$X Y_{2}$ dissociates as,
$X Y_{2} (g) ⇌ X Y (g) + Y (g)$
Initial pressure of $X Y_{2}$ is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming volume of system to remain constant, the value of $K_{P}$ is

X Y_{2}

dissociates as:

X Y_{2} (g) ⇌ X Y (g) + Y (g)

Initial pressure of

X Y_{2}

600 m m

H g

. The total pressure at equilibrium is

800 m m

H g

. Assuming volume of system to remain constant, the value of

K_{p}

is:

X Y_{2}

dissociates as

X Y_{2} (g) ⇌ X Y (g) + Y (g)

when the initial pressure of

X Y_{2}

is 600 mm of Hg, the total equilibrium pressure is 800 mm of Hg. FInd the value of Equilibrium constant for the reaction assuming that the volume of the system remains unchanged.

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XY2 dissociation as-XY2(g)⇌XY(g)Y(g)Initial pressure of XY2 is 600 mm Hg.The total pressure st equilibrium is 800 mm Hg. Assuming volume of system to remain constant. the value of Kp is:

${X Y}_{2}$ dissociation as-
${X Y}_{2} (g) ⇌ X Y (g) Y (g)$
Initial pressure of ${X Y}_{2}$ is $600 m m H g$
.The total pressure st equilibrium is $800 m m H g$ .
Assuming volume of system to remain constant. the value of $K_{p}$ is: