Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now
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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 ...


4

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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You must first calculate dilution factors. Assuming the volumes are additive, we can calculate dilution using $M_1V_1=M_2V_2$ equation. Also, assuming $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated, we can say: $$\ce{[Fe(NO3)3] = [Fe^3+]} \text{ and } \ce{[KSCN] = [SCN-]}$$ Thus, initial $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ are $0.20$ and $\pu{...


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Your confusion probably stems from the definition of a reversible path between two states: an infinite sequence of steps along a continuum of equilibrium states, such that at each step the equivalence condition of the second law of thermodynamics, $\mathrm dS_\text{total}=0$, is satisfied. However, there is no requirement that a path between two states ...


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This problem is relatively easy to solve if you remember that the vapor pressure of the products is independent of the amount of solid, provided there is enough solid for it to be in equilibrium with the gases. Therefore: $$K_p=p_{\ce{NH3}}^2p_{\ce{CO2}}$$ Since $p_{\ce{NH3}}=2p_{\ce{CO2}}$ $$K_p=4p_{\ce{CO2}}^3$$ and $p_{\ce{CO2}}=\pu{0.0194 atm}$. This ...


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Since at equilibrium, Moles of CO= Moles of Zn= moles of CO2 (as no information about initial moles are given) If no information is given, you should just give the three amounts names and treat them as unknowns. ZnO is exposed to pure CO at 1300 K This means that initially there is no carbon dioxide and no elemental zinc. Carbon dioxide and elemental ...


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Since battery scientists got Nobel this week, this question is worth pondering. It is not a trivial question. However consider this question: A car battery has 12 V, whereas 9 V cells are also common. If we connect two 1.5 V cells and one 9 V cell on series, one can generate 1.5+1.5+9 =12 V, yet this arrangement cannot start a car despite producing the same ...


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Water autohydrolyzes, or autodissociates, or autoionizes, to produce a very small amount of H+ and OH- ions. The product of their concentrations is 10e-14. Most salts that dissolve in water dissociate into ions. Often, one of these ions will have a strong attraction for H+ or OH- and tends to reduce the concentration of that ion in the solution. But since ...


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You can relate an equilibrium constant $K_p$ written in terms of partial pressures to one written in terms of mole fractions $K_\chi$, as follows: $$K_\chi=K_p (p/p_0)^{-\Delta \nu}\tag{1}$$ where $\Delta \nu$ is the difference in the stoichiometric coefficients for a molar amount of reaction. $K_p$ is independent of the pressure $p$ of the system but ...


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For your reaction, $\Delta n_g = 3-(1+1)=1$ (assuming all A,B,C are gases). $K_p = \frac{[C]^3}{[A][B]} (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$, where $\Sigma n$ is the total number of moles of gases in the reaction mixture(even of those species which are not taking part in the reaction). Here, it is a constant. When pressure is decreased($P_{total}$), $ ...


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I will first try to answer your second question. There's absolutely no mistake you made! As you assumed first step is RDS(rate determining step), You would expect $K=K_{f1}$. Continuing from where you left, $$K=K_{r2}\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$$ $$=\frac{\left[\ce{B}\right]^{b}\...


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The equilibrium constant normally written as $K_p$ for a gas phase reaction is independent of pressure but does depend on temperature. This means that if the pressure is changed then the amount of dissociation of reactants and products changes (and so does their partial pressure) until equilibrium is restored. $K_p$ is defined as the ratio of the partial ...


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I gather that the equilibrium constant K = 1 implies that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured. There is nothing special about K = 1. At equilibrium, no matter the value of K, there is no net reaction. Forward and backward reactions proceed at the same speed. But if pressure of the system ...


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