What is the half-life of a first-order reaction with arate constant of 4.00*10 -

Question

What is the half-life of a first-order reaction with arate constant of 4.00*10-4 s-1? What is the rate constant of a first-order reaction that takes9.8 mins for the reactant concentration to drop to half of itsinitial value? A certain first-order reaction has a rate constant of8.70*10-3 s-1. How long will it take for the reactantconcentration to drop to 1/8 of its initial value? What is the half-life of a first-order reaction with arate constant of 4.00*10-4 s-1? What is the rate constant of a first-order reaction that takes9.8 mins for the reactant concentration to drop to half of itsinitial value? A certain first-order reaction has a rate constant of8.70*10-3 s-1. How long will it take for the reactantconcentration to drop to 1/8 of its initial value? What is the rate constant of a first-order reaction that takes9.8 mins for the reactant concentration to drop to half of itsinitial value? A certain first-order reaction has a rate constant of8.70*10-3 s-1. How long will it take for the reactantconcentration to drop to 1/8 of its initial value?

Explanation / Answer

What is the half-life of a first-order reaction with arate constant of 4.00*10-4 s-1? The relation between half life , t 1/2 & rateconstant(k) in a first order reaction is t 1/2 =0.693 / k So t 1/2 = 0.693 /   k             = 0.693 / 4.00*10-4 s-1              =1732.5 s What is the rate constant of a first-order reaction that takes9.8 mins for the reactant concentration to drop to half of itsinitial value? Half life is tha time taken for the reactant to change itsamount to half So the given time is tha half life, t 1/2 =9.8 min t 1/2 = 0.693 / k ===> k = 0.693 / t 1/2 = 0.693 / k              = 0.693 / 9.8 min             = 7.0714 * 10^-2 min^-1 A certain first-order reaction has a rate constant of8.70*10-3 s-1. How long will it take for the reactantconcentration to drop to 1/8 of its initial value? The expression for a first order reaction is given by a =ao e -kt Where a o = initial amount            a= amount left after time t = a o / 8            k = rate constant = 8.7 * 10^-3 s^-1            t = time taken Plug the values we get a = ao e-kt                         taking natural log on both sides we get                                ln a = ln  ao   -   kt                               kt = ln ( ao / a )                                 = ln ( 8 )                                  =2.079                               t = 2.079 / 8.7 * 10^-3                                 = 239 s Where a o = initial amount            a= amount left after time t = a o / 8            k = rate constant = 8.7 * 10^-3 s^-1            t = time taken Plug the values we get a = ao e-kt                         taking natural log on both sides we get                                ln a = ln  ao   -   kt                               kt = ln ( ao / a )                                 = ln ( 8 )                                  =2.079                               t = 2.079 / 8.7 * 10^-3                                 = 239 s

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