Calculate the value of equilibrium constant `K_(p)` for the reaction:
`O_(2(g))+O_((g))hArrO_(3(g))` …(`i`)
if `NO_(2(g))hArrNO_((g))+O_((g))`,
`K_(P_(1))=6.8xx10^(-49)` …(`ii`)
and `O_(3(g))+NO_((g))hArrNO_(2(g))+O_(2(g))`,
`K_(P_(2))=5.8xx10^(-34)`....(`iii`)
If initial pressure of an equimolar mixture of the reactants in change (`i`) is `4.0 bar`, what are the equilibrium parital pressure of the reactants and products?

Explanation

`K_(p_(1))=(P_(NO)xxP_(O))/(P_(NO_(2)))`, `K_(p_(2))=(P_(NO_(2))xxP_(O_(2)))/(P_(O_(3))xxP_(NO))`
`K_(p)=(P_(O_(3)))/(P_(O_(2))xxP_(O))=(1)/(K_(p_(2)))xx(1)/(K_(p_(1)))`
`=(1)/(6.8xx10^(-49)xx5.8xx10^(-34))`
`=2.5xx10^(81)`
The reaction thus goes to almost completion.
Also `underset(0)underset(P)(O_(2(g)))+underset(0)underset(P)(O_((g)))hArrunderset(0)underset(P)(O_(3(g)))`
Given `2P=4 :. P=2`
Now `underset(2-P')underset(2)(O_(3(g)))hArrunderset(P')underset(0)(O_(2(g)))+underset(P')underset(0)(O_((g)))`
`:. (1)/(2.5xx10^(81))=((P')^(2))/(2-P')` [`:' P'ltlt2`]
`implies ((2-P')^(2))/(P')=4xx10^(-82)`
or `P'=2xx10^(-41)"bar"`