An equilibrium mixture
`CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g)`
present in a vessel of one litre capacity at `815^(@)C` was found by analysis to contain `0.4` mol of CO, `0.3` mol of `H_(2)O, 0.2` mol of `CO_(2)` and `0.6` mol of `H_(2)`.
a. Calculate `K_(c)`
b. If it is derived to increase the concentration of CO to `0.6` mol by adding `CO_(2)` to the vessel, how many moles must be addes into equilibrium mixture at constant temperature in order to get this change?

Answer

A, B, C

Explanation

`{:(CO(g),+,H_(2)O(g),hArr,CO_(2)(g),+,H_(2)(g)),(0.4,,0.3,,0.2,,0.6),(,,,,,,"moles at equilibrium"):}`
a. `:. K_(C)=([CO_(2)][H_(2)])/([CO][H_(2)O])=(0.2xx0.6)/(0.4xx0.3)=1`
`( :' Deltan=0, :. "Volume terms are not needed")`
b. Now it is desired to increase the concentration of CO by `0.2` at equilibrium by forcing `CO_(2)` into equilibrium mixture. Suppose a mol of `CO_(2)` are forced in vessel at equilibrium, by doing so reaction proceeds in backward direction, i.e.,
`{:(CO_(2),+,H_(2)(g),hArr,CO,+,H_(2)O),((0.2+a),,0.6,,0.4,,0.3),(,,,,,,("Addition at initial equilibrium")),((0.2+a-0.2),,(0.6-0.2),,(0.4+0.2),,(0.3+0.2)),(=a,,=0.4,,=0.6,,=0.5),(,,,,,,"moles at new equilibrium"):}`
`:. 1/K_(C)=([CO][H_(2)O])/([CO_(2)][H_(2)])=(0.6xx0.5)/(axx0.4)`
`:. a=0.75 "moles"`