Some `PH_(3)(g)` is introduced into a flask at `600^(@)C` containing an intert gas. `PH_(3)` proceeds to decompose into `P_(4)(g) and H_(2)(g)` and the reaction goes to complection. Total pressure is given below as a fucntion of time. Find the order of the reaction and calcualate the rate constant. `{:("Time (sec)",0,60,120,oo),("Pressure (mm Hg)",262.40,272.90,275.51,276.51):}`

Explanation

`4PH_(3)(g)+"Inert gas " rarrP_(4)(g)+6H_(2)(g)+"Inert gas"`
`{:(T=0,P,P_(i),0,0,0),(t=t,(p-x),P_(i),x//4,6P//4,P_(i)),(t=oo,0,P_(i),P//4,6P//4,P_(i)):}`
At `t=0 " " 262.40 =P+P_(i)`
`t=60sec " " 272.90 =P_(PH_(3)("left"))+P_(P_(4)(g))+P_(H+_(4)(g))+P_(H_(2)(g))+P_(i)`
`t=120sec " " 275.51 =P_(PH_(3)("left"))+P_(P_(4)(g))+P_(H+_(4)(g))+P_(H_(2)(g))+P_(i)`
`t=oo " " 276.90 =P_(P_(4)(g))+P_(H_(2)(g))+P_(i)`
`276.40=(P)/(4)+(6P)/(4)+P_(i)`
`276.40xx4=7P+4P_(i)" ....(iii)`
At `t=0, " " 262.2=P+P_(i)" " ....(iii)`
Or solving eqs.(ii) and (iii), we get
`P=18.6mm,P_(i)=243.74 mm`
At 60 sec `" " 272.90 =(P-x)+P_(i)+(x)/(4)+(6x)/(4)`
`272.90=18.66-z+243+(7x)/(4)`
`k=(2.303)/(t)log. ((a)/(a-x))=(2.303)/(60)log. (18.66)/(18.66-14)`
`k=2.32xx10^(-2) sec`
Similarly at 120 sec `" " k=2.30xx10^(-2)sec^(-1)` (do yourself)
Since, `:.` of 'k' are same hence the reaction belongs to first order :