Question

A homogeneous ideal gaseous reaction ₎ ${\mathrm{AB}}_2\left(\mathrm{g}\right)\rightleftharpoons \mathrm{A}\left(\mathrm{g}\right)+2\mathrm{B}\left(\mathrm{g}\right)$ is carried out in a $25\mathrm{liter}$ flask at $27°\mathrm{C}$. The initial amount of ${\mathrm{AB}}_2\mathrm{was}1\mathrm{mole}$and the equilibrium pressure was $1.9\mathrm{atm}$. The value of ${\mathrm{K}}_{\mathrm{p}}\mathrm{is}\mathrm{x}\times {10}^{–2}$. The value of x here is ______. $[\mathrm{R}=0.08206{\mathrm{dm}}^3\mathrm{atm}{\mathrm{K}}^{–1}{\mathrm{mol}}^{–1}]$

Answer 1

A homogeneous ideal gaseous reaction is carried out in a $25\mathrm{liter}$ flask at $27°\mathrm{C}$. The initial amount of ${\mathrm{AB}}_2\mathrm{was}1\mathrm{mole}$and the equilibrium pressure was $1.9\mathrm{atm}$. The value of ${\mathrm{K}}_{\mathrm{p}}\mathrm{is}\mathrm{x}\times {10}^{–2}$. The value of x is $\mathbf{72}$. $[\mathrm{R}=0.08206{\mathrm{dm}}^3\mathrm{atm}{\mathrm{K}}^{–1}{\mathrm{mol}}^{–1}]$

Step 1: Given: $\mathrm{R}=0.08206{\mathrm{dm}}^3\mathrm{atm}{\mathrm{K}}^{–1}{\mathrm{mol}}^{–1}$, equilibrium pressure = $1.9\mathrm{atm}$, $\mathrm{V}=25\mathrm{liter}$,T=$27°\mathrm{C}$

Step 2: Formula used: ${\mathrm{K}}_{\mathrm{p}}=\left({\mathrm{P}}_{\mathrm{A}}\right)\times {\left({\mathrm{P}}_{\mathrm{B}}\right)}^2/{\mathrm{P}}_{\mathrm{AB}}$,$\mathrm{PV}=\mathrm{nRT}$

${\mathrm{K}}_{\mathrm{p}}=\left({\mathrm{X}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{T}}\right)\times {\left({\mathrm{X}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{T}}\right)}^2/{\mathrm{X}}_{\mathrm{AB}}{\mathrm{P}}_{\mathrm{T}}$

Step 3:

$$\begin{gathered} {\mathrm{AB}}_2\left(\mathrm{g}\right)\rightleftharpoons \mathrm{A}\left(\mathrm{g}\right)+2\mathrm{B}\left(\mathrm{g}\right) \\ \mathrm{Initial}1-\mathrm{x}\mathrm{x}2\mathrm{x} \\ \mathrm{At}\mathrm{equillibrium}1/1+2\mathrm{x}1/1.9 \end{gathered}$$

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ 1.9\times 25={\mathrm{n}}_{\mathrm{T}}\times 0.082\times 300 \\ {\mathrm{n}}_{\mathrm{T}}=1.93 \\ 1+2\mathrm{x}=1.93 \\ 2\mathrm{x}=0.93 \\ \mathrm{x}=0.465 \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=\left({\mathrm{X}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{T}}\right)\times {\left({\mathrm{X}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{T}}\right)}^2/{\mathrm{X}}_{\mathrm{AB}}{\mathrm{P}}_{\mathrm{T}} \\ {\mathrm{K}}_{\mathrm{p}}=(0.465\times 1.9/1.93)\times {(0.93\times 1.9/1.93)}^2/(0.535\times 1.9/1.93) \\ {\mathrm{K}}_{\mathrm{p}}=0.072{\mathrm{atm}}^2 \\ \mathrm{In}\mathrm{different}\mathrm{terms} \\ {\mathrm{K}}_{\mathrm{p}}=0.072\times 100/100{\mathrm{atm}}^2 \\ {\mathrm{K}}_{\mathrm{p}}=72\times {10}^{-2}{\mathrm{atm}}^2 \end{gathered}$$

Hence, a homogeneous ideal gaseous reaction ₎ ${\mathrm{AB}}_2\left(\mathrm{g}\right)\rightleftharpoons \mathrm{A}\left(\mathrm{g}\right)+2\mathrm{B}\left(\mathrm{g}\right)$ is carried out in a $25\mathrm{liter}$ flask at $27°\mathrm{C}$. The initial amount of ${\mathrm{AB}}_2\mathrm{was}1\mathrm{mole}$and the equilibrium pressure was $1.9\mathrm{atm}$. The value of ${\mathrm{K}}_{\mathrm{p}}\mathrm{is}\mathrm{x}\times {10}^{–2}$. The value of x here is 72.