# Tag Info

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At 77 K, both HCl and Ar are solids. So if you cool those gases separately and mix the two solids, you obtain a simple mixture of two powders. If you mix Ar and HCl at room temperature, you obtain a mixture of gases. By cooling this mixture slowly, HCl will first liquefy at -83°C (density 1.194), then solidify at -112°C. Later Argon will liquefy at -185°C ...

6

The metabolism of living creatures keeps the dynamic equilibrium of their $\ce{^{14}C/^{12}C}$ ratio with the enviromental $\ce{^{14}C/^{12}C}$ ratio via photosynthesis, breath, food and excrements. It is not just about one time building, but also about continuous recycling of the body content. Such an equilibrium means continuous resetting of the $t=0$ ...

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There are plenty of good sources online explaining the principle behind radiocarbon dating. For instance, the wikipedia explains: During its life, a plant or animal is in equilibrium with its surroundings by exchanging carbon either with the atmosphere, or through its diet. It will therefore have the same proportion of $\ce{^14C}$ as the atmosphere, or in ...

0

Although the comment by Ezze is a valid and helpful answer, according to chegg.com, the order of reaction with respect to the $\ce{H+}$ ion is $\ce{0}$ and therefore it can be omitted from the rate expression as $\ce{[H+]^0= 1}$.

3

As pointed out in the comments, the concept of pH as a thermodynamic property of nanodroplets does not quite make sense unless the nanodroplets are in equilibrium (stable). If they are in equilibrium you can still apply the equation $$\textrm{pH}=-\log_{10}(a_{\ce{H+}})$$ but now you have to be aware that the activity will be influenced by surface (and ...

-2

The answer is No ! Solid have no vapor pressures. And if they still have, this pressure is a constant at the equilibrium conditions. So it does not change between the beginning and the end of the reaction. So it is included in the equilibrium constant.

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The equilibrium constant depends on temperature, the only thing it depends on. For an exothermic reaction the equilibrium constant will decrease ie shift towards the reactants side Also consider kinetics as described above

3

There are two separate effects we need to consider here: thermodynamic and kinetic. Let's assume you are only providing the energy thermally. So providing more energy means increasing the temperature (T). Thermodynamically, if a reaction is exothermic, and you increase T, the reaction becomes less favorable (assuming it stays exothermic over that ...

3

The crystalline property that would allow to cut the "diamond creature" (or whatever those things are in your universe) you are looking at is cleavage (a writer might also use a term in a wordplay). In crystallography, cleavage is the ability of crystals to crack or split in certain crystallographic directions. This mechanical property of crystalline media ...

1

SAPT(O) is an amazing tool for partitioning the interaction energies, it yields meaningful values and data. ETS-NOCV is also very good for partitioning the interaction energy, but it does it in a much different way, it yields values that correspond more to the orbital interactions, and more so the sigma and pi orbital mixing (although orca will not partition ...

5

Note: this is a preliminary description, which I think answers the question, but disagrees in one detail (see asterisk). Tin(IV) iodide is completely hydrolyzed by water [1, p. 120] to colorless hexahydroxostannic acid: $$\ce{SnI4 + 6 H2O <=> H2[Sn(OH)6] + 4 HI}$$ Hydrogen peroxide can not oxidize $\ce{Sn^4+}$ further (in fact, hexahydroxostannate(...

