The following data were obtained during the first thermal decompoistion 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) decreases by 2x atm. As two moles of `N_(2)O_(5)` decomposes to give two moles of `N_(2)O_(4)`(g) and mole of `O_(2)` (g), the pressure of `N_(2)O_(4)` (g) increase 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,2xatm,xatm):}`
`P_(t) = P_(N_(2)O_(5)) + P_(N_(2)O_(4)) + P_(O_(2))`
`=(0.5 -2x) + 2x +x = 0.5 +x`
`x = p_(t) -0.5`
`P_(N_(2)O_(5)) = 0.5 - 2x`
at t =100 s, `P_(t) = 0.512` atm
`P_(N_(2)O_(5)) = 1.5 -2 xx 0.512 = 0.476` atm
But `K = 2.303/t log P_(i)/P_(f) = 2.303/100 log 0.5/0.476`
`=2.303/100 xx 0.0216 = 4.98 xx 10^(-4)s^(-1)`