Energetics & Equilibria 4: Equilibrium

solids: trick question, solid species don’t contribute to expression

Chemical potentials of solids/liquids = standard chemical potentials I.e. µ = µo, so the term that solid A with coefficient vA = (µ/µo)vA = 1. Therefore the contribution is to multiply the overall expression by 1 – no change.

∆rG = 0 Effectively, ∆rG is the slope of a graph of Gibbs energy against composition, composition being the ratio of products to reactants. See image

∆rGo = -RTlnK

Assume generic reaction A + B ⇔ P + Q For a mixture of gases (NB you could do equiv calculation assuming solutions instead): ∆rG = vpμp(pP) + vQμQ(pQ) - vAμA(pA) - vBμB(pB) Assume all species are gaseous μ(pi) = μoi + RT ln(pi/po) Then sub in chemical potential expressions ∆rG = vp(μoP + RT ln(pp/po)) + vQ(…) - vA(…) - vB(…) Gather together the standard chemical potentials ∆rG = (vpμoP + vQμoQ - vAμoA - vBμoB) + vpRT ln(pp/po) + vQ(…) - vA(…) - vB(…) Quantity in bracket = ∆rGo. Put the stoichoimetric coefficients onto the ln terms So ∆rG = ∆rGo + RT ln(pp/po)vp + …etc Then gather the ln terms, from here on see image

ΔrG = µ(CO2) + µ(CaO) - µ(CaCO3) For the solids, µ = µo. For CO2 gas, write in terms of partial pressure, recalling µi(pi) = µoi + RTln(pi/po) ΔrG = µo(CO2) + µo(CaO) - µo(CaCO3) + RTln(pCO2/po) ΔrG = [µo(CO2) + µo(CaO) - µo(CaCO3)] + RTln(pCO2/po) Quantity in [] = ΔrGo ΔrG = ΔrGo + RTln(pCO2/po) At equilibrium, ΔrG = 0, so: ΔrGo = -RTln(pCO2,eq/po) This demonstrates that, although solids/liquids don’t contribute to the expression for the eq constant K, they do still contribute to ΔrGo, and so they affect the value of K indirectly

• removing product increases proportion of reactant in mixture • graph moves from eq position to position marked a • to return from a to eq is associated with a decrease in gibbs energy, therefore spontaneous • so some reactants reform products to counteract the change

Derivation Recall ΔrGo = -RTlnK and also ΔrGo = ΔrHo - TΔrSo So -RTlnK = ΔrHo - TΔrSo Divide through by -RT lnK = -(ΔrHo/R)(1/T) + (ΔrSo/R) Interpretation Assume ΔrHo and ΔrSo are temperature-independent over a modest temperature range. • For an endothermic reaction, ΔrHo > 0, so -(ΔrHo/R)(1/T) is negative, so increasing T makes the term less negative, so lnK and thus K increase. Shifts eq towards products (increased K) • For an exothermic reaction, ΔrHo rHo/R)(1/T) is positive, so increasing T makes the term less positive, so lnK and thus K decrease. Shifts eq towards reactants (decreased K) NB this approach is only applicable for reactions which go to completion

d/dT(G/T) = -H/T2 (constant pressure)

d/dT(G/T) = -H/T2 (constant pressure) Replace with standard quantities d/dT(ΔrGo/T) = -ΔrHo/T2 (constant pressure) Sub in ΔrGo = -RTlnK –> ΔrGo/T = -RlnK d/dT(-RlnK) = -ΔrHo/T2 Divide through by -R dlnK/dT = ΔrHo/RT2

• For a gas phase reaction, Kp and ∆rGo are related via ∆rGo = −RT lnKp, and ∆rGo is defned under standard conditions (pressure = 1 bar), so Kp does not vary with pressure • Composition of eq mixture changes in order to keep Kp constant

Let α = degree of dissociation of dimer A2, i.e. the fraction which have dissociated, where α = 0 means no dissociation and α = 1 means full dissociation. n0 moles of A2 are allowed to reach eq at pressure ptot, giving degree of dissociation α. So neq,A2 = (1-α)no, and neq,A = 2αno, and therefore neq,total = (1+α)n0. Partial pressure of a species i: pi = xiptot Then write out the partial pressures of each species and derive the eq constant, then rearrange for α (image from here on).

• Couple an unfavourable reaction (ΔrGo > 0) to a favourable one (ΔrGo​ > 0) such that overall (ΔrGo​ > 0), so the reaction can occur • Coupling occurs physically through connecting reactions closely, eg via an enzyme • Examples: • ATP hydrolysis is often the favourable reaction used to drive another, e.g. protein synthesis • ATP synthesis is therefore unfavourable; it is coupled to glucose oxidation • Photosynthetic production of glucose is not favourable, so is coupled to light energy