If `K_1` and `K_2` are the respective equilibrium constants for the two reactions
`XeF_6(g)+2HF(g)hArrXeOF_4(g)+2HF(g)`
`XeO_4(g)+XeFe_6(g)hArrXeOF_4(g)+XeO_3(g)`
The equilibrium constant of the reaction, `XeO_4(g)+2HF(g)hArrXeO_3F_2(g)+H_2O(g)` will be

Options

A. `K_(1)K_(2)^(2)`
B. `K_(1)-K_(2)`
C. `K_(2)//K_(1)`
D. `K_(2)//K_(2)`

Answer

3

Explanation

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