Half-life (kinetics) for First Order Reactions Half-life equation for first-orde

Question

Half-life (kinetics) for First Order Reactions Half-life equation for first-order reactions: The integrated rate law allows chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached The integrated rate law for a first-order reaction is [A] [Aloe-kt where t1/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s-1) Where 1/2 IS the nal-lle in seconds (s), and Is the rate constant In Inverse seconds (s Part A What is the half-life of a first-order reaction with a rate constant of 8.10x104 s-1 Now say we are particularly interested in the time it would take for the concentration to become one- half of its inital value. Then we could substitute Ao for [A] and rearrange ft forA and rearrange the equation to 0.693 3150 This equation caculates the time required for the reactant concentration to drop to half its initial value. In other words, it calculates the half-life Submit Hints My Answers Give Up Review Part Incorrect Try Again, 4 attempts remaining

Explanation / Answer

Part A:

For the first order reaction; the rate constant is related with the half time as follows:

t ½ = 0.693 / k

here rate constant k=8.10*10^-4 s-1

t ½ = 0.693 / k

t ½ = 0.693 / 8.10*10^-4 /s

t ½ =855.6 s or 14.26 min

Part B

For the first order reaction; the rate constant is related with the half time as follows:

t ½ = 0.693 / k

here t ½ =3.60 min

k= 0.693 / t ½

k = 0.693 / 3.60 min

k= 0.1925 min-1

part C

ln [A]t = -kt +ln [A]0

assume that if initial concentration is 1.0 so [A]t = 1/8

given that; k = 5.60*10^-3 s-1

ln [A]t = -kt +ln [A]0

ln 1/8= - 5.60*10^-3 s-1 *t +ln 1.0

-2.079 = -5.60*10^-3 s-1 *t +0.0

t= 371.25 s or 6.19 min

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