1.

If K_1 and K_2 are the respective equiliberium constant for the two reactions XeF_6 (g) + H_2O(g) = XeOF_4 (g) + 2HF(g) XeO_4 (g) + XeF_6 (g) ⇌ XeOF_4 (g) XeO_3F_2(g) , The equiliberium constant for the reaction,XeO_4 (g) + 2HF (g) ⇌ XeO_3F_2 + H_2O (g) is :

Answer»

`K_1 K_2`
`(K_1) /(K_2^2)`
`(K_2)/(K_1)`
`(K_1)/(K_2)`

ANSWER :C


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