6E.1 The equilibrium constant for the attachment of a sub- strate to the active site of an enzyme was measured as 200. In a separate experiment, the rate constant for the second- order attachment was found to be 1.5 × 108 dm³ mol‍¹ s¯¹. What is the rate constant for the loss of the unreacted sub- strate from the active site?

Derive an expression for the time dependence of the degree of polymerization for the stepwise polymerization of a hydroxyacid HO−R−COOH for which the rate law is d[A]/dt = −kr[A]2[OH], where A denotes the carboxylic acid group.

6c3. Please solve the following problem, give the relevant equation(s) used and explanation.

E17B.1(b) In a study of the enzyme-catalysed oxidation of ethanol, the molar concentration of ethanol decreased in a first-order reaction from 220 mmol dm3 to 56.0 mmol dm in 1.22 x 10 s. What is the rate constant of the reaction?

The equilibrium constant for the attachment of a substrate to the active site of an enzyme was measured as 200. In a separate experiment, the rate constant for the secondorder attachment was found to be 1.5 x 108 dm3 mol-1 s- 1. What is the rate constant for the loss of the unreacted substratefrom the active site?

The protein catalase catalyzes the reaction 2H,0,(aq) – 2 H,0O(0) + 0,(g) and has a Michaelis–Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s-1. The total enzyme concentration is 0.016 µM and the initial substrate concentration is 6.58 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax ) for this enzyme. Rmax mM-s- Calculate the initial rate, R (often written as Vo), of this reaction. R = mM-s-

The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. Vmax [S] U = Km + [S] where v is the velocity, or rate, Vmax is the maximum velocity, Km is the Michaelis-Menten constant, and [S] is the substrate concentration. A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (vo) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. 300 V Next, move the line labeled 1/2 Vmax to its correct position. max 275 Then, move the line labeled Km to its correct position. 250 225 200 175 Michaelis–Menten curve 150 K a 125 m 100 1/2 V 75 max 50 25 10 20 30 40 50 60 70 80 90 100 [S] (µM) Incorrect Estimate the values for Vmax and Km. Vmax 175 µM/min Km 14 µM v (µM/min)

Suppose that, in the absence of a catalyst, a certain biochemical reaction occurs x times per second at normal body temperature 137 °C2. In order to be physiologically useful, the reaction needs to occur 5000 times faster than when it is uncatalyzed. By how many kJ>mol must an enzyme lower the activation energy of the reaction to make it useful?

The free energy change, AG, in the homogenous cross-exchange ET reaction of [Ru(bpy)2]* and ferricytochrome c is -104.6 kJ mol¯1. The second-order activation-limited rate constant, kobs, is 2.5 × 108 M-1 s-1. Taking KDA = 1.0 M¹, N = 1 × 1011 s-1, and x = 1, estimate the values of AG* and 2. In what kinetic region is this reaction? Hint: By substituting the Marcus equation into the expression for the rate constant of an adiabatic homogenous ET reaction, the rate constant of a cross-exchange ET reaction may be written as (AG+2)² Ket = KVN exp - 4λRT and, by rearrangement, In Ket KV 1 (AG+ 2)² 1 AG2 1 AG == + constant 42 RT 42 RT 2 RT

The adsorption of a gas is described by the Langmuir isotherm with α = 0.548 kPa−1 at 25 °C. Calculate the pressure at which the fractional surface coverage is (i) 0.20, (ii) 0.75.