Our body obtains energy from the combustion of glucose. During this process, glucose is transformed first into fructose. Thisreaction has an activation energy close to 55 kJ inside our body, where the process is catalyzed by enzimes. Without thesecatalysts, the activation enerrgy is close 97 kJ. How many times does the catalyst increase the rate of this reaction in our body ata normal temperature close to 98.6 F (37 °C)?Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.abСd1.8 times1.2 x 107 times2.3 x 10 times780 timesUnanswered. 2 attempts leftSubmit

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Our body obtains energy from the combustion of glucose. During this process, glucose is transformed first into fructose. This reaction has an activation energy close to 55 kJ inside our body, where the process is catalyzed by enzimes. Without these catalysts, the activation enerrgy is close to 97 kJ. How many times does the catalyst increase the rate of this reaction in our body at a normal temperature close to 98.6 F (37 °C)? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b с d 1.8 times 1.2 x 107 times 2.3 x 10 times 780 times

Our body obtains energy from the combustion of glucose. During this process, glucose is transformed first into fructose. This reaction has an activation energy close to 55 kJ inside our body, where the process is catalyzed by enzimes. Without these catalysts, the activation enerrgy is close to 97 kJ. How many times does the catalyst increase the rate of this reaction in our body at a normal temperature close to 98.6 F (37 °C)? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b с d 1.8 times 1.2 x 107 times 2.3 x 10 times 780 times

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A) The activation energy of a certain reaction is 36.9 kJ·mol−1. At 35 °C, the rate constant is 0.0135 s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?. B) The initial rate constant is 0.0135 s−1 at an initial temperature of 35 °C, what would the rate constant (in s−1) be at a temperature of 225 °C for the same reaction described in Part A?
2a) Briefly explain, in your own words, the difference between the "activation energy" of a reaction and the "net free energy change" of a reaction.
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Please answer question #23. Thank you!
NEED AS SOON AS POSSIBLE A catalyst decreases the activation energy of a particular exothermic reaction by 20 kJ/mol, to 36 kJ/mol. Assuming that the mechanism has only one step, and that the products are 51 kJ lower in energy than the reactants, sketch approximate energy-level diagrams for the catalyzed and uncatalyzed reactions. What is the activation energy for the uncatalyzed reverse reaction?
Consider the following reaction: 3 Y(g) → 2 Z (g) The half life of this reaction is 12.16 days at 407.0 K and it is independent of the initial concentration of Y. If the activation energy is 139.0 kJ/mol, what is the half-life (in seconds) at 742.5 K?
For an exothermic reaction, what does the activation energy of the reverse reaction equal? the activation energy of the forward reaction minus the heat of reaction of the forward reaction the heat of reaction of the forward reaction plus the activation energy of the forward reaction the activation energy of the forward reaction the heat of reaction of the forward reaction minus the activation energy of the reverse reaction the heat of reaction of the forward reaction minus the activation energy of the forward reaction
Rank the bold-faced hydrogens for the following compounds from most acidic to least acidic. LOH X CF3 COOH I H II A) III >> V > II > IV B) V > III > | > |I > IV C) I>II> III > IV> V D) V >II> IV > III > I E) III > V > II > I > IV H III H₂C 220 H₂C IV CH3
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. If an enzyme increases the rate of the hydrolysis by a factor of (7.07x10^0)E6, how much is the activation energy of the catalyzed reaction? Assume the frequency (A) factors for the catalyzed and the uncatalyzed reactions are identical and the temperature is 25 deg. C. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer units
Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the reaction equation 2 H, О, (aq) 2 H,О() + О,(g) The activation energy for this reaction is 75 kJ-mol¬'. The enzyme catalase, found in blood, lowers the activation energy to 8.0 kJ-mol-'. At what temperature would the non-catalyzed reaction need to be run to have a rate equal to that of the enzyme-catalyzed reaction at 25 °C?
What is the activation energy for a reaction if k = 1.89 × 10–1 s–1 at 541 Kelvin and k = 7.31 s–1 at 601 Kelvin, in kJ/mol? Report an integer value, without units.
Given the same reactant concentrations, the below reaction at 257. °C is 94.6 times as fast as the same reaction at 163. °C. Calculate the activation energy for this reaction. Assume that the frequency factor is constant. Round your answer to 2 significant digits. co(g) + Cl₂(g) - COCL, (g) Note: Reference the Fundamental constants table for additional information. E₁ - 1 kJ mol x10 X