A decomposition reaction has a rate constant of 0.0012 yr^-1. (a) What is the half-life of the reaction? (b) How long does it take for [reactant] to reach 12.5% of its original value?
A decompoition reaction has a rate constant of 0.008. What is it's half-life value? (Show how you got it)
2007-02-01 12:55:31 · 2 answers · asked by Cupcake 2 in Science & Mathematics ➔ Chemistry
ln{[A]o/[A]}=kt
t 1/2= 0.693/k
2007-02-01 13:16:47 · update #1
You already have the equations so you just need to plug-in the values.
t1/2 =0.693/0.0012 = 577.5 years
If 12.5% remains that means that A/A0=0.125, so A0/A=1/0.125 and
ln(1/0.125)= 0.0012t =>
t=1732.9 years
You're not providing the units for the k=0.008. Let's call it u^-1 where u could be s, min, h,year or whatever unit of time.
Then t1/2 =0.693/0.008 =86.6 u
2007-02-02 00:43:30 · answer #1 · answered by bellerophon 6 · 0⤊ 0⤋
don't you need to know what order the reaction is in first?
2007-02-02 11:16:41 · answer #2 · answered by Marty Kay 2 · 0⤊ 0⤋