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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.

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$K$, 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
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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
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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
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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% ($0^oC$): https://www.cdc.gov/niosh/npg/npgd0092.html solubi …
answered Nov 8 '18 by Buck Thorn
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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
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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
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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
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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
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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
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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 $$K_\mathrm{a} = \frac{[\ce{A-} …
answered May 1 by Buck Thorn
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There are a few separate issues here to keep in mind: $K_c$ (the equilibrium constant in terms of concentrations) is defined as $$K_c = \prod {c_i}^{\nu_i} \tag{1}$$ Agreement between $K_c$ and … the product of forward/back rate constants ($k_{\pu{fwd}}/k_{\pu{rev}}$ in the OP example) is expected only if the mechanism is correct, assuming some proportional rate law, and at equilibrium. When …
answered Jul 3 by Buck Thorn
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In my version of the book the original assumption (step 7a) is shown to lead to a contradiction when later checked (logically, this form of proof is called reductio ad absurdum or proof by contradicti …
answered Feb 20 by Buck Thorn
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, and volume $$ [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
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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
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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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