Question

$X{Y}_2$ 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. Assuming volume of system to remain constant, the value of $K_P$ is

• A. 50
• B. 100
• C. 20
• D. 400

Answer 1

The correct option is

A

100

$X{Y}_2\left(g\right)$ $\rightleftharpoons$ $XY\left(g\right)$ $+$ $Y\left(g\right)$
$t=0$ 600 mm Hg 0 0
At equilibrium 600 – P P P
$P_{total}=600-P+P+P=600+P$

$\therefore 600+P=800$

$\Rightarrow P=200mmHg$

$\therefore K_P=\frac{P_{XY.P_Y}}{P_{X{Y}_2}}=\frac{200\times 200}{400}=100$