Delta G^\circ<0$ must have a larger $K$ at a higher temperature because $\Delta G^\circ$ is more negative. If that were true, we would have a relationship of the form $\Delta G^\circ = -cK$, where $c$ is a ...
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Note: This answer has been heavily edited, because the original answer I posted contained some assumptions and was very limited (and incorrect) in the analysis. I would like to highlight the patience of @Buck Thorn, @Andrew and @Karsten Theis in pointing out my errors, expanding the discussion and going into a lot of detail.
The trouble is not with any ...
1
The dependence of an equilibrium constant on temperature is given by the van't Hoff equation:
$$\left(\frac{\partial{\log(K)}}{\partial{T}}\right)_p=\frac{\Delta H^\circ}{RT^2}$$
Therefore for an exothermic reaction ($\Delta H^\circ<0$) you expect a decrease in $K$ with temperature, as Le Châteliers principle would predict.
To give an example, the ...
2
My question is: Where does the thermodynamic energy come from to lower a mass of ice and water to a significantly lower temperature?
In order to decrease the temperature of a sample, you don't need any energy. It is the opposite, you have to lower the kinetic energy. So the question becomes where does this energy go.
The answer is given by Poutnik: Some ...
3
The minimum temperature reached should be $-21^\mathrm{o}$ C at $29$ % NaCl. Initially suppose that the ice, water and water vapour are at $0^\mathrm{o}$ C in a thermally insulated container. Say point A in the figure and some salt is now added. However, the mixture is now not at equilibrium, and some ice melts and dissolves some salt. The resulting ...
2
In contrary to mixing/dissolving salts with/in water, doing the same with ice instead works with much less total thermal energy.
Note that melting ice needs as much energy as heating the same mass of water by $\pu{ 80 ^{\circ}C}$
Water molecules in saturated solution at ice/salt contact have lower chemical potential than in ice, so ice is dissolving with ...
3
If the heat transfer in fluid 2 is fast, the thickness of the boundary layer wall - fluid 2 is very small (the 'not so linear' part of the temperature profile on that side)
One can then approximate that the temperature of fluid 2 directly at the wall is equivalent to the bulk temperature.
So the temperature profile through the wall would be steeper, ...
2
Answering "does every reaction have a reverse reaction?" (are there more endothermic or exothermic reactions):
You've actually hit on a topic people have been exploring for nearly a century:
Lewis. A New Principle of Equilibrium. PNAS, 1925, 11(3),179-183
The Law of Entire Equilibrium — Thus I am led to propose a law which in its general form ...
0
Does a Gibbs energy maxima correspond to equilibrium state or not?
(a) If yes, doesn't this violate the Second Law which implies that Gibbs energy should be minimized whenever possible? No matter which direction you move, your Gibbs energy will always decrease.
There are a number of central concepts in thermodynamics that are relevant here: equilibrium,...
5
G vs. $\xi$ does not have a maximum
If all reactants and products are pure liquids and solids, G vs. $\xi$ is linear. If some of the species are in mixtures, the entropy of mixing is responsible for the "sagging" shape of the curve. If the curve has an extreme value, it will be a minimum.
Analogy to mechanics
In mechanics, this situation would be called ...
3
Let's look at the $\ce{^9_4Be}$ isotope and apply the definitions:
Relative atomic mass
The relative atomic mass of $\ce{^{12}C}$ is, by definition, 12. Looking at the periodic table, we find that the relative atomic mass of beryllium is 9.0121831(5). That makes sense because $\ce{^{12}C}$ has a mass number of 12 (6 protons and 6 neutrons) while beryllium ...
12
It is the same reason that a dozen doesn't depend on whether you are counting grapes or elephants. That is how the mole is defined: it is a number, nothing else.
The confusion, I suspect, is because of how we measure that number (or, strictly, how we originally measured it as the definition changed recently). The intention of the unit was always to define a ...
8
I'm not an expert but this is how I would do it.
