For the reaction, `H_(2)(g)+CO_(2)(g) hArr CO(g)+H_(2)O(g)`, if the initial concentration of `[H_(2)]=[CO_(2)]` and x mol `L^(-1)` of `H_(2)` is consumed at equilibrium, the correct expression of `K_(p)` is:

Options

A. `(x^(2))/((1-x)^(2))`
B. `((1+x)^(2))/((1-x)^(2))`
C. `(x^(2))/((2+x)^(2))`
D. `(x^(2))/(1-x^(2))`

Answer

A

Explanation

`{:(,H_(2(g)),+,CO_(2(g)),hArr,CO_((g)),+,H_(2)O_((g))),("Initial conc.",1,,1,,0,,0),("At equilibrium",(1-x),,(1-x),,x,,x):}`
`K_(p) = (PCO.PH_(2)O)/(PH_(2).PCO_(2)) = (x.x)/((1-x)(1-x)) = (x^(2))/((1-x)^(2))`