The thermal decomposition of sulfuryl chloride to sulfur dioxide and chlorine SO_{2}2Cl_{2}2(g) \underrightarrow{\Delta}Δ SO_{2}2(g) + Cl_{2}2(g) is a first order reaction with a half-life of 2300 seconds. How many seconds are required for 66% of the original SO_{2}2Cl_{2}2 sample to remain?
Given:
Half life = 2.3*10^3 s
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(2.3*10^3)
= 3.013*10^-4 s-1
we have:
[A]o = 100 (Let initial concentration be 100)
[A] = 66 (66% of initial remains)
k = 3.013*10^-4 s-1
use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln(66) = ln(100) - 3.013*10^-4*t
4.19 = 4.605 - 3.013*10^-4*t
3.013*10^-4*t = 0.4155
t = 1.38*10^3 s
Answer: 1.38*10^3 s
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