New answers tagged physical-chemistry
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How can I calculate the enthalpy for those balanced reaction given the data of $\ce{NO +O2 -> NO2}$ enthalpy?
The way the enthalpy data is presented is sloppy (is this the original text of the exercise?). For some types of enthalpy, it is unnecessary to specify the chemical reaction equation because there is a convention. For example, for the enthalpy of ...
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Supposing the dissociation process is isochoric, the total volume $V$ required for calculation of molarity at equilibrium of $i$th component $c_{\mathrm{eq},i}$
$$c_{\mathrm{eq},i} = \frac{n_i}{V}$$
can indeed be found from the ideal gas law using initial amount of undissociated sulfur trioxide:
$$V = \frac{n_0(\ce{SO3})RT}{p} = \frac{\pu{1.00 mol}\cdot\...
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For a quick general answer, this equation represents the charge balance. In a solution, the charges of all ions should add up to zero. You charge balance reads:
$$\ce{[H+] = [Cl-] + [OH-]}$$
If you had divalent ions, you would have to correct for that, e.g. if the solution also contained magnesium ions:
$$\ce{[H+] + 2 [Mg^2+] = [Cl-] + [OH-]}$$
How ...
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Let's take the degree of dissociation of water be $\alpha$ and the $[\ce{H2O}] = c$, then at equilibrium the $[\ce{H^+}] = [\ce{OH^-}] = c\alpha$. And since $\ce{HCl}$ is a strong acid it will completely dissociate. So $\ce{HCl}$ is a then the $[\ce{H^+}] = [\ce{Cl^-}] = a$. So $[\ce{H^+}]_{\textrm{net}} = a+ c\alpha$, which is equal to $[\ce{OH^-}] + [\ce{...
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Nitrogen is a 5 group element hence has valency 5 and AZIDE ion has an negative charge it means it gained an electron hence as AZIDE HAS 3 NITROGENS and a negative charge it will be 5 multiplied by 3 = 15 + one electron gained = 16
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If you mix CaCO3 + S and heat the mix, the first thing to occur will be that sulfur melts. Second, CaCO3 loses CO2 and becomes CaO. Third, sulfur boils off (or has already boiled off).
A thought that perhaps CaS + O2 would form can be eliminated on the basis that roasting metal sulfides in air produces SO2 + the metal oxide. So the metal sulfide is unlikely....
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As you mentioned, Gaussian can do it, but is proprietary.
I believe ezSpectrum (https://github.com/iopenshell/ezSpectrum) might be able to do what you are looking at but I have not used it myself so far.
There also used to be a program available called FCClasses which would run the Franck-Condon computations for you based on (I think) Gaussian output, but ...
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The change in Gibbs energy has to be negative for a process to happen (at constant pressure in the absence of work). Gibbs energy is a combination of entropy and enthalpy.
I do not understand why this phenomenon occurs as a counter example to the natural laws of diffusion along the concentration gradient.
Diffusion is driven by entropy changes, ...
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Chemical kinetic describes both reversible and irreversible reactions.
The reaction scheme and rate expression do not refer to the same thing. The reaction scheme is an overall one whereas the rate equation refers to a particular multi-step reaction. Two examples are given below:
Ozone can be destroyed by UV light in the $240$ to $\pu{320 nm}$ region with ...
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Even though this question 1) has an answer with multiple upvotes (and I was the first upvote), 2) the OP has accepted the answer and 3) I have great respect for @Karsten Theis, having co-taught a quant class with him back in 2008 and knowing, first hand, that he is an excellent scientist and teacher, nonetheless, I have several problems with this trick exam ...
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How to start solving the problem since the $E^\circ$ of individual half reactions is not given?
This is a concentration cell, i.e. the half reactions at the anode and at the cathode are the same (except for the direction).
AgCl(s) electrode
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
$$\ce{Ag+(aq) + e- -> Ag(s)} $$
AgBr(s) electrode
$$\ce{Ag(s) -> ...
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For a single phase, enthalpy is a function of temperature and pressure, and, at constant pressure, $\mathrm dH = C_p\,\mathrm dT.$
However, at a change of phase, enthalpy (per unit mass) undergoes a jump change, even at constant temperature and pressure. If the phase change is from liquid to vapor, for example, this jump change in enthalpy is called the ...
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The rate of electrolysis (decomposition of water) is defined by the current that passes. If you apply a constant voltage, factors arise as suggested in the comment by MaxW.
At constant current, the rate of hydrogen (and oxygen) evolution would be somewhat different initially because on a microscopic level, hydrogen atoms have to cover the surface to some ...
