Consider the following reactions in which all the reactants and the products are in gaseous state.
`2PQ hArr P_(2)=Q_(2), K_(1)=2.5xx10^(5)`
`PQ +1//2 R_(2) hArr PQR, K_92) =5xx10^(-3)`
The value of `K_(2)` for the equilibrium
`1//2P_(2)+1//2Q_(2) +1//2R_(2) hArr PQR,` is

Options

A. `2.5xx10^(-3)`
B. `2.5xx10^(3)`
C. `1.0 xx 10^(-5)`
D. `5xx10^(3)`

Answer

C

Explanation

`2PQ hArr P_(2) +O_(2), K_(1)=2.5 xx 10^(5)`
`PQ + 1//2 R _(2) hArr PQR , K_(2)=5xx10^(-3)`
`1//2 P_(2)+1//2 Q_(2) hArr PQ ,`
`K'_(1)=(1)/(sqrt(K_(1)))=(1)/(sqrt(2.5xx10^(5)))=(1)/(5xx10^(2))`
`:. 1//2 P_(2)+1//2Q_(2)+1//2 R_(2)hArr PQR`
`K_(3)=K_(2)xxK'_(1)=(5xx10^(-3))/(5xx10^(2))=10^(-5)`