The relation between `K_(p)` and `K_(c)` is `K_(p)=K_(c)(RT)^(Deltan)` unit of `K_(p)=(atm)^(Deltan)`, unit of `K_(c)=(mol L^(-1))^(Deltan)`
The equilibrium constant of the following reactions at `400 K` are given:
`2H_(2)O(g) hArr 2H_(2)(g)+O_(2)(g), K_(1)=3.0xx10^(-13)`
`2CO_(2)(g) hArr 2CO(g)+O_(2)(g), K_(2)=2xx10^(-12)`
Then, the equilibrium constant K for the reaction
`H_(2)(g)+CO_(2)(g) hArr CO(g)+H_(2)O(g)`
is

Options

A. `2.04`
B. `20.5`
C. `0.85`
D. `1.4`

Answer

D

Explanation

From reaction (i) and (ii), we get
`2H_(2)(g)+2CO_(2)(g) hArr 2CO(g)+2H_(2)O(g),`
`K=1.7xx10^(-12)`
And hence `H_(2)+CO_(2) hArr CO+H_(2)O,`
`K=(1.7xx10^(-12))/(2)=0.85xx10^(-12)`