The following data were obtained during the first thermal decomposition of `N_(2)O_(5) (g)` at constant volume.
`2N_(2)O_(5) (g) rarr 2N_(2)O_(4) (g)+O_(2)(g)`
`|{:("S.No.","Time (s)","Total pressure (atm)"),(i.,0,0.5),(ii.,100,0.512):}|`
Calculate the rate constant.

Explanation

Let the pressure of `N_(2)O_(5)` (g) decrease by 2x atm . As two moles of
`N_(2)O_(5)` decompose to give two moles of `N_(2)O_(4)` (g) and one mole of `O_(2)` (g),
the pressure of `N_(2)O_(4)` (g) increases by 2x atm and that of `O_(2)`(g)
increases by x atm.
`2N_(2)O_(5) (g) to 2N_(2)O_(4)(g) + O_(2) (g)`
`{:("Start "t=0," "0.5" atm",0 " atm",0" atm"),("At time t",(0.5-2x)" atm",2x" atm",x" atm"):}`
`p_(t)=P_(N_(2)O_(5)+P_(O_(2)))`
=`(0.5-2x)+2x+x=0.5+x`
`x=P_(t)-0.5`
`P_(N_(2)O_(5) =0.5-2x`
`=0.5-2(P_(1)-0.5)=1.5-2p_(t)`
At `t=100S,P_(1)=0.512` atm
`P_(N_(2)O_(5) =1.5-2xx 0.512=0.476 " atm" `
Using equation (4.16)
`k=2.303/t"log"P_(i)/P_(A)=2.303/(100s)"log"(0.5atm)/(0.476atm)`
= `2.303/(100s)xx0.0216=4.98xx10^(-4)S^(-1)`