# Tag Info

21

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 ...

12

Chemical equilibrium is a type of dynamic equilibrium, but not every dynamic equilibrium is a chemical equilibrium. In a chemical equilibrium there is no change on the macroscopic scale. That means that if you look at the system it seems like nothing is happening, but at molecular scale there are reactions going on and the rate of forward reaction = rate of ...

12

A chemical equilibrium concerns chemical reactions. There should be at least a forward- and backward reaction between two species but more complex systems with multiple individual reactions may occur. The important observation is that there is no macroscopic change to the chemical constituents of the system, i.e. the concentrations of all reaction partners ...

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\...

8

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

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

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. [1]. 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 [2]. However, thirty years later (late 1980s) there's been another study by Licht [3], which showed significant ...

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, ... 6 Current definition implies that \mathrm{pH} is a function of relative activity. Originally, the amount concentration of \ce{H+} in \pu{mol L-1} was proposed, which is also often used these days as an approximation [1]. \mathrm{pH} was originally defined by Sørensen in 1909 … in terms of the concentration of hydrogen ions (in modern nomenclature) ... 6 [Comment by Poutnik] Important is also en.wikipedia.org/wiki/Grotthuss_mechanism for the proton interchange. Mobility of H3O+ and OH- gives a hint it must be fast. If you compare the diffusion coefficient of hydroxide (\pu{5.270e9 m^2/s}) to that of fluoride (\pu{1.460e9 m^2/s}), you might be surprised to see such a difference despite their comparable ... 5 It depends on the circumstances, but mostly not Under most normal atmospheric circumstances water in an open vessel is not in equilibrium with water vapour in the atmosphere. We know this because water usually evaporates, albeit slowly. One way to measure whether water is in equilibrium is to measure the humidity which is a measure of the maximum ... 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) <...

5

Strictly speaking, the answer depends on conditions and, as a consequence, states of aggregation. Reaction (1) can be reversible if iron(II) sulfide is subsequently heated above $\pu{700 °C}$ in vacuum to prevent oxidation of the elements [1, p. 422]: $$\ce{FeS(s) ->[\pu{700 °C}][vac] Fe(s) + S(g)}\tag{1a}$$ Reactions \eqref{rxn:2a} and \eqref{rxn:3a} ...

5

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 ...

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