# Tag Info

1

The Lattice and hydration energies applies to cases of dissolution of ionic compounds like $\ce{NaCl}$. For covalent liquid compounds with polar bonds like water, the lattice energy is replaced by the bond dissociation energy. The value of the dissociation energy ( enthalpy ) -- involving hydration -- can be indirectly determined from the temperature ...

0

If the system is at lower pressures where the ideal gas law is close to valid, then the partial pressure of the volatile species (mole fraction times total pressure) can be used in place of the vapor pressure with the Clausius Clapeyron equation. At higher pressures, the free energy of the liquid is affected by the higher overall pressure, as is the partial ...

3

I've seen that Gibbs free energy is defined as the maximum amount of work that a system can do. From this I gather, this is how much energy it will release - right? No, this conclusions is incorrect, the Gibbs energy is not how much energy a process releases. A system can exchange energy with the surrounding in the form of heat or work. The Gibbs energy is ...

5

Suppose you have an ideal gas in a container in mechanical and thermal equilibrium with its surroundings (same temperature and pressure; say, $\pu{298 K}$ and $\pu{1 atm}$). It can't do any work on the surroundings. Now compress it isothermally to, say, $\pu{100 atm}$. Now it certainly can do expansion work on the surroundings. What change in ...

1

The freezing but liquid concentrated brine may have better thermal conductivity, so cooling down could be initially faster. But brine would be warming itself up immediately. Ice in the salt and ice slurry would have ( more or less) melt before the resulting brine would start to warm up. Ice+salt would be warming up slower. Energy needed to melt ice is equal ...

1

Note that activity coefficient takes into account the non-ideality due to electrostatic interactions. For instance, you can look up for the activity coefficient of ions in the Debye-Hückel theory. This indicates that the electric part of the hydronium ion potential is already included in its activity coefficient. So, since the activity coefficient of a ...

1

Here is a plot of p on a logarithmic scale vs 1/T on a linear scale. It is almost a perfect straight line. The slope is $-\Delta H/R$

-4

Yes, one can calculate the specific heat capacity of the compound from its constituent elements that is just some of its constituent element's specific heat capacity. As it is known that specific heats of elements remain unchanged when they enter into compounds. However, the density/volume/mass may change and so the heat capacity.

23

To convert this into a generic linear algebra problem you'd rewrite it in the form $Ax=b$ where $A$ is a matrix of stoichiometric coefficients of size $m \times n$; $x$ is a vector of length n of unknown fitting parameters (the heats you want to determine) and $b$ a known vector of length n. For the example problem the set of equations can be written in ...

6

Two water drops 1 mm apart do not attract each other, if I omit gravity and electrostatic charges. Are they hydrophobic ? No, they are not. The relevant intermolecular forces are contact ones, decreasing very fast with high power order of distance. Water molecules in a gaseous phase have still strong attraction during their collisions, what reflects in vapor ...

4

Yes. The universal condition for equilibrium is that, given the current constraints on the system, no further spontaneous change can take place. I.e., that the entropy of the universe (system + surroundings) is maximized. At constant temperature and pressure, a maximization of the entropy of the universe corresponds to a minimization of the Gibbs free ...

1

The van der Wals expression can be shown graphically at constant temperature by the curve ABCDEFG in following diagramm, if the temperature is lower thant the critical temperature. Try to follow the line. Start at A with a large volume of gas and a small pressure. Increase the pressure. The volume decreases from A to B and C. But after C, along CDE, the ...

2

Fundamentally, the problem is in the problem statement. The proper form for the Van der Waals equation is $[P+(a/V_m^2)](V_m\color{blue}{-b})=RT$ which would give a cubic equation when we try to isolate the molar volume. It is this cubic equation that usually has one or three roots that can be correlated with the possible phases, as described in the other ...

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