Question

For the given equation, calculate the value of Kp, if the initial pressure and the total pressure are given. Also, assume that the volume of the system is constant.

The given equation is:

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

The initial pressure of $X{Y}_2$ is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg.

Answer 1

Step1: Given data

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

The initial pressure of $X{Y}_2$ is $600\mathrm{mm}\mathrm{Hg}$. The total pressure at equilibrium is $800\mathrm{mm}\mathrm{Hg}$.

Step2: Calculation value of $K_p$

Let us assume that, to reach equilibrium $\mathrm{x}\mathrm{mm}\mathrm{Hg}\mathrm{of}\mathrm{Y}$d dissociates. $\mathrm{x}\mathrm{mm}\mathrm{Hg}\mathrm{of}\mathrm{Xy}\mathrm{and}\mathrm{x}\mathrm{mm}\mathrm{Hg}\mathrm{of}\mathrm{Y}$will be formed at equilibrium.

As given, $600-\mathrm{x}\mathrm{mm}\mathrm{Hg}\mathrm{of}{\mathrm{XY}}_2$ will be present at equilibrium. ……….equation ($i$)

Therefore, total pressure = $(600-\mathrm{x})+\mathrm{x}+\mathrm{x}=600+\mathrm{x}\mathrm{mm}\mathrm{Hg}$ ……..equitation ( $ii$ )

But given total pressure $=800\mathrm{mm}\mathrm{Hg}$. ………equation ($iii$)

Subsitutuing equation ($ii$) and ($iii$)

$$\begin{gathered} 600+x=800 \\ x=200 \end{gathered}$$

Therefore the equilibrium pressure is ( using equation($i$) )

$$\begin{gathered} {\mathrm{P}}_{{\mathrm{XY}}_2}=600-\mathrm{x} \\ {\mathrm{P}}_{{\mathrm{XY}}_2}=600-200[\mathrm{x}=200] \\ {\mathrm{P}}_{{\mathrm{XY}}_2}=400\mathrm{mm}\mathrm{Hg} \end{gathered}$$

$$\begin{gathered} {\mathrm{P}}_{\mathrm{XY}}=\mathrm{x}=200\mathrm{mm}\mathrm{Hg} \\ {\mathrm{P}}_{\mathrm{Y}}=\mathrm{x}=200\mathrm{mm}\mathrm{Hg} \end{gathered}$$

Given that the volume of the system remains constant, therefore,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{\mathrm{XY}}{\mathrm{P}}_{\mathrm{Y}}}{{{\mathrm{P}}_{\mathrm{XY}}}_2} \\ {\mathrm{K}}_{\mathrm{p}}=\frac{200\times 200}{400}[\mathrm{putting}\mathrm{value}\mathrm{of}{\mathrm{P}}_{\mathrm{XY}},{\mathrm{P}}_{\mathrm{Y}},\mathrm{and}{\mathrm{P}}_{{\mathrm{XY}}_2}] \\ {\mathrm{K}}_{\mathrm{p}}=100\mathrm{mm}\mathrm{Hg} \end{gathered}$$

Hence, the value of $K_p$ is $100\mathrm{mm}\mathrm{Hg}$.