The equilibrium constants for the reactions X = 2Y and Z = P + Q are $K_{1} a n d K_{2}$ , respectively. If the initial concentrations and the degree of dissociation of X and Z are the same, the ratio $\frac{K_{1}}{K_{2}}$ is.

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In the system X+2Y⇌Z, the equilibrium concentrations are, [X]=0.06molL−1,[Y]=0.12molL−1,[Z]=0.216molL−1. Find the equilibrium constant of the reaction.

The correct option is A 250X+2Y⇌ZThe equilibrium constant Kc=ZX×(Y)2Kc=0.2160.06×(0.12)(0.12)Kc=0.2160.000864=250

In the system $X + 2 Y ⇌ Z$ , the equilibrium concentrations are,
$[X] = 0.06 m o l L^{- 1}, [Y] = 0.12 m o l L^{- 1}$ ,
$[Z] = 0.216 m o l L^{- 1}$ . Find the equilibrium constant of the reaction.

The correct option is A $250$
$X + 2 Y ⇌ Z$

The equilibrium constant $K_{c} = \frac{Z}{X \times (Y)^{2}}$
$K_{c} = \frac{0.216}{0.06 \times (0.12) (0.12)}$

$K_{c} = \frac{0.216}{0.000864} = 250$