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