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.
How much time is required for 3/4th of `PH_3` to decompose?

Explanation

`t_(1//2) = 37.9 s `
Since the reaction follows first order kinetics,
` k=( 0.693 )/( t_(1//2)) = ( 0.693 )/( 37.9 ) s^(-1) = 0.0183 s^(-1)`
` t_( 0.75 ) = ( 2.303 )/(k ) log ""(1)/(1- 0.75 ) = ( 2.303 )/( k ) log 4`
`=( 2.303 )/(0.0183 ) s^(-) xx 0.6020 = 75.8 s `
Thus, time required for 3/4th of `PH_3` to decompose is 75.8s.