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The decomposition of ethane into two methyl radicals has a first order rate constant of...

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lak-86 | Student, Undergraduate | Salutatorian

Posted August 2, 2013 at 10:59 AM via web

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The decomposition of ethane into two methyl radicals has a first order rate constant of 5.5*10^(–4) sec^(–1)  at 700 °C. What is the half-life for this decomposition in minutes?

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted August 2, 2013 at 12:35 PM (Answer #1)

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The chemical reaction for the cleavage of ethane to two methyl radicals can be expressed as:

`CH_3-CH_3 -> 2 CH_3 * `

`Rate = k [CH_3-CH_3] `

Half-life for the first law is expressed as:

`t_(1/2) = (ln 2)/(k) ` ` `

where:

`k =5.5*10^(-4) sec^(-1) `

Solve for `t_(1/2)`

`t_(1/2) = (ln 2)/(k) `

`t_(1/2) = (ln 2)/(5.5*10^( – 4) sec^( – 1)) `

`t_(1/2) =1260.268 s e conds * (1 m i n ute)/(60 s e conds) `

`t_(1/2) = 21 m i n utes` -> answer

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