Equilibrim constants are given for the following two equilibria.
(i) `A_(2)(g)+B_(2)(g)hArr2AB(g)`,
`K=2xx10^(-4)L"mol"^(-1)`
(ii) `2AB(g)+C_(2)(g)hArr2ABC(g)`,
`K=2xx10^(-2)L"mol"^(-1)`
Calculate the equilibrium constant for the following equilibrium.
`ABC(g)hArr1/(2)A_(2)(g)+1/(2)B_(2)(g)+1/(2)C_(2)(g)`

Options

A. `500"mol"^(1//2)L^(1//2)`
B. `4xx10^(-6)"mol"^(1//2)L^(1//2)`
C. `500"mol"^(1//2)L^(-1//2)`
D. `200"mol"^(1//2)L^(-)(1//2)`

Answer

C

Explanation

`A_(2)(g)+B_(2)(g)hArr2AB(g)`
`K_(1)=2xx10^(-4)"L/mol"`
`K_(1)=(|AB|^(2))/([A_(2)][B_(2)])` …(i)
`2AB(g)+C_(2)(g)hArr2ABC(g)`
`K_(2)=2xx10^(-2)"L/mol"`
So, `K_(2)=([ABC]^(2))/([AB]^(2)[C_(2)])` …(ii)
`ABChArr1/(2)A_(2)(g)+1/(2)B_(2)(g)+1/(2)C_(2)(g)`
So, `K_(3)=([A_(2)]^(1//2)[B_(2)]^(1//2)[C_(2)]^(1//2))/([ABC])` ...(iii)
Eq. (i) and Eq. (ii) are added, reversed and divided by 2 then we get Eq. (iii).
Therefore, `K_(3)=1/(sqrt(K_(1)xxK_(2)))`
`=1/(sqrt(2xx10^(-4)xx2xx10^(-2)))`
`=1000/(2)=500"mol"^(1//2)L^(-1//2)`