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Concept 05 / 08Equilibrium

Reaction Quotient and Equilibrium Calculations

The reaction quotient Q and predicting the direction of reaction, setting up ICE tables, solving for equilibrium concentrations, degree of dissociation, and finding the extent of dissociation from vapour density measurements.

Figure 1
Reaction quotient. For any mixture (not necessarily at equilibrium), QcQ_{c} is written exactly like KcK_{c} but with the actual concentrations at that moment. Q<KQ < K: too few products; the reaction proceeds forward. Q>KQ > K: too many products; the reaction proceeds backward. Q=KQ = K: the system is at equilibrium (see figure). Equilibrium calculations (ICE table). Write the Initial amounts, the Change in terms of a single unknown xx using stoichiometry, and the Equilibrium amounts; substitute into KK and solve. Convert moles to concentrations by dividing by volume (if Δn=0\Delta n = 0 the volume cancels). If KK is very small, xx is often negligible compared with initial concentrations, which simplifies the algebra; check afterwards that xx is less than about 5%5\% of the quantity it is subtracted from. Degree of dissociation (α\alpha). The fraction of a reactant that dissociates. For PCl5⇌PCl3+Cl2\mathrm{PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2}} starting with 1 mol1\,\mathrm{mol} PCl5\mathrm{PCl_{5}}: at equilibrium 1−α1 - \alpha, α\alpha, α\alpha; total moles 1+α1 + \alpha. With total pressure PP, each partial pressure is its mole fraction times PP: Kp=(α1+αP)21−α1+αP=α2P1−α2K_{p} = \frac{\left(\frac{\alpha}{1 + \alpha}P\right)^{2}}{\frac{1 - \alpha}{1 + \alpha}P} = \frac{\alpha^{2}P}{1 - \alpha^{2}}. If α≪1\alpha \ll 1, α≈KpP\alpha \approx \sqrt{\frac{K_{p}}{P}}: raising pressure suppresses dissociation. Vapour density method. Dissociation increases the number of moles, so the observed density (and apparent molar mass) of the equilibrium mixture falls. If one molecule gives nn particles, α=D−d(n−1)d\alpha = \frac{D - d}{(n - 1)d}, where DD is the theoretical vapour density (no dissociation) and dd the observed one. (For association, such as dimerisation, the same idea gives d>Dd > D.) Key Takeaways: - Q<KQ < K: forward; Q>KQ > K: backward; Q=KQ = K: equilibrium. - ICE tables in one unknown; approximate only when KK is small, then verify. - PCl5\mathrm{PCl_{5}}: Kp=α2P1−α2K_{p} = \frac{\alpha^{2}P}{1 - \alpha^{2}}. - α=D−d(n−1)d\alpha = \frac{D - d}{(n - 1)d} from vapour densities. - Pitfall: in KpK_{p} problems, partial pressure uses the total moles including products, so the denominator 1+α1 + \alpha must appear. Brain-Teaser 1: For H2(g)+I2(g)⇌2HI(g)\mathrm{H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)}, Kc=50K_{c} = 50 at about 720 K720\,\mathrm{K}. 1 mol1\,\mathrm{mol} each of H2\mathrm{H_{2}} and I2\mathrm{I_{2}} are heated in a 1 L1\,\mathrm{L} flask. Find the equilibrium amount of HI. Solution: At equilibrium: 1−x1 - x, 1−x1 - x, 2x2x. (2x)2(1−x)2=50\frac{(2x)^{2}}{(1 - x)^{2}} = 50, so 2x1−x=50=7.07\frac{2x}{1 - x} = \sqrt{50} = 7.07, giving x=7.079.07=0.780x = \frac{7.07}{9.07} = 0.780. HI =2x=1.56 mol= 2x = 1.56\,\mathrm{mol}; H2\mathrm{H_{2}} and I2\mathrm{I_{2}} are 0.22 mol0.22\,\mathrm{mol} each. Brain-Teaser 2: The vapour density of PCl5\mathrm{PCl_{5}} at a certain temperature is observed to be 6262. Find its degree of dissociation. Solution: M(PCl5)=208.5M(\mathrm{PCl_{5}}) = 208.5, so D=104.25D = 104.25. One molecule gives 22 particles: α=104.25−6262=0.68\alpha = \frac{104.25 - 62}{62} = 0.68. About 68%68\% has dissociated. Brain-Teaser 3: For N2+3H2⇌2NH3\mathrm{N_{2} + 3H_{2} \rightleftharpoons 2NH_{3}}, Kc=0.5 L2 mol−2K_{c} = 0.5\,\mathrm{L^{2}\,mol^{-2}} at some temperature. A mixture has [N2]=1[\mathrm{N_{2}}] = 1, [H2]=1[\mathrm{H_{2}}] = 1 and [NH3]=2 mol L−1[\mathrm{NH_{3}}] = 2\,\mathrm{mol\,L^{-1}}. In which direction will it react? Solution: Q=221×13=4>0.5=KQ = \frac{2^{2}}{1 \times 1^{3}} = 4 > 0.5 = K. There is excess product, so ammonia decomposes (backward reaction) until QQ falls to 0.50.5.