Question

In a first order decomposition, the constant is 0.00354 sec-1. What percentage of the compound has...

In a first order decomposition, the constant is 0.00354 sec-1. What percentage of the compound has decomposed after 43.3 seconds?

Homework Answers

Answer #1

t1/2   = 0.693/k

         = 0.693/0.00354

           =195.76sec

t   = 43.3sec

K   = 2.303/t log[A]/[A0]

0.00354 = 2.303/43.3log100/x

log100/x     = 0.00354*43.3/2.303

log(100/x)    = 0.0665

   100/x     = 10^0.0665

100/x       = 1.165

x             = 100/1.165 = 85.83

               = 85.83%

decomposition = 100-85.83   = 14.17%

no of half lives = total time/half life time

                       = 43.3/

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