6e1: In addition to solving the problem please give brief explanation for concepts and solution if able. 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?

Concepts Explained:1. Equilibrium Constant (Keq): This is a ratio of the rate constant for the…

Answered: 6E.1 The equilibrium constant for the…

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter11: Chemical Kinetics: Rates Of Reactions

Section: Chapter Questions

Problem 81QRT

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List two ways that enzyme catalysis of a reaction is superior to normal conditions.
E17C.1(b) The equilibrium constant for the binding of a drug molecule to a protein was measured as 200. In a separate experiment, the rate constant for the binding process, which is second order overall, was found to be 1.5 x 10 dm mol s. What is the rate constant for the first-order dissociation of the drug molecule from the protein- drug complex?
The degradation of the antibiotic clindamycin stored at 343 K in aqueous solution at pH 4 is found to be first order with a rate constant of 2.49 x 10−7 s −1. Over the temperature range 320 K to 360 K theactivation energy was found to be 123.3 kJ mol−1. (a) Calculate the rate constant at 325 K.(b) The threshold for product safety is 1% degradation. At 295 K the time taken for 1% of the antibiotic to degrade is found to be close to 0.01/ k. Comment on the shelf-life of the drug.
20B.2(b) The rate constant for the first-order decomposition of a compound A in the reaction 2 A→P is krr=3.56×10−7 s−1 at 25 °C. What is the half-life of A? What will be the pressure, initially 33.0 kPa after (i) 50 s, (ii) 20 min after initiation of the reaction?
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.
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?
A reaction has Keq' = 120. The rate constant of this reaction was found to be 1.11x 101s-1. If the catalytic power of an enzyme for this reaction is 1.25 x 10', what is the rate constant, in s1, for the reverse direction of the catalyzed reaction?
Calculate the velocity of the reaction (M · s–1) of an α-amino group in a blood protein at 37°C if its concentration is 0.60 mM and the partial pressure of carbon dioxide is 40 torr. Use 2 significant figures and scientific notation (example: 1000 would be entered as 1.0e+3). The rate constant k for this reaction is 4950 M–1·s–1.
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-
Calculate the magnitude of the diffusion-controlled rate constant at 320 K for the recombination of two atoms in water, for which η = 0.89 cP. Assuming the concentration of the reacting species is 1.5 mmol dm−3 initially, how long does it take for the concentration of the atoms to fall to half that value? Assume the reaction is elementary.
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)

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