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))`

Options

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

Answer

D

Explanation

equation b-a = c `:. K_(3)=(K_(2))/(K_(1))`