1.

15 moles of H_(2) and 5.2 moles of I_(2) are mixed and allowed to attain equilibrium at 500^(@)C. At equilibrium, the concentration of HI is found to be 10 moles. The equilibrium constant for the formation of HI is

Answer»

50
15
100
25

Solution :`UNDERSET(15-5)underset(15)(H_(2))+underset(5.2-5)underset(5.2)(I_(2))hArrunderset(10)underset(0)(2HI)`
`K_(C)=[HI]^(2)/[[I_(2)])=(10xx10)/(10xx0.2)=50`


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