The following data were obtained during the first order thermal decomposition of `N_(2)O_(5)(g)` at constant volume.
`2N_(2)O_(5)(g) to 2N_(2)O_(2)(g) + O_(2)(g)`

Calculate the rate constant for the gaseous reaction.

Explanation

The initial pressure of `N_(2)O_(5)` is 0.5 atm. Since two moles of `N_(2)O_(5)` (g) are involved, let the decrease in pressure be 2x atm. Let us find the value of x ie.e. decrease in pressure.
`2N_(2)O_(5)(g) to 2N_(2)O_(4)(g) + O_(2)(g)`
At time(t) = 0 , 0.5 atm, 0 atm, 0 atm
At time t=100s , `(0.5-2x)`atm, 2x atm, xatm
Pressure after time t (pt)`=(0.5 -2x+2x+x)` atm = (0.5 + x) atm or x`=(p_(t)-0.5)` a tm
After time t, `p_(N_(2)O_(5)) = (0.5-2x) atm=0.5 -2(p_(t)-0.5)atm`
`p_(N_(2)O_(5)) = 0.5 -2(0.512-0.5) = (0.5-1.024 + 1.0)` atm
`=(1.5 - 1.024) = 0.476 atm`
For the first order reaction,
`k=2.303/t log p_(i)/p_(A) = 2.303/(100s)log(0.5 atm)/(0.476 atm)`
`=2.303/(100s)log 1.0504 = 2.303/(100s) xx 0.0213.6 = 4.91 xx 10^(-4)s^(-1)`