0

At any given temperature, the change from a liquid to a vapor occurs at constant pressure. What we are trying to find with the Clapeyron equation is how this constant pressure changes as the given temperature changes. At equilibrium at a given temperature and constant pressure, we have $$G_{vapor}=G_{liquid}$$ and $$S_{vapor}-S_{liquid}=\frac{(H_{vapor}-H_{... 0 The magnitude of heat capacity depends on the pressure and volume, especially in the cases of properties of gases. The temperature of a gm -mole of gas raised by one degree at constant volume. Again in another operation same raise of temperature allowing the volume to very. The quantity of heat in the two operations would be different. Hence, in mentioning ... 0 This happens because the equilibrium constant K_p is independent of the pressure, although it does depend on temperature, which we will assume to be constant. The equilibrium constant can be written as$$\displaystyle K_p =\frac{p_A^2}{p_Np_H^3}$$where for simplicity I have abbreviated the chemical names. The partial pressure p can be written in ... 1 It appears that you are correct in your thought process. Determining by how much the activation energy of the reaction is reduced requires you to basically find$$Δ(\%) = \frac{E_\mathrm{a1} - E_\mathrm{a2}}{E_\mathrm{a1}} × 100\% = \left(1 - \frac{E_\mathrm{a2}}{E_\mathrm{a1}}\right) × 100\%$$There is a system of three independent equations$$ \begin{...

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From the definition of the Gibbs free energy $G=H-TS$, $$\Delta G = \Delta H - T \Delta S \tag{constant T}$$ For a pure substance, equilibrium between two phases (such as for a point on a coexistence line) reduces the degrees of freedom to one, such that p is constrained by the choice of T, or vice-versa. Since the condition for equilibrium at constant T ...

1

Yes, you can talk about viscoelastic solids and viscoelastic liquids, and it is possible to distinguish the two by oscillatory mechanical tests (e.g. with a rheometer). You usually apply a sinusoidal deformation over time to your specimen and measure the time-dependent response. In case of a true elastic solid, the deformation will induce a force which ...

3

There are two basic models that describe viscoelastic behaviour: The Maxwell model is made up of a spring and a dashpot in series: For fast deformations, the dashpot does not really move, and your spring gives pure elastic behaviour. For slow deformations, the spring stays relaxed, and the damper fluid in your dashpot turns all energy into heat. The Voigt-...

1

Thus we can see that the Al is oxidised but there are no Cu2+ ions to be reduced also Na+ reduction would lead to a negative Ecell standard and a positive Gibbs. Is it an air cathode where O2 is being reduced? Well, there is no free lunch in this world. This is not a conventional cell in which each electrode is sitting in its own ions and one can easily ...

2

NIST Webbook Chemistry is a reference which I would recommend as a first (as in initial) resource to consult for enthalpies of small organic molecular and inorganic materials. The entry about methanol lists for one temperature the value in question, the experimental method of acquisition, and the primary literature reference. The later may contain more ...

1

If fuel is oxidized at high temperature, the conductivity of the fuel itself is irrelevant -- only the conductivity of the combustion products, which are ionized because of the high temperature and are therefor conductive. Brittanica states that temperatures greater than ~2,500 K are needed for sufficient ionization. Alkali metals can be used to increase ...

1

The above reasons to use LLE are focused on the compound thermal similarities and instabilities. But it does not take into the account the amount of compounds. While distillation separation can be great for major components, it may fail for minor one, and definitely for trace ones. The LLE is frequently used for separation and content enrichment of trace ...

0

Required energy $E$ can be found as a product between voltage $U,$ current $I$ and time $t$: $$E = UIt\label{eqn:1}\tag{1}$$ Unknown multiplier $It$ can be found using united Faraday law: $$m = \frac{MIt}{zF}\quad\implies\quad It = \frac{mzF}{M}\label{eqn:2}\tag{2}$$ where $m$ is the mass of the substance liberated at the electrode; $M$ is the molar mass;...

1

The key point to keep in mind in this type of problem is that the energy of an ideal gas is only a function of temperature, not of volume or pressure$^\ast$. We also assume that the dependence of the energy of the condensed phases on p and V is negligible. Therefore the final pressure or volume is not going to affect $\Delta U$ for the reaction, provided n ...

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I think what you are asking is this: Equilibria for chemical reactions typically* (see note at end) require specific ratios of products to reactants (as expressed by the equilibrium constant). By contrast, equilibria for phase transitions don't require specific ratios of products to reactants. [For instance, at the phase transition between ice and water, ...

