Passage: Whenall the coefficients ina balancedchemical equationare multipliedbya constant factor X the equilibrium constant (originallyK) becomes K^J.Similarly,whenbalancedequations are addedtogether, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for eachstep.Equilibrium constant of the reversedreactionis numericallyequal tothe 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))

Passage: Whenall the coefficients ina balancedchemical equationare mu

1.	Passage: Whenall the coefficients ina balancedchemical equationare multipliedbya constant factor X the equilibrium constant (originallyK) becomes K^J.Similarly,whenbalancedequations are addedtogether, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for eachstep.Equilibrium constant of the reversedreactionis numericallyequal tothe 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))
Answer» `K_(1) // K_(2)^(2)` `K_(1) // K_(2)` `K_(1)^(2) // K_(2)` `K_(2) // K_(1)` SOLUTION :equation b-a = C `:. K_(3)=(K_(2))/(K_(1))`