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 $Hg$, the total equilibrium pressure is 800 mm $Hg$. Calculate $K$ for the reaction assuming that the volume of the system remains unchanged.

• A. 50
• B. 100
• C. 166.6
• D. 400

Answer 1

The correct option is A 50

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

The initial pressure of $X{Y}_2$ is 600 mm Hg.

Let $x$ mm Hg be the change in pressure of $X{Y}_2$ to reach equilibrium.

The equilibrium pressures of $X{Y}_2$, $XY$ and $Y$ are $(600-x)$ mm Hg, $x$ mm Hg and $x$ mm Hg respectively.

Total pressure at equilibrium $=(600-x)+x+x=600+x$ mm Hg. But it is equal to 800 mm Hg.

$(600+x)=800$

$x=200$ mm Hg
The equilibrium pressures of $X{Y}_2$, $XY$ and $Y$ are :

$(600-200)=400$ mm Hg, $200$ mm Hg and $200$ mm Hg respectively.
The equilibrium constant $K_p=\frac{\left[P_{XY]}\right[P_Y]}{\left[P_{XY2}\right]}$
The equilibrium constant $K_p=\frac{200\times 200}{400}=100$