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

`2N_(2)O_(5)(g)rarr2N_(2)O_(4)(g)+O_(2)(g)`.
`{:(At t=0,0.5 "atm",0 "atm",0 "atm"),("At time t",(0.5-2x),2x atm,x atm):}`
Let the pressure of `N_(2)O_(5)(g)` decreases by 2x atm. As two moles of `N_(2)O_(5)(g)` 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.
`P_(T)=P_(N_(2)O_(5))+P_(O_(2))`
`= (0.5-2)+(2x)+(x)`
`= 0.5+x`
or `x = P_(T)-0.5`
Partial pressure of `N_(2)O_(5)` at any time 't' is `P_(t)` which is given as
`P_(t)= .5 - 2x`
Substituting the value of x, we get
`P_(t)= 0.5-2(P_(T) - 0.5)`
`= 1.5 - 2P_(T)`
At time `t = 100` s, we have
`P_(T) = 0.512 atm`
Thus `P_(t)(N_(2)O_(5)= 1.5 - 2(0.512)`
`= 1.5 - 1.024`
`= 0.476 atm`
Integrated rate law in term of pressures is
`k = (2.303)/(t) log (P_(0))/(P_(t))`
`= (2.303)/(100s) log (0.5atm)/(0.476 atm)`
`= (2.303)/(100s)xx0.0216`
`= 4.98xx10^(-4)s^(-1)`