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What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?A. `K_(c)=([P_(4)O_(10)])/([P_(4)][O_(2)]^(5))`B. `K_(c)=(1)/([O_(2)]^(5))`C. `K_(c)=[O_(2)]^(5)`D. `K_(c)=([P_(4)O_(10)])/(5[P_(4)][O_(2)])` |
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Answer» Correct Answer - B In the heterogeneous equilibrium `P_(4)(s)+5O_(2)(g) hArr P_(4) O_(10)(s)` `K_(c)=([P_(4)O_(10)(s)])/([P_(4)(s)][O_(2)(g)]^(5))` Here `[P_(4)O_(10)(s)]=[P_(4)(s)]=1` ( By convention ) `:. K_(c ) =(1)/([O_(2)]^(5))` |
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