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20

Because equilibrium is dynamic not static There are many things in the world of chemistry which could be simpler but are not. Equilibrium is one of them. There are few chemical processes where equilibrium is equivalent to the situation where everything just stops. Equilibrium normally occurs when the rates of processes going forward are the same as the ...

19

Yes, equilibrium and steady-state are distinct concepts. A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant. Steady-state implies a system that is not ...

15

The textbook is precisely correct. The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures. In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$. One must be very ...

13

$\Delta G^\circ_m$ is the difference in molar Gibbs free energy between the reagents and products in their standard states (in the case of $\ce{AgI(s)}$, the standard state for the reagent is the pure solid and that for the product is the solute in a $\pu{1 molal}$ ideal solution, all at $p^\circ =\pu{ 1 bar}$). If $\Delta G^\circ_m>0$ you cannot ...

8

The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants. Here is the ...

8

I think the easy way out is to invoke $S_\mathrm m = R \ln \Omega$. If we assume that for a generic complex $\ce{MA_{n}B_{$N-n$}}$, $$\Omega = {N \choose n} = \frac{N!}{n!(N-n)!} \quad \left[ = {N \choose N-n} \right]$$ and that for the individual molecules $\ce{A}$ and $\ce{B}$, $\Omega = 1$, then the equilibrium constant $K$ for $$\ce{MA_{n}B_{N-n} + ... 8 I'm not an expert but this is how I would do it. Determine the reaction rate formula from the reaction equation$$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\...

7

This is the nature of the quantum world. An action, at some level, either takes place or does not... there is no half-way state. Another example is the photoelectric effect. Planck and Einstein explained the requirement for at least a minimum energy of a photon before it can raise an electron to a higher energy level. A million photons just under that energy ...

7

\begin{align} \ce{HNO2 + H2O &<=> NO2- + H3O+} &\quad K &= K_\mathrm{a} \tag{1}\\ \ce{NO2- + H2O &<=> HNO2 + OH-} &\quad K &= K_\mathrm{b} \tag{2}\\ \ce{H2O + H2O &<=> H3O+ + OH-} &\quad K &= K_\mathrm{w} \tag{3} \end{align} Above are the reactions associated with equilibrium constants commonly ...

7

The statement is inaccurate. A better statement would be something like: A chemical equilibrium can not be established if one of the products is continuously removed (i.e. its concentration always decreases). If one of the products is a gas and can mix with earth's entire atmosphere, that would make its concentration (or partial pressure) drop to almost ...

7

Your first question: Before the second law was understood it used to be thought that the maximum amount of work that could be extracted from a reaction was $-\Delta H$ but many experiments showed that this was not the case. It is true that, according to the first law, (the law of conservation of energy), that external work done must be equal to the loss of ...

6

The $K_\mathrm{sp}(\ce{Co2S3})$ value of magnitude of $10^{-124}$ appears in paper by Goates et al. . The refined "thermodynamic" value of $\pu{2.6e-124}$ that you've listed and is used by numerous textbook up to these days has been proposed in . However, thirty years later (late 1980s) there's been another study by Licht , which showed significant ...

6

In my version of the book the original assumption (step 7a) is shown to lead to a contradiction when later checked (logically, this form of proof is called reductio ad absurdum or proof by contradiction). The point is that you should 1) learn not to be afraid to apply assumptions, but that you should then 2) verify that the original assumption was indeed ...

6

Molar concentration can be expressed via mass $m$ and volume $V$ all right: $$C_i = \frac{n_i}{V_i} = \frac{m_i}{M_iV_i}$$ Analyte doesn't change during the extraction and its molar mass $M$ remains the same $(M_i = \text{const})$ so the distribution coefficient $D$ can be rewritten as such: $$D = \frac{C_\mathrm{s}}{C_\mathrm{w}} = \frac{m_\mathrm{s}V_\... 6 See my comment above. As a numerical example, take acetic acid (\ce{AcOH}), which has K_\mathrm{a} = 1.8 \cdot 10^{-5}. This means that:$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]}$$And the total nominal concentration of acid is:$$C_\mathrm{a} = [\ce{AcOH}] + [\ce{AcO-}]$$Combining these two equations, you can see that the % of ... 6$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with? You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.$$\ce{H3O+ + OH- <=> 2 H2O}$$If you add up the two reaction, you get a third one:$$\ce{CH3COOH + OH- <=> H2O + ...

