Given
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g),K_(1)`
`N_(2)(g)+O_(2)(g)hArr2NO(g),K_(2)`
`H_(2)(g)+(1)/(2)O_(2)hArrH_(2)O(g),K_(3)`
The equilibrium constant for
`2NH_(3)(g)+(5)/(2)O_(2)(g)hArr2NO(g)+3H_(2)O(g)`
will be

Options

A. `K_(2)K_(3)^(3)//K_(1)`
B. `K_(2)K_(3)//K_(1)`
C. `K_(2)^(3)K_(3)//K_(1)`
D. `K_(1)K_(3)^(3)//K_(2)`

Answer

A

Explanation

`N_(2) + 3H_(2)hArr2NH_(3)" "K_(1) rarr` (1)
`N_(2) + O_(2)hArr2NO" "K_(2)rarr` (2)
`H_(2) + (1)/(2)O_(2)rarrH_(2)O" "K_(3)rarr` (3)
Now, equation (2) + 3 x equation (3) - equation (1)
`2NH_(3) + (5)/(2)O_(2)hArr 2NO + 3H_(2)O`
`implies K =K_(2.K_(3)^(3))/(K_(1))` , Option (1)
In equilibrium reaction when equations are added 'k' are multiplied and if equations are subracted 'K' is divided.