If `K_(1)` and `K_(2)` are respective equilibrium constants for two reactions `:`
`XeF_(6)(g) +H_(2)O hArr XeOF_(4)(g) +2HF_(g)`
`XeO_(4)(g)+XeF_(6)(g)hArr XeOF_(4)(g)+XeO_(3)F_(2)(g)`
Then equilibrium constant for the reaction
`XeO_(4)(g)+2HF(g) hArr XeO_(3)F_(2)(g)+H_(2)O(g)` will be

If K_(1) and K_(2) are the respective equilibrium constants for the two reactions : XeF_(6)(g)+H_(2)O(g)hArr XeOF_(4)(g)+2HF(g) XeO_(4)(g)+XeF_(6)(g)hArr XeOF_(4)(g)+XeO_(3)F_(2)(g) The equilibrium constant for the reaction : XeO_(4)(g)+2HF(g)hArr XeO_(3)F_(2)(g)+H_(2)O(g) will be :

If K_(1) and K_(2) are the equilibrium constant for the two reations XeF_(4(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 reaction XeO_(4(g))+2HF_((g))hArrXeO_(3)F_(2(g))+H_(2)O_((g)) is

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 :

If K_1 and K_2 are the respective equilibrium constants for the two reactions (i) XeF_6(g)+H_2O(g)iffXeOF_4(g)+2HF(g) (ii) XeO_4(g)+XeF_6(g)iffXeOF_4(g)+XeO_3F_2(g) The equilibrium constant for the reaction XeO_4(g)+2HF(g)iffXeO_3F_2(g)+H_2O(g) will be

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 :

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

Passage: When all the coefficients in a balanced chemical equation are multiplied by a constant factor X the equilibrium constant (originally K) becomes K^J . Similarly, when balanced equations are added together, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for each step. Equilibrium constant of the reversed reaction is numerically equal to the reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) Consider the two reactions: XeF_(6(g))+H_(2)O_((g)) harr XeOF_(4(g))+2HF_((g)), K_(1)" "XeO_(4(g))+XeF_(6(g)) harr XeOF_(4(g))+XeO_(3)F_(2(g)), K_(2) Then the equilibrium constant for the following reaction will be XeO_(4(g))+2HF_((g)) harr XeO_(3)F_(2(g))+H_(2)O_((g))