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.

Options

A. `3.39xx10^(-4)S^(-1)`
B. `1.39xx10^(-5)S^(-1)`
C. `5.45xx10^(-4)S^(-1)`
D. `4.91xx10^(-4)S^(-1)`

Explanation

When t=0 `" "` 0.5 atm
At time t `(0.5-2x)""atm" " "2xatm " "xatm"`
Total pressure `(p_(t))=P_(N2O5)+ p_(N_2O_5)+P_(O2)`
`=0.5-2x+2x+x`
`p_(t)=0.5+x`
`x=p_(t)-0.5`
`p_(N_2O_5)=0.5-2x`
`=0.5-2(p_(t)-0.5)=0.5-2p_(t)+1`
`p_(N_2O_5)=1.5-2p_(t)`
At `t=100_(s),p_(t)=0.512"atm"`
`p_(N_2O_5)=1.5-2xx0.512`
`=1.5-1.024=0476"atm"`
`k=(2.303)/(t)"log"[(p_(0)(N_(2)O_5))/(p_(A)(N_(2)O_(5)))]`
`=(2.303)/(100)"log"(0.5)/(0.476)=(2.303)/(100)xx0.02136`
`=4.92xx10^(-4)S^(-1)`