Concept explainers
Interpretation:
The factor at which the rate of an enzyme catalyzed reaction changes when the substrate concentration is changed from
Concept Introduction:
Catalysts are the substances that alter the rate of
The relationship between rate and initial concentration of reactant is described by rate law. It is an experimentally determined equation and cannot be found out theoretically from the stoichiometry of the reactants and products.
Where,
For a zeroth order kinetics rate can be expressed as follows,
Where,
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Chapter 11 Solutions
Chemistry: The Molecular Science
- 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_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_forwardMany biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and X is the reactant) is Step 1: Step 2: Step 3: Derive the rate law from this mechanism. Determine the order of reaction with respect to HA. Determine how doubling the concentration of HA would affect the rate of the reaction.arrow_forward
- The 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_forwardList two ways that enzyme catalysis of a reaction is superior to normal conditions.arrow_forward
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