Harpers

ENZYMES: KINETICS / 61For the reaction A + B → P+Q—and for reaction (5)P QK eq = [ ][ ][ A][ B]A+ A → ← PPK eq = [ ][ A] 2—∆G 0 may be calculated from equation (3) if the concentrationsof substrates and products present at equilibriumare known. If ∆G 0 is a negative number, K eqwill be greater than unity and the concentration ofproducts at equilibrium will exceed that of substrates. If∆G 0 is positive, K eq will be less than unity and the formationof substrates will be favored.Notice that, since ∆G 0 is a function exclusively ofthe initial and final states of the reacting species, it canprovide information only about the direction and equilibriumstate of the reaction. ∆G 0 is independent of themechanism of the reaction and therefore provides noinformation concerning rates of reactions. Consequently—andas explained below—although a reactionmay have a large negative ∆G 0 or ∆G 0′ , it may neverthelesstake place at a negligible rate.THE RATES OF REACTIONSARE DETERMINED BY THEIRACTIVATION ENERGYReactions Proceed via Transition StatesThe concept of the transition state is fundamental tounderstanding the chemical and thermodynamic basisof catalysis. Equation (7) depicts a displacement reactionin which an entering group E displaces a leavinggroup L, attached initially to R.E+ R−L → ← E −R+L(7)Midway through the displacement, the bond betweenR and L has weakened but has not yet been completelysevered, and the new bond between E and R is as yetincompletely formed. This transient intermediate—inwhich neither free substrate nor product exists—istermed the transition state, ERL. Dotted linesrepresent the “partial” bonds that are undergoing formationand rupture.Reaction (7) can be thought of as consisting of two“partial reactions,” the first corresponding to the formation(F) and the second to the subsequent decay (D) ofthe transition state intermediate. As for all reactions,(4)(5)(6)characteristic changes in free energy, ∆G F, and ∆G Dareassociated with each partial reaction.E+ R− L → ← ELLR L ∆GFELL R L → ← E −R+L ∆G DE+ R − L ←→ E − R + L ∆ G = ∆ GF+ ∆ GD(8)(9)(8-10)For the overall reaction (10), ∆G is the sum of ∆G Fand∆G D. As for any equation of two terms, it is not possibleto infer from ∆G either the sign or the magnitudeof ∆G For ∆G D.Many reactions involve multiple transition states,each with an associated change in free energy. For thesereactions, the overall ∆G represents the sum of all ofthe free energy changes associated with the formationand decay of all of the transition states. Therefore, it isnot possible to infer from the overall G the numberor type of transition states through which the reactionproceeds. Stated another way: overall thermodynamicstells us nothing about kinetics.∆G F Defines the Activation EnergyRegardless of the sign or magnitude of ∆G, ∆G Ffor theoverwhelming majority of chemical reactions has a positivesign. The formation of transition state intermediatestherefore requires surmounting of energy barriers.For this reason, ∆G Fis often termed the activation energy,E act , the energy required to surmount a given energybarrier. The ease—and hence the frequency—withwhich this barrier is overcome is inversely related toE act . The thermodynamic parameters that determinehow fast a reaction proceeds thus are the ∆G Fvalues forformation of the transition states through which the reactionproceeds. For a simple reaction, where means“proportionate to,”−EactRate ∝ e RT(11)The activation energy for the reaction proceeding in theopposite direction to that drawn is equal to −∆G D.NUMEROUS FACTORS AFFECTTHE REACTION RATEThe kinetic theory—also called the collision theory—of chemical kinetics states that for two molecules toreact they must (1) approach within bond-forming distanceof one another, or “collide”; and (2) must possesssufficient kinetic energy to overcome the energy barrierfor reaching the transition state. It therefore follows