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) rarr4NO_(2)(g)+O_(2)(g)`
`{:(Time ("in s"),0,52,103,205,309),("Total pressure (in mm)",117.04,163.40,197.6,239.4,258.4):}`
Calculate the rate constant of the reaction.

Explanation

`" "2N_(2)O_(5)rarr4NO_(2) +O_(2)`
`{:(t=0:,P_(i),0,0),(t=t:,P_(i)-x,2x,(x)/(2)):}`
`k=(2.303)/(t) log.(a)/(a-x), "At " t = 0 " total pressure " = P_(i) . So , a prop P_(i) . " Again, at " t = t `
Total pressure `P_(t)=P_(i)-x+2x+(x)/(2)=P_(i)+(3x)/(2)`
`therefore x = (2)/(3)(P_(t)-P_(i))`
At time t, concentration of `N_(2)O_(5)`
`(a-x)prop{P_(i)-(2)/(3) (P_(t)-P_(i))}prop((5P_(i)-2P_(t))/(3))`
`therefore k = (2.303)/(52)log.(a)/(a-x)=(2.303)/(t)log.(3P_(i))/(5P_(i)-2P_(t))`
By putting the values of `P_(i) and P_(t)` (obtained at different times ),
t = 52s,
`k = (2.303)/(52)log.(3xx117.04)/(5xx117.04-2xx163.4)=5.89xx10^(-3)s^(-1)`
t = 103 s,
`k = (2.303)/(103)log.(3xx117.04)/(5xx117.04-2xx197.6)=5.96xx10^(-3)s^(-1)`
t = 205,
`k = (2.303)/(205)log.(3xx117.04)/(5xx117.04-2xx239.4)=5.82xx10^(-3)s^(-1)`
t = 309s,
`k = (2.303)/(309)log.(3xx117.04)/(5xx117.04-2xx258.4)=5.29xx10^(-3)s^(-1)`
Average value of `k = (1)/(4)(5.89+5.96+5.82+5.29)xx10^(-3)`
`=5.74xx10^(-3)s^(-1)`