1.

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)])`

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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