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184 chemistry Let x mol per litre of PCl be dissociated, Taking antilog of both sides, we get, 5 K = e –∆G/RT (6.23) At equilibrium: (3-x) x x Hence, using the equation (6.23), the reaction spontaneity can be interpreted in K c [PCl 3[Cl 2/[PCl 5  1.8 = x / (3 – x) terms of the value of ∆G  . • If ∆G  < 0, then –∆G  /RT is positive, x2 + 1.8x – 5.4 = 0  and e ∆DG  /RT>1, making K >1, which x = [–1.8 ± √(1.8)2– 4(–5.4)]/2 implies a spontaneous reaction or the reaction which proceeds in the forward x = [–1.8 ± √3.24 + 21.6]/2 x = [–1.8 ± 4.98]/2 direction to such an extent that the products are present predominantly. x = [–1.8 + 4.98]/2 = 1.59 • If ∆G  > 0, then –∆G  /RT is negative, and [PCl5] = 3.0 – x = 3 –1.59 = 1.41 M  ∆G </RT 1, that is , K < 1, which implies e [PCl3] = [Cl2] = x = 1.59 M a non-spontaneous reaction or a reaction which proceeds in the forward direction 6.7 Re lationsh ip betwee n to such a small degree that only a very Equilibrium Constant K, minute quantity of product is formed. Reac tion Quo tient Q and Gibb s Energy G Problem 6.10 The value of K cfor a reaction does not depend The value of ∆G  for the phosphorylation of on the rate of the reaction. However, as you glucose in glycolysis is 13.8 kJ/mol. Find have studied in Unit 5, it is directly related the value of K ct 298 K. to the thermodynamics of the reaction and in particular, to the change in Gibbs energy, Solution ∆G. If, ∆G  = 13.8 kJ/mol = 13.8 × 10 J/mol • ∆G is negative, then the reaction is  Also, ∆G  = – RT lnK c spontaneous and proceeds in the forward Hence, ln K = –13.8 × 10 J/mol direction. c –1 –1 (8.314 J mol K × 298 K) • ∆G is positive, then reaction is considered ln K = – 5.569 non-spontaneous. Instead, as reverse c reaction would have a negative ∆G, the K = e–5.569 c products of the forward reaction shall be Kc= 3.81 × 10 –3 converted to the reactants. Problem 6.11 • ∆G is 0, reaction has achieved equilibrium; at this point, there is no longer any free Hydrolysis of sucrose gives, energy left to drive the reaction. Sucrose + H O Glucose + Fructose 2 A mathematical expression of this Equilibrium constant K for the reaction is thermodynamic view of equilibrium can be 2 ×10 13at 300K. Calculate ∆G  at 300K. described by the following equation: Solution ∆G = ∆G + RT lnQ (6.21) ∆G  = – RT lnK where, G is standard Gibbs energy. ∆G  = – 8.314J mol K ×1 –1 13 At equilibrium, when ∆G = 0 and Q = K c 300K × ln(2×10 ) the equation (6.21) becomes, ∆G  = – 7.64 ×10 J mol –1  ∆G = ∆G + RT ln K = 0  6.8 FACTORS AFFECTIN G EQUILIBRI A ∆G = – RT lnK (6.22) One of the principal goals of chemical lnK = – ∆G / RT synthesis is to maximise the conversion of the Reprint 2026-27