Question

The thermal decomposition of sulfuryl chloride to sulfur dioxide and chlorine SO_{2}2​Cl_{2}2​(g) \underrightarrow{\Delta}Δ SO_{2}2​(g) + Cl_{2}2​(g)...

The thermal decomposition of sulfuryl chloride to sulfur dioxide and chlorine SO_{2}2​Cl_{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}2​Cl_{2}2​ sample to remain?

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Answer #1

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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