(i) Find `K_(p)` partial pressures of the component species and the volume of the container for the reaction:
`H_(2)(g)+CO_(2)(g)hArrH_(2)O(g)+CO(g)`
The equilibrium for the given reaction is established in an evacuated vessel at 734 K starting of `CO_(2)`. Equilibrium mixture contains 10 per cent mole of water vapour, and equilibrium presure is 0.5 atm.
(ii) Adding solid CoO and solid Co in the flask, two new equilibria are established.
`CoO(s)+H_(2)(g)hArrCo(s),K_(1)+H_(2^(@))(g)`
`CoO(s)+CO(s)hArrCo(s),K_(2)+CO_(2)(g)`
The new equilibrium mixture contains 30 percent mole of water vapour. Find the equilibrium constants for the new equilibria & report the ratio `(K_(1))/(K_(2))`

Answer

`0.0756`

Explanation

`underset((0.1-x))(P_(H_(2))(g)+underset((0.2-x))(CO_(2)(g))hArr)underset(x)(H_(2)O(g))+underset(x)(Co(g))`……………
`Sigman=0.3`
`x/0.3=(X_(H_(2)O))=0.1`
`:.x=0.03`
Final `implies0.07` moles 0.17 moles,0.03 moles 0.03 moles ………..ii
`P_(eq)=0.5` atm
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`implies` Due to addition of CoO(s) and Co(s) new equilibria are developed and mole `H_(2)O(g)=0.3(30%)`
`CoO(s)+underset((0.07-a))(H_(2)(g))overset(K_(1))(hArr)CO(s)+underset((0.03+a))(H_(2)O(g))`
`larr{"moles(Final)"............(iii)`
`CoO+underset((0.03-b))(Co(g))overset(K_(2))(hArr)Co(s)+underset((0.17+b))(Co_(2)(g))`
`larr{"Moles (Final)".......(V)`
a and b are different because `K_(1)` and `K_(2)` are different
`X_(H_(2))O=0.3=(0.03+a)/0.3implies0.09=0.03+a`
`:.a=0.06`
`K_(1)=([H_(2)O])/([H_(2)])=0.09/0.01=9`
(iii) -(iv) gives reaction I)
`K=(K_(1))/(K_(2))` But `IK_(C)=(0.03xx0.03)/(0.07xx0.17)=0.0756=119`