3

[OP] Why are melting and boiling considered equilibrium processes [...] They should not be considered equilibrium processes. If melting is defined as the process where there is a net change from solid to liquid phase, this is not an equilibrium. If boiling is defined as the process where liquid turns into vapor (rolling boil with bubbles forming below the ...

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Two different phases of a substance in contact with each other in a closed system at some uniform temperature and pressure (thermal and mechanical equilibrium) will be in equilibrium if the chemical potential of the substance is the same in both phases. It turns out that at its boiling point, a liquid has the same chemical potential as its vapor at that ...

0

Mercury metal exposed to the atmosphere will (dependent on the specific location) soon gather dust particles upon it's surface if left undisturbed. In this state and at 23C the mercury will not evaporate into the atmosphere as the combination of dust and temperature will work against this. I'm talking about a very small spill, that one would have to get down ...

2

Let $\Phi, \Psi$ be normalized wave functions. Then $0 \le|\langle\Phi\vert\Psi\rangle| \le 1$, with $\langle\Phi\vert\Phi\rangle = 1$ and $\langle\Phi\vert\Psi\rangle = 0$ for orthogonal $\Phi$ and $\Psi$. Since we can argue that any wave function completely contains itself and for two orthogonal wave functions none of them contains a part of the other, ...

0

Water is often described as the universal solvent because most things will dissolve in water if you have enough time and water. For a compound that is only slightly soluble in water adding more water will indeed increase the number of electrolytes formed. It may help to think about your reaction in this format: $$\ce{A + H2O <=> C + D}$$ If ...

