Question

$X{Y}_2$ dissociates as $X{Y}_2\left(g\right)\rightleftharpoons XY\left(g\right)+Y\left(g\right)$ 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.

• A. 50 mm Hg
• B. 100 mm Hg
• C. 166.6 mm Hg
• D. 400 mm Hg

Answer 1

The correct option is B 100 mm Hg

$X{Y}_2$ dissociates as:
$X{Y}_2\left(g\right)\rightleftharpoons XY\left(g\right)+Y\left(g\right)$
Initial pressure of $X{Y}_2$ is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg.

Let x mm Hg of $X{Y}_2$ dissociates to reach equilibrium. x mm Hg of XY and x mm Hg of Y will be formed at equilibrium. 600 - x mm Hg of $X{Y}_2$ will be present at equilibrium.

Total pressure $=(600-x)+x+x=600+xmmHg$

But total pressure = 800 mm Hg
$600+x=800orx=200$

Thus, the equilibrium pressures are
$P_{X{Y}_2}=600-x=600-200=400mmHg$
$P_{XY}=x=200mmHg$
$P_Y=x=200mmHg$

Assuming volume of system to remain constant, the value of $K_p=\frac{P_{XY}{P}_Y}{P_{X{Y}_2}}$

$K_p=\frac{200\times 200}{400}$
$K_p=100mmHg$