6

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

The second definition probably refers to the intensive properties of the phases, not the extent of the phases, being invariant, making the two definitions you present equal. The confusion within the first question you link to has to do with the definition of "boiling" and its application to open systems. Phase transitions of the sort being discussed here ...


3

I would approach this a little differently. I would integrate the pressure between $V_D$ and $V_B$, such that the average pressure is 17.76 bars: $$\frac{\int_{V_D}^{V_B}{PdV}}{(V_B-V_D)}=17.76$$or, $$17.76(V_B-V_D)=RT\ln{\frac{(V_B-b)}{(V_D-b)}}-a\left[\frac{(V_B-V_D)}{(V_B-b)(V_D-b)}\right]$$ This would provide an equation for expressing a in terms of b. ...


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 ...


3

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 ...


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 ...


2

First of all, it is useful to envision an adiabatic process: add ice at the normal melting temperature to water or cold brine slightly above the MP in a perfectly insulated container (zero heat transfer to the outside) at constant pressure. What do you expect to happen? If the salt concentration is nil (pure water) then ice will melt and the temperature of ...


2

Phase change is not at constant temperature, phase equilibrium is [OP] it's a phase transition so the temperature of the ice should remain constant Here is a counter example: If you add ice cubes to hot water, the ice will melt, cooling down the hot water. In this system, there is no thermal equilibrium, so not all of the components are at the melting ...


2

Generally speaking you can write that $$\Delta H^\circ(T) = \Delta H^\circ(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^T \Delta C_p \,\mathrm dT \tag{1}$$ where $T_\mathrm{ref}$ might be $\pu{25 ^\circ C}$ and the pressure is 1 bar (in your case, approximately 1 bar). If your temperature window is not too large you may get away with assuming $\Delta H^\circ$ ...


1

Your target equation is: $\ce{Mg(s) + 1/2O2 -> MgO(s)}$. You have given following set of equations to get your target's enthalpy of formation: $$\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)} \quad ΔH = \pu{-1300 kJ} \tag {1}$$ $$\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \quad ΔH = \pu{-602 kJ} \tag {2}$$ $$\ce{H2(g) + 1/2O2 -> H2O(l)} \quad ΔH = ...


1

What this ultimately boils down to is that this is a system of equations which we're trying to solve. Since we have two unknowns, we need two equations. Luckily, we have 3. $$17.76 atm = \frac{RT}{V_m - b} - \frac{a}{V_m^2}$$ where our three equations come from having three different known values of the molar volume. As is typical with systems of ...


1

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 ...


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 ...


1

The gas pressure equals to the external pressure during isothermal irreversible expansion. To maintain P(gas)=P(ext), heat is supplied reversibly. By this reason delta S(surr) must be calculated above. Am I clear?


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