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