The decomposition of `O_(3)(g)` follows first order kinetics and is given by
`O_(3)(g) rarr O_(2)(g)+O(g)`
The rate constant for this reaction is `1.0xx10^(-3)s^(-1)`. The initial pressure of `O_(3)(g)` is 100 atm.
What will be the partial pressure ( in atm ) of `O_(3), O_(2), O` respectively after 38.38 minutes ?

Options

A. 95, 5, 5
B. 10, 90, 0
C. 10, 90, 90
D. 10, 0, 90

Answer

C

Explanation

Decomposition of `O_(3)`,
`O_(3)(g)rarrO_(2)(g)+O(g) `
`{:("Initial pressure",100 "atm",0,0),("Partial pressure",100-x,x,x):}`
after `38.38 m` in
`p_(i)` ( initial pressure) `=100` atm
`p_(i)` ( final pressure) after `38.38`m.
`=100-x`
From first order reaction,
Given, `k=1xx10^(-3)s^(-1), t=38.38m`.
`k=(2.303)/(t) log. (p_(i))/(p_(f))`
`1xx10^(-3) s^(-1)=(2.303)/(38.38xx60) log. (100("atm"))/(100 -x("atm"))`
`log. (100)/(100-x)=(1xx10^(-3)s^(-1)xx38.38xx60)/(2.303)`
`x=90` atm
Pressure of `O_(3)` after 38.38 m.
`po_(3)=p_(f)=100-x=100-90"atm"=10`atm
Pressure of `O_(2)(g)(x)=90` atm
Pressure of `O(g)(x)=90` atm
Hence, option (c) is correct.