The following data were obtained during the first order thermal decomposition of `N_(2)O_(5) (g)` at constant volume:
decpmposition 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)),(1.,0,0.5),(2.,100,0.512):}`
Calculate the rate constant.

Explanation

Let the pressure of `N_(2)O_(5) (g)` decreases by `2x atm`. As `2mol` of `N_(2)O_(5)` decompoistion to give `2 mol` of `N_(2)O_(4) (g)` and `1 mol` 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), rarr, 2N_(2)O_(4)(g) +, O_(2)(g)), (When t = 0,0.5 atm,,"0 atm","0 atm"),("At time t",(0.5 - 2x) atm,,"2x atm","x atm"):}`
Total pressure `(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 = 0.5-2(p_(t) - 0.5) = 1.5 - 2p_(t)`
At `t = 100 s, p_(t) = 0.512 atm`
`p_((N_(2)O_(5))) = 1.5 - 2 xx 0.512 = 0.476 atm`
Uisng Eq. (i) above,
`k = (2.303)/(t) log.((p_(0))/(3p_(0) - 2p_(t)))`
`= (2.303)/(t) log.((p_(0)(N_(2)O_(5)))/(p_(A)(N_(2)O_(5))))`
`= (2.303)/(100s) log.((0.5 atm)/(0.476 atm))`
`= (2.303)/(100 s) xx 0.0216 = 4.98 xx 10^(-4) s^(-1)`