At 500 K, equlibrium constant, `K_(c)` for the following reaction is 5.
`1//2 H_(2)(g)+ 1//2(g)hArr HI (g)`
What would be the equilibrium constant `K_(c)` for the reaction `2hi(g)hArrH_(2)(g)+l_(2)(g)`

Options

A. 0.04
B. 0.4
C. 25
D. 2.5

Answer

A

Explanation

For the reaction `1/2 H_(2) (g)+1/2 I_(2) (g) Leftrightarrow HI(g)`
`K_(c)=([HI])/([H_(2)]^(1//2) [I_(2)]^(1//2))=5`
Thus, for the reaction `2HI(g) Leftrightarrow H_(2) (g)+I_(2) (g)`
`K_(c_(1))=([H_(2)] [I_(2)])/([HI]^(2))=(1/(KC))^(2)=(1/5)^(2)=0.04`
In the reaction. `N_(2) (g)+3H_(2) (g) Leftrightarrow 2NH_(3) (g)`. If the total pressure at which the equilibrum is established, is increased without changing the temperature, K will remain same. K changes only with change in temperature.