Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 13, Problem 13.20QE
Interpretation Introduction
Interpretation:
The reason to why enzymatic reactions are zero order in the substrate has to be explained.
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Check out a sample textbook solutionChapter 13 Solutions
Chemistry: Principles and Practice
Ch. 13 - Prob. 13.1QECh. 13 - Prob. 13.2QECh. 13 - What is the difference between the integrated and...Ch. 13 - Prob. 13.4QECh. 13 - Explain why half-lives are not normally used to...Ch. 13 - Derive an expression for the half-life of a...Ch. 13 - Prob. 13.7QECh. 13 - Prob. 13.8QECh. 13 - Prob. 13.9QECh. 13 - Prob. 13.10QE
Ch. 13 - Prob. 13.11QECh. 13 - Prob. 13.12QECh. 13 - Prob. 13.13QECh. 13 - Prob. 13.14QECh. 13 - Prob. 13.15QECh. 13 - Prob. 13.16QECh. 13 - Prob. 13.17QECh. 13 - Prob. 13.18QECh. 13 - Prob. 13.19QECh. 13 - Prob. 13.20QECh. 13 - Prob. 13.21QECh. 13 - Prob. 13.22QECh. 13 - Nitrogen monoxide reacts with chlorine to form...Ch. 13 - Prob. 13.24QECh. 13 - Prob. 13.25QECh. 13 - Prob. 13.26QECh. 13 - Prob. 13.27QECh. 13 - Prob. 13.28QECh. 13 - Prob. 13.29QECh. 13 - Prob. 13.30QECh. 13 - Prob. 13.31QECh. 13 - Prob. 13.32QECh. 13 - Prob. 13.33QECh. 13 - Write a rate law for NO3(g) + O2(g) NO2(g) +...Ch. 13 - Prob. 13.35QECh. 13 - Prob. 13.36QECh. 13 - Prob. 13.37QECh. 13 - Rate data were obtained at 25 C for the following...Ch. 13 - Prob. 13.39QECh. 13 - Prob. 13.40QECh. 13 - Prob. 13.41QECh. 13 - Prob. 13.42QECh. 13 - Prob. 13.43QECh. 13 - Prob. 13.44QECh. 13 - Prob. 13.45QECh. 13 - Prob. 13.46QECh. 13 - Prob. 13.47QECh. 13 - Prob. 13.48QECh. 13 - When formic acid is heated, it decomposes to...Ch. 13 - Prob. 13.50QECh. 13 - The half-life of tritium, 3H, is 12.26 years....Ch. 13 - Prob. 13.52QECh. 13 - Prob. 13.53QECh. 13 - Prob. 13.54QECh. 13 - Prob. 13.55QECh. 13 - Prob. 13.56QECh. 13 - The decomposition of ozone is a second-order...Ch. 13 - Prob. 13.58QECh. 13 - Prob. 13.59QECh. 13 - Prob. 13.60QECh. 13 - A reaction rate doubles when the temperature...Ch. 13 - Prob. 13.62QECh. 13 - Prob. 13.63QECh. 13 - Prob. 13.64QECh. 13 - Prob. 13.65QECh. 13 - The activation energy for the decomposition of...Ch. 13 - Prob. 13.67QECh. 13 - Prob. 13.68QECh. 13 - Prob. 13.69QECh. 13 - Prob. 13.70QECh. 13 - Prob. 13.71QECh. 13 - Prob. 13.72QECh. 13 - Prob. 13.73QECh. 13 - Prob. 13.74QECh. 13 - Prob. 13.75QECh. 13 - Prob. 13.76QECh. 13 - Prob. 13.77QECh. 13 - Prob. 13.78QECh. 13 - Prob. 13.79QECh. 13 - Prob. 13.80QECh. 13 - The gas-phase reaction of nitrogen monoxide with...Ch. 13 - Prob. 13.82QECh. 13 - Prob. 13.83QECh. 13 - A catalyst reduces the activation energy of a...Ch. 13 - Prob. 13.85QECh. 13 - Prob. 13.86QECh. 13 - Prob. 13.87QECh. 13 - Prob. 13.88QECh. 13 - Prob. 13.89QECh. 13 - Prob. 13.90QECh. 13 - Prob. 13.91QECh. 13 - Prob. 13.92QECh. 13 - Prob. 13.93QECh. 13 - Prob. 13.94QECh. 13 - Prob. 13.95QECh. 13 - Prob. 13.96QECh. 13 - Prob. 13.98QECh. 13 - Prob. 13.99QE
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- When enzymes are present at very low concentration, their effect on reaction rate can be described by first-order kinetics. Calculate by what factor the rate of an enzyme-catalyzed reaction changes when the enzyme concentration is changed from 1.5 107 M to 4.5 106 M.arrow_forwardAccount for the relationship between the rate of a reaction and its activation energy.arrow_forwardThe hydrolysis of the sugar sucrose to the sugars glucose and fructose, C12H22O11+H2OC6H12O6+C6H12O6 follows a first-order rate equation for the disappearance of sucrose: Rate =k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.) (a) In neutral solution, k=2.11011s1 at 27 C and 8.51011s1 at 37 C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature). (b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65107M . How long will it take the solution to reach equilibrium at 27 C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible. (c) Why does assuming that the reaction is irreversible simplify the calculation in pan (b)?arrow_forward
- Label the elementary processes for the reaction between H2 and O2 see section 20.7 as initiation, propagation, branching, or termination reactions.arrow_forwardCompare the functions of homogeneous and heterogeneous catalysts.arrow_forwardThe hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose. C12H22O11(aq)+H2O(l)C6H12O6(aq)+C6H12O6(aq) Rate =k[C12H22O11] In neutral solution, k=2.11011 at 27 C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C6H12O6, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36105 at 27 C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1.arrow_forward
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