Question

For the given reaction at equilibrium, moles of ${\mathrm{Cl}}_2\left(\mathrm{g}\right)$ are equal to the moles of $\mathrm{Cl}\left(\mathrm{g}\right)$ and the equilibrium pressure is $1\mathrm{atm}$. If the ${\mathrm{K}}_{\mathrm{p}}$ of this reaction is $\mathrm{x}\times {10}^{-1}$ find $\mathrm{x}$.

${\mathrm{Cl}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{Cl}\left(\mathrm{g}\right)$

Answer 1

The above question can be solved as:

${\mathrm{Cl}}_2\rightleftharpoons 2\mathrm{C}[{\mathrm{P}}_{\mathrm{total}}=1\mathrm{atm}]$

Step 1:

At equilibrium ,

Moles of $\mathrm{Cl}=$Moles of ${\mathrm{Cl}}_2$

Step 2:

Given : ${\mathrm{K}}_{\mathrm{P}}=\mathrm{x}\times {10}^{-1}$

Step 3:

${\mathrm{n}}_{\mathrm{Cl}}={\mathrm{n}}_{{\mathrm{Cl}}_2.}$

Therefore, ${\mathrm{P}}_{{\mathrm{Cl}}_2}={\mathrm{P}}_{\mathrm{Cl}}$

Hence, $$\begin{gathered} {\mathrm{P}}_{{\mathrm{Cl}}_2}={\mathrm{P}}_{\mathrm{Cl}} \\ =0.5\mathrm{atm} \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}={\left({\mathrm{P}}_{\mathrm{Cl}}\right)}^2/{\mathrm{P}}_{{\mathrm{Cl}}_2} \\ =0.5 \\ =5\times {10}^{-1} \\ \mathrm{x}=5 \end{gathered}$$

Therefore, the value of $\mathrm{x}$ is $5$