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 ...

6

The van't Hoff equation describes how the equilibrium constant changes with temperature: $$\ln \frac{K_2}{K_1} = −\Delta H_R^\circ (\frac{1}{T_2} − \frac{1}{T_1})$$ So if the reaction enthalpy is zero for a given temperature interval, the equilibrium constant will not change. This also means that the activation energy in either direction is the same, at ...

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, ...

5

I think the author of this problem forgot to add units to $K_c$ since it's not a dimensionless entity: $$K_c = \frac{[\ce{NO}]^2}{[\ce{N2O}][\ce{O2}]^{0.5}}$$ and should be $K_c = \pu{1.7e-13 mol^{0.5} L^{-0.5}}$. In general $$[K_c] = \mathrm{dim}(c)^{Δn}$$ where square brackets denote the dimensions of the quantity $K_c$, $c$ is concentration and $Δn$ ...

5

For one thing, it would be hugely more unlikely to exist. We are more likely to have worlds where there is a little more deuterium in the hydrogen, not 100%. Most of the hydrogen in our Universe is protium (hydrogen-1) because the vast majority of unfused nuclear particles in stars are protons. Neutrons must either fuse with something or they decay to ...

5

In most of these type problems chemists look for ways to simply the mathematical expression by using approximations based on a knowledge of the chemistry. The gist is that because of the significant figures you assume that the reaction essentially goes to completion. So $\ce{Ag(CN)_2^− \approx 0.0300}$ and $\ce{CN− \approx 0.0400}$. Then you solve for [Ag+]. ...

5

The problem is a bit strange since the details of the absorbance measurement aren't detailed very well. I'll assume the book defines the relationship as: $$A = \epsilon bc$$ where: A=absorbance $\epsilon$ = molar attenuation coefficient $b$ = path length $c$ = concentration So the concentration, c, of the $\ce{[Fe(SCN)]^2+}$ is: $$c = \dfrac{A}{\epsilon ... 5 To start with, its often helpful to make an ICE table when dealing with problems of chemical equilibrium.$$\begin{array}{cccc} \begin{array}{c|ccc} \hline & \ce{Fe^3+} & \ce{SCN-} & \ce{[Fe(SCN)]^2+} \\ \hline \mathrm{initial} & 1.32\times10^{-5} & 8.40\times10^{-6} & 0\\ \mathrm{change} & -x & -x &+x \\ \mathrm{equil} &...

5

I think the answer may just come down to a simple counting of available sites. The equilibrium constant for a reaction is equal to the ratio of the forward and reverse reaction rates. For the forward reaction, there are $N-n$ sites available at which a ligand can replace an $\ce{H2O}$. Conversely, for the reverse reaction, there are $n+1$ ligand sites at ...

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 ...

5

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 ...

5

Does this mean that it only applies to homogeneous reactions or that it can also be applied to homogeneous reactions among other types? The latter. If you consider the distribution of a solute S in a mixture of water and octanol, this is a heterogenous system (in this case, two immiscible liquids). The equilibrium reaction (or process) is: $$\ce{S(oct) <... 4 The equilibrium constant K_p is dimensionless as it is defined, ultimately, in terms of the ratio of activities, which are themselves dimensionless. In practice we often use partial pressure instead of activities and then we use the numerical value by effectively dividing each partial pressure by 1 unit of pressure, say 1 atm. In cases when a mole ... 4 There are many examples of such reactions, for instance:$$ \begin{align} \ce{S (s) + O2 (g) &<=> SO2 (g)}\\ \ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\\ \ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)} \end{align}  However, your assumption all that is considered is which side the solid/liquid is on, not the moles of gas present on ...

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