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The equilibrium constant for the followi...

The equilibrium constant for the following two reactions are `K_(1)` and `K_(2)` respectively.
`XeF_(6)(g)+H_(2)O(g) hArrXeOF_(4)(g)+2HF(g)`
`XeO_(4)(g)+XeF_(6)(g)hArrXeOF_(4)(g)+XeO_(3)F_(2)(g)`
The equilibrium constant for the given reactiono is `:`

A

`K_(1)K_(2)^(2)`

B

`K_(1)-K_(2)`

C

`K_(2)//K_(1)`

D

`K_(2)//K_(2)`

Text Solution

Verified by Experts

The correct Answer is:
3
Substracting `1^(st)` reaction from second, we will get desired reaction
`3^(rd)`. So `K=K_(2)//K_(1)`
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