1.

Equilibrium constant for the following reactions have been determined at 823K. CaO(s)+H_(2)(g)hArrCO(s)+H_(2)O(g) K_(1)=60 CaO(s)+CO(g)hArrCO(S)+CO_(2)(g) K_(2)=400 Using this information, calculate, equilibrium constant (at the same temperature) for :- CO_(2)(g)+H_(2)(g)hArrCO(s)+H_(2)O(g)K_(3)=? CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g)K_(4)=?

Answer»

`K_(3)=0.15,K_(4)=6.66`
`K_(3)=1.5,K_(4)=66.6`
`K_(3)=15,K_(4)=666`
None

Solution :`K_(3)=(K_(1))/(K_(2))`
`K_(4)=(K_(2))/(K_(1))`


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