For the decomposition of dinitrogen pentoxide at `200^(@)C`,
`N_(2)O_(5)(g)toN_(2)O_(4)(g)+(1)/(2)O_(2)(g)`,
if the intial pressure is 14 mm and after 25 minutes of the reaction, total pressure of the gassous mixture is 133 mm, calculate the average rate of reaction in (a) atm `min^(-1)" "(b)" mol "L^(-1)s^(-1)`.

Options

A. `0.002, 8.58 xx 10^(-7)`
B. `0.001, 8.58 xx 10^(-7)`
C. `0.002, 8.58 xx 10^(-4)`
D. `0.0001, 8.58 xx 10^(-3)`

Answer

A

Explanation

`underset(a-x)(2N_(2)O_(5)) to underset(x)(2N_(2)O_(4)) + underset(x//2)(O_(2)`
Pressure due to a moles = 114 mm
Pressure due to `a+x/2` moles = 133 mm
Hence pressure due to `x/2` moles =133-114 and due to x moles `19 xx 2=38` mm
`therefore` Pressure due to (a-x) moles = 114-76=38

Average rate `=38/(25 xx 760) = 0.002 Atm m^(-1)`
Again PV = nRT
`therefore n/V = 0.002/(0.0821 xx 473 xx 60)`