The decomposition of `N_(2)O_(5)` according to the equation, `2N_(2)O_(5) (g) rarr 4NO_(2) (g) + O_(2) (g)` is a first order reaction, After 30 minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be 284.5 mm Hg and on completion, the total pressure is 584.5 mm Hg. Calculate the rate constant of the reaction.

Options

A. `5.2 xx 10^(-3) "min"^(-1)`
B. `10.1 xx 10^(-3) "min"^(-1)`
C. `8.3 xx 10^(-3) "min"^(-1)`
D. `2.2 xx 10^(-3) "min"^(-1)`

Explanation

`2N_(2)O_(5) rarr 4NO_(2) + O_(2)`
On decomposition of 2 moles of `N_(2)O_(5)`, 4 moles of `NO_(2)` and 1 mole of `O_(2)` are produced. Thus, the total pressure after completion corresponds to 5 moles and initial pressure to 2 moles.
Initial pressure of `N_(2)O_(5), p_(0) = (2)/(5) xx 584.5 = 233.8 mm Hg`
After 30 minutes, the total pressure = 284.5 mm Hg
`{:(2N_(2)O_(5),rarr,4NO_(2),+,O_(2)),(p_(0) - 2p,,4p,,p):}`
or `p_(0) + 3p = 284.5` or `3p = 284.5 - 233.8 = 50.7 mm Hg` or `p = (50.7)/(3) = 16.9 mm Hg`
Pressure of `N_(2)O_(5)` after minutes `= 233.8 - (2 xx 16.9) = 200 mm Hg`
`k = (2.303)/(30)log_(10)""(233.8)/(200.0) = 5.2 xx 10^(-3) "min"^(-1)`