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Thermodynamic properties are functions of composition, which can be represented as $(\chi _1,...,\chi _N)$ where $\chi _i$ is the mole fraction of component $i$ present in the system. Now imagine a change in the composition such that the amount of component $i$ is increased by a very small quantity $dn _i$ (alternately, imagine the system to be infinitely ...
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If you can't wrap your head around getting the answer directly, just calculate the enthalpy of reaction for the two bond dissociation energies you are considering.
Assume bond dissociation energy is 141 kJ/mol
$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot \pu{141 kJ/mol} - 6 \cdot \pu{272 kJ/mol} = \text{negative value}$$
Assume bond dissociation energy is ...
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For hydrogen to burn or explode in air, its concentration needs to be at least 4%.
Unless you're using very high power levels, you're unlikely to reach that concentration in the room's air. You could ignite bubbles of hydrogen as soon as they pop at the surface, but if you don't, the hydrogen (with about 1/8 the density of air) will rise quickly, diffusing ...
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I asked a somewhat similar question a few weeks ago and Poutnik made a smart suggestion that I simply mix the H2 with nitrogen to keep it safely below explosive concentrations, which I set out to do prior to extracting it outside with a fan. So the nitrogen ensure that concentrations are low enough for the fan engine's sparks not to light it on, and the fan ...
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The key idea behind zero padding or zero filling is that the "resolution" or the step size of the frequency axis in the Discrete Fourier Transform is dependent on the number of points you have in the time domain. Zero filling is done before doing the Fourier transform. Thus after the FFT the resolution appears to be improved because you have more points in ...
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I think the fact you are missing is that after combustion, water vapors will be given away and the volume lost during combustion will be equal to the amount of water lost. $\ce{KOH}$ will absorb $\ce{CO2}$ and the volume absorbed by $\ce{KOH}$ will be equal to volume of $\ce{CO2}$ in the solution. Thus, you will have two equations and you will be able to ...
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Boiling water in a closed container essentially requires you to add heat to a constant volume. As noted above, this increases pressure, and ways to get around this basically open the system.
However, heat pipes are small closed systems (really closed, and really small), and liquids are boiled in them to transfer heat from one end to the other. Pressure ...
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I am being told that boiling can't take place in closed containers. Can someone please explain why this is so.
Source: https://upload.wikimedia.org/wikipedia/commons/thumb/9/9c/Simple_distillation_apparatus.svg/2000px-Simple_distillation_apparatus.svg.png
A liquid can boil in a closed container under certain conditions. For example, you can close off a ...
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It is very analogical scenario as connecting capacitors or electronic DC sources ( with floating potentials ) in a serie.
With 2 cells in a serie, the same provided charge spends the doubled amount of electrode material. And vice versa for charging.
Even without looking into electrochemistry details, energy conservation law requires doubled voltage for 2 ...
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Voltage and (thermodynamic) electrode potential are slightly different things. Electrode potential refers to the maximum amount of work that can be extracted from a given single cell under very specific conditions and at a given concentration of metal ions.
Now forget about the cells shown in chemistry textbooks. First think of a hydraulic analogy. Think of ...
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There are a few separate issues here to keep in mind:
$K_c$ (the equilibrium constant in terms of concentrations) is defined as
$$K_c = \prod {c_i}^{\nu_i} \tag{1}$$
Agreement between $K_c$ and the product of forward/back rate constants ($k_{\pu{fwd}}/k_{\pu{rev}}$ in the OP example) is expected only if the mechanism is correct, assuming some proportional ...
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This is a complicated question among chemical educationists- with tons of arguments over arguments. Just search do equilibrium constant have units? I would say the "units" of K are in a grey area (just like in any real world science.). You should read the nice section on Dimensionality of the Equilibrium Constant, and I will just give a background:
Now, ...
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See if we have $K_c = \pu{100 mol dm-3}$ and the concentration of product changes to $\pu{50 mol}$ so the net change of concentration is $50$ means half of the concentration decrease and if we look at initial concentration so it was for $\pu{1 dm3}$ and we always talk about $\pu{1 dm3}$ so to get the same value of $K_c$ you will have also to divide the ...
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$G$ is defined as $H - TS$, and $H$ is $U + PV$.
Since in closed system you have: $\mathrm{d}U = T \mathrm{d}S - P \mathrm{d}V$, it follows that:
$\mathrm{d}H = T \mathrm{d}S + V \mathrm{d}P$
$\mathrm{d}G = -S \mathrm{d}T + V \mathrm{d}P$
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There are several databases of thermochemical properties, a number of them being listed on this Wikipedia page (at the bottom). Most of them are not freely available online, but require subscription. I've searched online but couldn't find that information freely available.
