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?
1 Answer
jerichorayel | College Teacher | (Level 2) Senior Educator
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