By Jagdish K. Vij

ISBN-10: 0471180831

ISBN-13: 9780471180838

Prigogine and Rice's hugely acclaimed sequence, *Advances in Chemical Physics*, presents a discussion board for severe, authoritative reports of present issues in each sector of chemical physics. Edited via J.K. Vij, this quantity makes a speciality of fresh advances in liquid crystals with major, up to date chapters authored via the world over well-known researchers within the box.

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13 on page 16. We can similarly define the Gibbs free energy of dilution or mixing by substituting this equation into the definition of ∆G°: C ∆G dil = ∆H dil – RT ln -----1- C 2 (24) If the substance in question forms an ideal solution with the other components, then ∆Hdil is by definition zero, and we can write1 C2 ∆G dil = RT ln ------ C1 (25) These relations tell us that the dilution of a substance from an initial concentration C1 to a more dilute concentration C2 is accompanied by a decrease in the free energy, and thus will occur spontaneously.

Those that did not adapt to this new environment have literally “gone underground” and constitute the more primitive anaerobic bacteria. The function of oxygen in respiration is to serve as an acceptor of the electrons that glucose loses when it undergoes oxidation. Other electron acceptors can fulfill the same function when oxygen is not available, but none yields nearly as much free energy. For example, if oxygen cannot be supplied to mammalian muscle cells as rapidly as it is needed, they switch over to an anaerobic process yielding lactic acid instead of CO2: C6H12O6 + 2 ADP → 2 CH3CH(OH)COOH ∆G° = –218 kJ mol–1 In this process, only (2 × 30 kJ) = 60 kJ of free energy is captured, so the efficiency is only 28% on the basis of this reaction, and it is even lower in relation to glucose.

So the sign of the entropy change determines whether the reaction becomes more or less allowed as the temperature increases. We often want to know how a change in the temperature will affect the value of an equilibrium constant whose value is known at some fixed temperature. Suppose that the equilibrium constant has the value K1 at temperature T1 and we wish to estimate K2 at temperature T2. Expanding Eq. 35 in terms of ∆H° and ∆S°, we obtain –RT1 ln K1 = ∆H ° – T1 ∆S° and –RT2 ln K2 = ∆H ° – T2 ∆S° Dividing both sides by RT and subtracting, we obtain Free energy: the Gibbs function Page 34 ∆H° ∆H° ln K 1 – ln K 2 = – ----------- + ----------- RT 1 RT 2 (39) Which is most conveniently expressed as the ratio K1 ∆H° 1 1 ln ------ = – ----------- ----- – ----- K2 R T1 T2 (40) Do you remember the Le Châtelier Principle?

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