The decomposition of phosphine, `PH_3` proceeds according to the following equation:
` 4PH_3 (g) to P_4 (g) + 6H_2 (g)`
It is found that the reaction follows the following rate equation:
Rate `=k[PH_3] `
The half-life of `PH_3 ` is 37.9s at `120^@ `C.
What fraction of the original sample of `PH_3` remains behind after 1 minute?

Explanation

`t_(1//2) = 37.9 s `
Let initial amount of phosphine be 100, i.e., a = 100, (a-x) =?
For a first order reaction,
` k= ( 2.303 )/( k ) log ""(a)/((a-x))`
` or 0.0183 s^(-1) = ( 2.303 )/(60 s) log "" (100)/( (a-x))`
` or log "" (100)/((a-x)) = ( 0.0183 xx 60 )/( 2.303 ) = 0.4768`
` or log 100 - log (a-x) = 0.4768 `
` or log (a-x) = 2 - 0.4768 = 1.5232 `
taking antilogs,` (a-x) = 33.35 `
Thus, fraction of `PH_3` left after 1 minute = 33.35%