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What, if anything, can be said about the...

What, if anything, can be said about the magnitude of the equilibrium constant K for the following equilibrium ?

A

`K = 1 `

B

`K lt 1`

C

`K gt 1`

D

No estimate of K can be made

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The correct Answer is:
B

equilibrium favours reactant so `K lt 1`.
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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))

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) The equilibrium constants for the following reactions at 1400 K are given: 2H_2O_((g)) harr 2H_(2(g))+O_(2(g)) , K_1=2.1 xx 10^(-13) 2CO_(2(g)) harr 2CO_((g))+O_(2(g)) , K_2=1.4 xx 10^(-12) Then the equilibrium constant K for the reaction, H_(2(g))+CO_(2(g)) harr CO_((g)) + H_2O_((g)) is

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