0

If our dummy dissociation degree is $\alpha=\frac{[\text{C}]}{[\text{A}]+[\text{C}]}=\frac{[\text{D}]}{[\text{A}]+[\text{D}]}$, and the total concentration of dissociated and undissociated species is $x=[\text{A}]+[\text{D}]=[\text{A}]+[\text{C}]$, the equilibrium constant, on which le Chatelier's principle is (well, should be) based is $K=\frac{[\text{C}][\... 3 To prevent the condensation, you need to create the conditions under which the vapor does not become saturated. The way to do that is to heat up the whole transfer line (the roof of the first flask and all the tubes that follow) - the higher the temperature, the higher the saturated vapor pressure. Heated transfer lines are often employed in commercial ... 0 Your book is right, On dilution, the degree of dissociation of an electrolyte increases. But your interpretation is wrong. If you increase the degree of dissociation of A, it means that there is less A, and more B and C in the solution. The degree of dissociation is by definition the fraction of A which are transformed into B and C. This fraction increases ... -2 In your flask, the water will always go from the drops of liquid water to the silica, and not in the opposite direction. If you want to remove the water from the silica, the only way to do it is to heat the hydrated silica in the air oder under a vacuum, and quickly close the hot silica when dry. 1 Just complementing the first answer of @Buck Thorn : Full rewriting of the van Der Waals equation in terms of$Z=f(p,V_\mathrm{m}): $$\left(p+\frac{a}{{V_\mathrm{m}}^2}\right)\left(V_\mathrm{m}-b\right)=RT$$ $$pV_\mathrm{m} - pb + a/V_\mathrm{m} - ab/{V_\mathrm{m}}^2=RT$$ $$pV_\mathrm{m} \left( 1 - \frac{b}{V_\mathrm{m}} + \frac{ a}{ p{V_\mathrm{m}}^2} -... 2 The van der Waals equation is$$\left(p+\frac{a}{V_m^2}\right)\left(V_m-b\right)=RT$$which for p \gg a/V_m^2 ^\ast can be rewritten as$$pV_m=RT+bp$$or$$pV=nRT+nbp \tag{1}$$The problem says that the tangent of this curve evaluated at the same T has an intercept (the value of pV when p goes to zero) of \pu{40 Latm} when n=2, which means ... 2 Raoult's and Henry's laws are limiting laws, generally applicable when the solute concentration goes to zero. In this limit the vapor pressure of any component in the solution depends linearly on its mole fraction, implying the absence of solute-solute interactions. Raoult's law describes the dependence of the vapour pressure of a solvent as a function of ... 2 Denbigh, in his book the Principles of Chemical Equilibrium, considers the case where "(a)the only heat transferred to the system is from a reservoir which remains at the constant temperature T; (b) the initial and final temperatures of the system are equal, and are equal to the temperature T of the reservoir; (c) apart from the system, the only other body ... 2 Compare and contrast As stated in the caption of the figure posted by the OP, Henry's law applies to the solute and Raoult's law applies to the solvent. If a solute is non-volatile (e.g. sucrose or sodium chloride), Raoult's law still works, but invoking Henry's law, while possible technically (by setting Henry's constant to zero), would be kind of ... 6 After further reflection I realized an initial comment I made was flawed*. To quote the Wikipedia: Isenthalpic processes on an ideal gas follow isotherms, since \mathrm dh = 0 = c_p\,\mathrm dT. That means that for an ideal gas isenthalpic is synonymous with isothermal, and this implies \Delta U = 0 and so q = -w = P_\mathrm{ext}\Delta V. For ... 2 Your question is how does the inequality |\mathrm dW_\mathrm{reversible}| > |\mathrm dW_\mathrm{irreversible}| follow from more general statements of the Second Law (a "corollary" being the Clausius inequality). Or, how is that inequality a definition of spontaneity? There are various established formulations of the Second Law. Clausius' statement is ... 3 (a) is not true, because, in addition to being adiabatic, the process must be reversible. (b) is not true, because the gas must be assumed to have constant heat capacity for this equation to apply 4 For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium:$$\ce{ZnO(s) + \underset{1}{CO(g)} <=> \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$Partial pressure of carbon monoxide can be found via its mole fraction x_1 and given total pressure p:$$p_1 = x_1p\tag{1}$$To find ... 2 Since at equilibrium, Moles of CO= Moles of Zn= moles of CO2 (as no information about initial moles are given) If no information is given, you should just give the three amounts names and treat them as unknowns. ZnO is exposed to pure CO at 1300 K This means that initially there is no carbon dioxide and no elemental zinc. Carbon dioxide and elemental ... 3 The answer is: it depends. Dissolution of a salt implies that the entropy gained exceeds the cost of breaking lattice interactions (the solution enthalpy, assuming it is positive). Electrostatic interactions compete with kT (thermal jostling). Under physiological conditions, long range interactions are strongly screened by intervening solvent molecules and ... 0 why is the enthalpy change not zero? When only PV work is possible, and pressure is constant, we can write$$q_p = \Delta H \tag{1}$$that is, the enthalpy change at constant pressure is equal to the heat exchanged. If the pressure is not constant all bets are off and you should resort to some other expression such as$$\Delta H = \int_1^2 d(U+PV)$$... 0 The indicator may slowly disappear after the titration because it is reacting with the air. As it reacts with chemicals in the air (CO 2 for example) which causes the indicator to be slowly disappear from the solution. 1 The enthalpy of combustion (aka heat of combustion) is defined as the enthalpy change in going from pure reactants (each at 1 bar pressure) to pure products (again, each at 1 bar pressure) at 298 K. In this problem, it is first of all necessary to assume ideal gas behavior to proceed. For ideal gases, the change in enthalpy in going from pure reactants ... 0 For an isobaric process$$q_p=\Delta H$$and, as you rightly write, for an isochoric process$$q_V=\Delta U$$and the relation between U and H is$$\begin{align}\Delta H&=\Delta U + \Delta(PV)\\&=\Delta U+RT\Delta n\end{align}\$ The last equality holds only if T is constant, which it evidently wasn't during the experiment, but for our ...

0

Enthalpy of combustion is the change in enthalpy. The enthalpy change of a combustion reaction is called ΔH (combustion). The heat supplied to the calorimeter represents the heat supplied by virtue of the reaction taking place. Normally, ΔH = ΔU + PΔv The calorimeter is constant volume, so ΔV = 0. Hence, ΔH = ΔU in this case. So, as per the given ...

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