Determine the reaction rate formula from the reaction equation
$$r = K[\ce{NO}]^2[\ce{O2}]$$
Changing the volume by a factor $2$ means the concentration of $\ce{NO}$ and $\ce{O2}$ will become half
$$r' = K\cdot\frac{[\ce{NO}]}{2}^2\cdot\frac{[\ce{O2}]}{2}$$
Compare the reaction rates
$$r' = K[\ce{NO}]^2\...
1
It's strange, but there are mechanisms whereby this type of thing can happen. Basically you have a rate determining step that involves an intermediate species, and that rate determining step must have the usual positive temperature dependence. But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence ...
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Thermodynamics works perfectly well by assuming that matter does not need to be made of discrete particles, i.e atoms and molecule are not necessary. It was developed before the structure of atoms and molecules were known so it cannot inform us about anything at a molecular level. It is concerned primarily about energy and entropy.
If you break a bond in a ...
-1
Sorry, completely misunderstood your question earlier.
Edit:-
From Bradbury Science Museum, Los Alamos National Laboratory
Explosives are materials that burn or decompose quickly, creating large quantities of gases, which take up much more space than the original materials.
From the United States Department of Labour: Explosives and Blasting Agents:...
5
An explosion is a type of spontaneous chemical reaction that, once initiated, is driven by both a large exothermic change (great release of heat) and a large positive entropy change (great quantities of gases are released) $\ce{^1}$ from reactants to products (a thermodynamically favorable process).
The energetic stability of the gaseous products and ...
2
What you write is correct about all reactions proceeding if starting with reactants only, these will decrease in concentration and products increase until equilibrium is reached. How long this takes is a matter for chemical kinetics not thermodynamics. Strictly speaking thermodynamics has nothing to say about this process as it deals only with equilibrium ...
1
$K$, the standard thermodynamic equilibrium constant, is computed from $\Delta G^\circ$, using
$$\Delta G^\circ = -RT\log K \tag1$$
Generally speaking, $K$ in equation (1) is unitless. Its value depends on the specified reference standard states and $T$ (and obviously on the equilibrium activities of reactants and products). Both $K$ and $\Delta G^\circ$ ...
1
Like the Van't Hoff equation, which relates change in enthalpy to equilibrium constant, is there a similar equation for the relation between change in entropy and equilibrium constant?
A relation between $K$ and $\Delta S^\circ$ can be obtained as follows:
$$\begin{align}T\log K &= -\frac{\Delta G^\circ}{R} \\
\left(\frac{\partial(T\log K)}{\partial T}...
2
Your interpretation of the equation is correct. $\Delta G^o$ gives you thermodynamic favorability of a reaction under standard conditions (Q=1) and even reactions with positive values of $\Delta G^o$ (unfavored under standard conditions) can be driven to proceed if the concentrations of the reactants and products are made extreme enough.
The best way to ...
1
The questions is a good one, and answering it winds up being central to a lot of key concepts in thermodynamics. A brief correction before starting, though I don't think it's central to the question, "When work is done on by the system it means that a part of system kinetic energy is used to do the work..."
The disconnect between your understanding and ...
4
After a lot of help, I have the following to suggest as an answer:
Imagine a reaction with $\Delta_\mathrm{r} G^{\circ} = -1000Jmol^{-1}$ at 298K. Using the following equations: $$\Delta_\mathrm{r} G^{\circ} = - RT\ln K$$ $$e^{-\frac{\Delta_\mathrm{r} G^{\circ}}{RT}} = K$$
This would give a K value of 1.50, which indicates the Q value on the curve where G ...
5
Logically, that would mean that absolutely pure reactants have infinite G (which doesn't seem right).
It's the slope (rate of change) that is infinite, not the actual property (Gibbs free energy).
$\Delta_r G$ (expressed sometimes as $\Delta G'$, sometimes you'll see just $\Delta G_m$ without subscripts indicating explicitly that this is the molar Gibbs ...
5
G is a finite quantity for ξ = 0 or max. The slope of the graph is vertical on the extremes, though. This is because the chemical potential for reagents approaches negative infinity on one side, and that of the product positive infinity on the other side. (A vertical or infinite slope does not mean that the function value has to be infinite - a half-circle ...
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