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Consider the following equilibrium `H_(2)O(g)+CO(g)hArrH_(2)(g)+CO_(2)(g),K_(1)=2` `FeO(s)+CO(g)hArrFe(s)+CO_(2)(g),K_(2)=4` Then K for reaction: `Fe(s)+H_(2)O(g)hArrFeO(s)+H_(2)(g)`A. 2B. 1C. `(1)/(3)`D. `sqrt(2)` |
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Answer» `H_(2)O(g)+CO(p)hArrH_(2)(g)+CO_(2)(g)`….(i) `FeO(s)+CO(g)hArrFe(s)+CO_(2)(g)`…(ii) Required reaction: `Fe(s)+H_(2)O(g)hArrFeO(s)+H_(2)(g)`…(iii) Equation iii=Equation i-Equation ii `K_(3)=(K_(1))/(K_(2))=(2)/(4)=(1)/(2)` |
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