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The state in which both reactants and products are present at concentrations which have no further tendency to change with time.

$K$, the standard thermodynamic equilibrium constant, is computed from $\Delta G^\circ$, using $$\Delta G^\circ = -RT\log K \tag1$$ Generally speaking, $K$ in equation (1) is unitless. Its value … depends on the specified reference standard states and $T$ (and obviously on the equilibrium activities of reactants and products). Both $K$ and $\Delta G^\circ$ change (and are certainly allowed to …
answered May 24 by Buck Thorn
A variety of models have been concocted to describe the dependence of gas-phase reaction rates on the energy of collisions between two molecules. At its simplest you have the empirical Arrhenius equat …
answered Apr 29 by Buck Thorn
run we are all dead" sort). This is of course a tricky point. We often need to distinguish between "real equilibrium states" and "kinetically trapped or metastable states". However it is useful to … be considered to be at equilibrium during that interval. This definition is particularly valuable since thermodynamics is an empirical science. Caveat: that doesn't mean you can be sure that the system …
answered Mar 29 by Buck Thorn
A cursory search suggests the solubilities in g/100 mL on the wikipedia are closer to a consensus, for what it's worth: solubility 0.2% ($0^oC$): solubi …
answered Nov 8 '18 by Buck Thorn
the comments. The following shows more formally how to derive the connection between Henry's law, an equilibrium constant and the standard Gibbs free energy for evaporation of the solute. At … equilibrium between the solute in the gas and solution phases, the chemical potential of the solute is equal in both phases, so that $$\begin{align} \mu(sol)&=\mu(g)\end{align}$$ In the gas phase (assuming …
answered May 24 by Buck Thorn
The difficulty with problems in kinetics is that you can envision a pool of intermediates such that, while B is consumed, you don't see significant formation of C, so that C is a poor reporter of the …
answered Feb 25 by Buck Thorn
For each phase (graphite or diamond) you can write $$ \mu = \mu^\circ+\int_{P^\circ}^{P} V_mdP $$ at constant T. We are asked to find the pressure $P=P_{eq}$ at which carbon coexists in the two phas …
answered Feb 23 by Buck Thorn
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 …
answered Jun 3 by Buck Thorn
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, spontaneity, reversibility, work …
answered May 30 by Buck Thorn
Can I use the Henderson–Hasselbalch equation on reactions that are not buffers? Yes, but you should still keep in mind some limitations. The equation you provide $$K_\mathrm{a} = \frac{[\ce{A-} …
answered May 1 by Buck Thorn
There are a few separate issues here to keep in mind: $K_c$ (the equilibrium constant in terms of concentrations) is defined as $$K_c = \prod {c_i}^{\nu_i} \tag{1}$$ Agreement between $K_c$ and … the product of forward/back rate constants ($k_{\pu{fwd}}/k_{\pu{rev}}$ in the OP example) is expected only if the mechanism is correct, assuming some proportional rate law, and at equilibrium. When …
answered Jul 3 by Buck Thorn
In my version of the book the original assumption (step 7a) is shown to lead to a contradiction when later checked (logically, this form of proof is called reductio ad absurdum or proof by contradicti …
answered Feb 20 by Buck Thorn
, and volume $$ [i] = \frac{n_i}{V} $$ b) Equilibrium constant in concentration units (when activity coefficients are ~1) and in terms of moles of each substance at equilibrium: $$ K_c = \frac{[C][DA … ]}{[A]^2[B]} = V\frac{n_{C,eq}n_{DA,eq}}{n_{A,eq}^2n_{B,eq}}$$ Strategy: i) Compute final (equilibrium) moles of A, and moles of A that reacted: $$ n_{A,eq} = [A]_{eq}V = 0.40 M \times 2.0 L =0.80 …
answered Feb 6 by Buck Thorn
As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$\Delta G=\Delta G^\circ+ RT\ln{1}=\Delta G^\circ$$ On the other hand at equilibrium $Q=K$ and so $$\Delta G … =\Delta G^\circ + RT\ln{K}$$ This of course leads to $\Delta G^\circ = -RT\ln{K}$ since at equilibrium $\Delta G=0$. So you might want to think of it as multiple statements: For the conversion of …
answered Mar 24 by Buck Thorn
The key point is that the entropy of the surroundings changes, not that of the system (at least for a small change in T), so only [B] and [D] are relevant, as you correctly determined. You can write …
answered Feb 22 by Buck Thorn

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