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Concept 04 / 08Equilibrium

Dynamic Equilibrium, the Law of Mass Action, Kc and Kp

Physical and chemical equilibria as dynamic states, the law of mass action and the equilibrium constant Kc, pressure constant Kp and the relation Kp = Kc(RT)^Δn, rules for combining, reversing and scaling equilibrium constants, and homogeneous versus heterogeneous equilibria.

Figure 1
Many processes do not go to completion. In a closed system they reach a state where the amounts of reactants and products stop changing. This is equilibrium. Dynamic nature. At equilibrium the forward and backward reactions continue at equal rates, so there is no net change (see figure). Evidence: if radioactive 14CO2\mathrm{^{14}CO_{2}} is added to a closed system of CaCO3\mathrm{CaCO_{3}}, CaO\mathrm{CaO} and CO2\mathrm{CO_{2}} at equilibrium, radioactivity soon appears in the solid. Equilibrium can be approached from either side and needs a closed system. Physical equilibria: ice ⇌ water at 0∘C0^{\circ}\mathrm{C}; liquid ⇌ vapour (constant vapour pressure); solid ⇌ solution (saturated solution); gas ⇌ gas dissolved in a liquid (Henry's law). Law of mass action (Guldberg and Waage). For aA+bB⇌cC+dDa\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D} at equilibrium, Kc=[C]c[D]d[A]a[B]bK_{c} = \frac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}}, where the brackets are equilibrium molar concentrations. KcK_{c} depends only on temperature (and on how the equation is written), not on initial amounts, pressure or catalysts. Kp for gases. Using partial pressures: Kp=pCcpDdpAapBbK_{p} = \frac{p_{\mathrm{C}}^{c}p_{\mathrm{D}}^{d}}{p_{\mathrm{A}}^{a}p_{\mathrm{B}}^{b}}. Since p=nVRT=[ ]RTp = \frac{n}{V}RT = [\,]RT, Kp=Kc(RT)ΔnK_{p} = K_{c}(RT)^{\Delta n}, where Δn=(c+d)−(a+b)\Delta n = (c + d) - (a + b) counts gaseous moles only. If Δn=0\Delta n = 0, Kp=KcK_{p} = K_{c} (for example H2+I2⇌2HI\mathrm{H_{2} + I_{2} \rightleftharpoons 2HI}). Use R=0.0821 L atm mol−1 K−1R = 0.0821\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}} with pressures in atm (or 0.08310.0831 with bar). Manipulating K. Reverse the equation: K′=1KK' = \frac{1}{K}. Multiply the equation by nn: K′=KnK' = K^{n} (halving gives K\sqrt{K}). Add two equations: K=K1K2K = K_{1}K_{2}. Heterogeneous equilibria. Pure solids and pure liquids have constant concentration (activity =1= 1) and are omitted from KK. For CaCO3(s)⇌CaO(s)+CO2(g)\mathrm{CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)}, Kp=pCO2K_{p} = p_{\mathrm{CO_{2}}}: at a given temperature the pressure of CO2\mathrm{CO_{2}} is fixed regardless of how much solid is present. Magnitude of K. K>103K > 10^{3}: products predominate, reaction nearly complete. K<10−3K < 10^{-3}: reactants predominate. In between, appreciable amounts of both. KK says nothing about how fast equilibrium is reached. Key Takeaways: - Equilibrium is dynamic: forward rate == backward rate. - Kc=[products]coeff[reactants]coeffK_{c} = \frac{[\text{products}]^{\text{coeff}}}{[\text{reactants}]^{\text{coeff}}}; omit pure solids and liquids. - Kp=Kc(RT)ΔngK_{p} = K_{c}(RT)^{\Delta n_{g}}. - Reverse: 1K\frac{1}{K}; multiply by nn: KnK^{n}; add: multiply the KK values. - KK depends only on temperature. - Pitfall: water is omitted from KK only when it is the solvent or a pure liquid; in a gas-phase reaction, H2O(g)\mathrm{H_{2}O(g)} must be included. Brain-Teaser 1: For 2SO2(g)+O2(g)⇌2SO3(g)\mathrm{2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)}, Kc=280 L mol−1K_{c} = 280\,\mathrm{L\,mol^{-1}} at 1000 K1000\,\mathrm{K}. Find KpK_{p}, and KcK_{c} for SO3⇌SO2+12O2\mathrm{SO_{3} \rightleftharpoons SO_{2} + \frac{1}{2}O_{2}}. Solution: Δn=2−3=−1\Delta n = 2 - 3 = -1. Kp=2800.0821×1000=3.41 atm−1K_{p} = \frac{280}{0.0821 \times 1000} = 3.41\,\mathrm{atm^{-1}}. The second equation is the first reversed and halved: Kc′=1280=0.0598 (mol L−1)1/2K_{c}' = \frac{1}{\sqrt{280}} = 0.0598\,\mathrm{(mol\,L^{-1})^{1/2}}. Brain-Teaser 2: Write KpK_{p} for NH4HS(s)⇌NH3(g)+H2S(g)\mathrm{NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)}. If the total pressure at equilibrium (starting from the pure solid) is 0.84 atm0.84\,\mathrm{atm}, find KpK_{p}. Solution: Kp=pNH3 pH2SK_{p} = p_{\mathrm{NH_{3}}}\,p_{\mathrm{H_{2}S}} (the solid is omitted). The gases form in equal amounts, so each has 0.42 atm0.42\,\mathrm{atm} and Kp=0.422=0.176 atm2K_{p} = 0.42^{2} = 0.176\,\mathrm{atm^{2}}. Brain-Teaser 3: Given N2+O2⇌2NO\mathrm{N_{2} + O_{2} \rightleftharpoons 2NO} (K1K_{1}) and 2NO+O2⇌2NO2\mathrm{2NO + O_{2} \rightleftharpoons 2NO_{2}} (K2K_{2}), find KK for 12N2+O2⇌NO2\mathrm{\frac{1}{2}N_{2} + O_{2} \rightleftharpoons NO_{2}}. Solution: Adding the two gives N2+2O2⇌2NO2\mathrm{N_{2} + 2O_{2} \rightleftharpoons 2NO_{2}} with K=K1K2K = K_{1}K_{2}. Halving: K=K1K2K = \sqrt{K_{1}K_{2}}.

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Dynamic Equilibrium, the Law of Mass Action, Kc and Kp — DPP

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