What is the rate constant of a first-order reaction that takes 490 seconds for the reactant concentration to drop to half of its initial value?Express your answer with the appropriate units.• View Available Hint(s)ZesetTemplates Symbols undo redlokeyboard shortcuts helpValueUnits The integrated rate law allows chemists to predict the reactant concentration after a certain amount oftime, or the time it would take for a certain concentration to be reached.The integrated rate law for a first-order reaction is:Half-life equation for first-order reactions:0.693t1/2 =[A] = [A]oe-ktwhere t/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s).Now say we are particularly interested in the time it would take for the concentration to become one-[A],half of its initial value. Then we could substitute " for [A] and rearrange the equation to:0.693t1/2This equation calculates the time required for the reactant concentration to drop to half its initial value.In other words, it calculates the half-life.

The time period in which the initial concentration of the reactants is reduced to half of its…

What is the rate constant of a first-order reaction that takes 490 seconds for the reactant concentration to drop to half of its initial value? Express your answer with the appropriate units. • View Available Hint(s) Templates Symbols undo redo Peset keyboard shortcuts help Value Units

What is the rate constant of a first-order reaction that takes 490 seconds for the reactant concentration to drop to half of its initial value? Express your answer with the appropriate units. • View Available Hint(s) Templates Symbols undo redo Peset keyboard shortcuts help Value Units

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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