Another possible way to get the information would be through a search of the ...
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There is an excellent treatment of electrolytic conductivity in Wikipedia. However, the OP’s main question involved a specific equation, pertaining to the conductance of a mixture of equal volumes of two electrolyte solutions, and this was not specifically discussed in the Wikipedia article:
$$\text{conductance} = \frac{1}{2}\left(\frac{1}{R_{1}} + \frac{1}{...
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"Since change in enthalpy ∆H = Q + VΔP, it seems that enthalpy and heat flow (Q) are not strictly the same thing."
First, note that, when you write "∆H = Q + VΔP," you are assuming the only type of work is PV-work. That's a reasonable assumption for these discussions. It's just important to make that assumption explicit.
Given this, the quoted statement ...
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The heat of reaction $\Delta H^\circ$ is defined as the difference in enthalpy between the pure products and pure reactants, all at 1 bar pressure and 25 °C. If this $\Delta H^\circ$ is negative, the reaction is regarded as exothermic.
If we are dealing with all ideal gas species, where the enthalpy is independent of pressure and the heat of mixing is zero,...
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The ratio
$$\frac{x}{m}= k P ^{1/n}$$
can be re-expressed in terms of the area $A$, surface-to-volume ratio $\lambda$, and density $\rho$ (note that $m=A\rho/\lambda)$:
$$\frac{x}{A}= k' P ^{1/n}$$
where
$$k'=k\rho/\lambda$$
The advantage of using $k$ as opposed to $k'$ is that you don't need to determine $\lambda$ to know how much (as in, what mass) ...
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First passage problems
This is a first passage problem, asking when a system reaches a certain final state for the first time . If you were to use a simulation to explore this, you would erase parts of trajectories right after reaching the final state before summing up results. The Backwards Master Equation is especially suited for first passage problems (...
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The Dulong–Petit law applies in the classical limit, i.e. when temperature is high enough that the quantisation of energy levels (as prescribed by quantum mechanics) is not readily apparent. The larger the gaps between energy levels, the higher the temperature needs to be such that you no longer "see" these gaps. The comparison is fundamentally between ...
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Not a 'Law' a such it is more like a 'rule of thumb' based on a limited set of data measure at high (i.e. room) temperatures. The heat capacity is not constant but varies with temperature and for metals reaches a limiting value of approximately $3R$ at high temperatures. (The limiting value is higher for molecules due to the many translational, vibrational ...
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First, a reminder that most scientific "laws" are only convenient mathematical approximations. They aren't guaranteed to work for every case, and they aren't guaranteed to have underlying physical meaning. As such, don't think of exceptions as "not obeying the law"; it's more correct to say that the law isn't obeying reality,
In the case of the Dulong-Petit ...
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Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality 'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another. Fermi golden rule is the usual ...
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So we know that
$$\ln\frac{k_{T_2}}{k_{T_1}} = \frac{E_\mathrm a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$
If we consider both were raised from $T_1$ to $T_2$, $$\frac{1}{T_1}-\frac{1}{T_2} = \text{constant}$$
So
$$\frac{\ln\frac{k_{T_2}}{k_{T_1}}}{E_\mathrm a} = \text{constant}$$
So the one with higher $E_\mathrm a$ has a higher value for $\ln\frac{k_{T_2}}{k_{...
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$$k_\mathrm{high} = A e^{(−E_a/RT_\mathrm{high})}$$
$$k_\mathrm{low} = A e^{(−E_a/RT_\mathrm{low})}$$
$$\frac{k_\mathrm{high}}{k_\mathrm{low}} = e^{-E_a \cdot \beta}$$
If the activation energy is zero, the ratio of rate constants will be one, i.e. there is no temperature dependency of the rate. The higher the activation energy, the larger the exponent, ...
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The activation energy is the kinetic energy required from the reactants to form the products.Now lets go to your second part.Since the first reaction has a bigger activation energy than the second reaction , if we add energy to the system the first reaction has better chances of happening than the second reaction because more molecules from the first ...
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TL;DR The $\mathrm{d}_{z^2}$ orbital is a result of solving the Schrödinger equation for the hydrogen atom in the most mathematically convenient way.
To properly understand this, it is necessary to go back to the fundamentals. The complex d-orbitals are obtained by solution of the Schrödinger equation. In general, these d-orbitals are made up of a radial ...
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