New answers tagged

1

Ignoring the math, "in equilibrium" means the forward and reverse reactions are proceeding at the same rate, so there is no gross change in the amount of reactants. In a closed system, if initially one reaction was faster, then it's product will start to dominate, and, by the law of mass action, the reverse action will "catch up" because there is now a ...


3

How do I even write equilibrium constant for solid solid equilibrium? First, let's write the expression for the reaction quotient: $$Q = 1$$ Q is a constant, always one, as long as all reactants and products are present. If you run out of one, it is zero or infinity. Now, there are two cases. If the reaction is at equilibrium, K is also 1. If not, you ...


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 ...


0

Note: This answer has been heavily edited, because the original answer I posted contained some assumptions and was very limited (and incorrect) in the analysis. I would like to highlight the patience of @Buck Thorn, @Andrew and @Karsten Theis in pointing out my errors, expanding the discussion and going into a lot of detail. The trouble is not with any ...


1

The dependence of an equilibrium constant on temperature is given by the van't Hoff equation: $$\left(\frac{\partial{\log(K)}}{\partial{T}}\right)_p=\frac{\Delta H^\circ}{RT^2}$$ Therefore for an exothermic reaction ($\Delta H^\circ<0$) you expect a decrease in $K$ with temperature, as Le Châteliers principle would predict. To give an example, the ...


1

I have found a paper which does this same experiment and provides a discussion of this phenomenon[1]. I will begin by just quoting from the discussion section: It has been stated in the introductory section that a maximum in the formal Cp° for a pair of thermally stable free ions can be ascribed to an endothermic change in the solvent shell around ...


1

Initial note The specific problem with your scenario is, that you need a buffer of buffer capacity progressively decreasing with increasing $\mathrm{pH}$, but capacity of simple buffers is for $\mathrm{pH} \lt \mathrm{p}K_\mathrm{a}$ increasing. As consequence, its "breaking" capacity is too big, compared to "withstanding capacity". General analysis Let ...


0

You can use the formula for the ideal gas $$\rho=\frac{p\cdot M_\mathrm{mean}}{R\cdot T}$$ As pressure, temperature and density are given, you get the mean molar mass. $$M_\mathrm{mean}=\frac{\rho RT}{p}$$ From that, you get molar fractions. $$x_{\ce{O2}}=\frac{ M_{\ce{O3}}-M_\mathrm{mean}}{ M_{\ce{O3}}- M_{\ce{O2}}}$$ $$x_{\ce{O3}}=1-x_{\ce{O2}}$$ ...


0

Does a Gibbs energy maxima correspond to equilibrium state or not? (a) If yes, doesn't this violate the Second Law which implies that Gibbs energy should be minimized whenever possible? No matter which direction you move, your Gibbs energy will always decrease. There are a number of central concepts in thermodynamics that are relevant here: equilibrium,...


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 ...


2

What is equal at equilibrium? The forward and reverse rates are equal at equilibrium. What is constant at equilibrium? The concentrations are constant, and the rates are constant. What is typically not equal at equilibrium? The concentrations of reactants and products are typically not equal at equilibrium. For a reaction with multiple species in ...


1

Reaction rate depends on concentration of reactants. Consider a simple reaction $$\ce{A -> B} $$ In this reaction, rate in forward direction is given by rate law $r_1 =k_1 \ce{[A] }$ And in backward direction $r_2 =k_2\ce{[B]} $ Then eventually a time will come when $r_1=r_2$, since concentration of both A and B are either decreasing or increasing, ...


0

I think your confusion is coming from the idea that the rate of the forward direction isn't permanently increased for example. As an example consider $$\ce{N2 + 3H2 <=>2NH3}$$ as this is a reaction you may be familiar with. If I had the reaction in a sealed container meaning an equilibrium will be established then the moles and thus concentration of ...


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\...


2

The figure shows the relationship between the solid and liquid solvent and a solution. Instead of freezing at the freezing point of the solvent $T_1$ a solution freezes at $T_2$, where the vapour pressure of the solution equals that of the pure solid solvent. Using the Clausius-Clapeyron equation we can show that the solution curve will always intersect ...


1

I'm following the outline from the comment by user21398. Initial concentrations after mixing The volume of the mixture is $V_\text{mix} = \pu{10 mL}$. Using the dilution law, I get the following concentrations: $$\ce{[Fe^3+]_\text{initial}} = \pu{2.00 mM} \cdot \frac{\pu{5 mL}}{\pu{10 mL}} = \pu{1.00 mM}$$ $$\ce{[SCN-]_\text{initial}} = \pu{2.00 mM} \...


Top 50 recent answers are included