New answers tagged equilibrium
1
The terms reactant-favored and product-favored
As discussed in the comments and in YB609's answer, these are not technical terms with precise definitions.
First possible definition
One straight-forward definition would be to call reactant-favored reactions those for which $K$ is smaller than one, and product-favored those for which $K$ is larger than one. ...
0
You can never have more than one species labeled as liquid unless the liquids are immiscible. For the given reaction, just the solvent (if taking part in the reaction) should be labeled as iquid, and the rest not (“aq” if the solvent is water, and a non-textbook label otherwise). The underlying issue is whether the concentration of the species change in the ...
1
The notation is strange, but the result is correct.
For example, if you have 100 dioxygen molecules (so 200 oxygen atoms), 10% of the atoms are of the 16-O isotope and we distribute them randomly, we get the following:
81 both 18-O
18 mixed species
1 both 16-O
To visualize this, take all the numbers from 00 to 99, and say that the digit zero represents the ...
2
The problem is you were starting with an expression for the mean activity coefficient for salts where both the cation and anion were monovalent, and then attempting to derive, from this, a general expression for salts of all empirical formulas. To correct this, you need to start with the general expression for the mean activity coefficient of salts of any ...
2
This is an approach to Question 2.
The reaction equilibrium condition for the system is: $$\frac{V}{RT}\frac{n_C}{n_A n_B}=K_P\tag{1}$$Let $n_{A0}$, $n_{B0}$, and $n_{C0}$ be the number of moles at equilibrium for each species when the volume is equal to $V_0$. Then we have
$$\frac{V_0}{RT}\frac{n_{C0}}{n_{A0} n_{B0}}=K_P\tag{2}$$. For a small change in ...
1
Assuming $\ce{A, B,}$ and $\ce{C}$ are gases and behave like ideal gases we can conclude following by applying $P_xV=n_xRT$ where $x$ is either $\ce{A, B,}$ or $\ce{C}$, and $V$ and $T$ are constants:
$$[\ce{A}] = \frac{n_A}{V}=\frac{P_A}{RT}; \quad [\ce{B}] = \frac{n_B}{V}=\frac{P_B}{RT}; \quad \text{and } \quad [\ce{C}] = \frac{n_C}{V}=\frac{P_C}{RT} $$
...
1
Your conception of the equilibrium constant is flawed. $K_\mathrm{p}$ is preferably used when the reaction only contains gaseous and solid components whereas $K_\mathrm{c}$ is used primarily for a reaction taking place in water.
In reality, however, both of these equilibrium constants are special cases of equilibrium. This is because in the actual scenario, ...
3
The $K_\mathrm{sp}$ of $\ce{Fe(OH)3}$ is varied from source to source, but a reliable source gives the value of $2.79 \times 10^{-39}$ at $\pu{25 ^\circ C}$. We'll use this value throughout the calculations.
Suppose, $s$ amount of $\ce{Fe(OH)3}$ dissolves some in water according to its $K_\mathrm{sp}$, but assume water is not ionized:
$$\ce{Fe(OH)3(s) <=&...
-1
The solubility product of $\ce{Fe(OH)3}$ is about $\ce{10^{-39}}$. So its influence on the real solubility of $\ce{Fe(OH)3}$ is not significant. The following equilibrium constants are much more important : $$\ce{[Fe(H_2O)_6]^{3+} <=> [Fe(H_2O)5(OH)]^{2+} + H+ ; \ ...\ K = 10^{-3.05}}$$ $$\ce{[Fe(H_2O)_5(OH)]^{2+} <=> [Fe(H_2O)4(OH)_2]^{+} + H+ ; ...
1
The process described in Atkins must refer to an isothermal process, since otherwise the appropriate expression to use in computing the change in entropy of the surroundings is
$$ \ce{\Delta}S_{surr}=\int\dfrac{dq_{surr}}{T_{surr}}=-\int\dfrac{dq}{T_{surr}}$$
where $dq$ represents an inexact differential.
This becomes equal to the statement you quote only if ...
0
Mixing equal volumes of 0.2M $\ce{HCl}$ and 0.6M $\ce{H2SO4}$ would yield a solution with nominal concentration of 0.1M $\ce{HCl}$ and 0.3M $\ce{H2SO4}$.
Also given that $K_\mathrm{a2}$ for $\ce{H2SO4}$ is $1.2\times10^{-2}$.
RANT- I have come to absolutlye hate problems that don't use significant figures consistently. Is the initial HCl concentration ...
1
As a purely mathematical definition, a first order reaction is one in which the rate is directly proportional to concentration of reactants. The thing you should look for is physical examples of this. This only happens if you have reactions where one compound is decreasing and the other one is abundance (pseudo - first order) , or , radioactive decay where ...
0
Yes, I agree with your teacher's comment that upon increasing the total pressure of a system, there are select reactions "more dependent on collisions", and such reactions have been a matter of study.
For example, per this 2008 work reported in 'The Journal of Physical Chemistry', The Temperature and Pressure Dependence of the Reactions H + O2 (+M)...
1
The problem is that both titration curves are wrong: at the first equivalence points, the pH does not equal $\mathrm{p}K_\mathrm{a1}$ and the pOH does not equal $\mathrm{p}K_\mathrm{b1}$. As well, the two provided dissociation constants for carbonate ion, in the OP's upper figure, are also wrong.
The figure below shows two titration curves: 1) for 10 mL of 0....
0
The (equivalence)points, which you have marked in the graphs, are where all of the acid present in the sample has been neutralized by the base added with the titrant –and vice versa.
But, pKa equals pH at the point where only half of the acid is neutralized. In other words: [HA] = [A-] -the concentration of acid equals the concentration of the corresponding ...
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