Grades for Andrea Brisby: Fi S. University of Oklahoma + O saplingleaming.com/ib

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Grades for Andrea Brisby: Fi S. University of Oklahoma + O saplingleaming.com/ibiscms/mod/ibi5Nie w.php?id-3871 620 Sapling Learning Jump to. Andrea Brisby macmillan learring Question 21 of 28 Available From: Due Date: Points Possible: Grade Category: 11/17/2017 06:00 PM 12/1/2017 11:55 PM 5 Homework 14 1 sapling Learning 15 0 16 1 17 0 18 0 19 0 After 500 min, 30.0% of a compound has decomposed, what is the half-life of this reaction assuming first-order kinetics? 5 attempt. 0% penalty. Number Solutions After Due Date 0 You can check your answers. You can view solutions after the due date You have five attempts per question. There is no penalty for incorrect answers. 20 21 0 eTextbook 22 3 23 OHelp With This Topic 24 Web Help & Videos 250 26 0 Technical Support and Bug Reports 27 3 100 281100 Previcus Check AnrNeExt Hint © 2011-2017 Sapling Learning, Inc. about u | careers | privacy policy | tams of use | contact us | help 7:13 PM Type here to search

Explanation / Answer

Using integrated rate law for first order reaction, A--->products for compound A

ln [A]=-kt +ln[Ao]

or,[A]=[Ao]exp(-kt)

t=time=30 min

[Ao]=initial concentration of compound

[A]=concentration of compound after time t

As 30 % of the compound decomposed after 30 min,remaining compound=100-30=70%

[A]=70%[Ao]=0.70[Ao]

So,[A]=[Ao] exp(-kt)

or,[A]/[Ao]=exp(-kt)

or,0.70[Ao]/[Ao]=exp(-k*30min)

or,ln0.70=-k*30min

or,0.357/30=k=0.0119 min^-1

for half -life,

[A]=[Ao]/2=0.5[Ao]

0.5[Ao]/[Ao]=exp(-k* t1/2)    where t(1/2)=half-life

ln0.5=-0.0119 min^-1*t(1/2)

t(1/2)=0.693/0.0119min-1=58.247 min=58.2 min(half-life)

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