
A moles of $PC{{l}_{5}}$ undergoes thermal dissociation as: \[PC{{l}_{5}}\leftrightarrows\ PC{{l}_{3}}+C{{l}_{2}}\], the mole fraction of $PC{{l}_{3}}$ at equilibrium is 0.25 and the total pressure is 2.0 atmosphere. The partial pressure of \[C{{l}_{2}}\] at equilibrium is:
a.) 2.5
b.) 1.0
c.) 0.5
d.) None of these
Answer
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Hint: Partial pressure of any component is product of pressure of pure component and its mole fraction. At equilibrium, reactants and products are in equilibrium so they have the same mole fraction depending upon mole ratio for the reaction.
Complete step by step answer:
-Mole fraction of any component is ratio of number of moles of that component to total number of moles of all components present in the system.
-As given \[PC{{l}_{5}}\leftrightarrows\ PC{{l}_{3}}+C{{l}_{2}}\]
As given in data initially moles of $PC{{l}_{5}}$ are $a$ , Initial moles of products will be zero.
At equilibrium, $x$ moles of products are formed so moles of $PC{{l}_{5}}$ at equilibrium will be a-$x $and moles of $PC{{l}_{3}}$ will be $x$ and moles of $C{{l}_{2}}$ will be $x$ at equilibrium.
Mole fraction of $PC{{l}_{3}}$ and \[C{{l}_{2}}\] at equilibrium are same as number of moles of both are equal at equilibrium.
\[{{X}_{C{{l}_{2}}}}={{X}_{PC{{l}_{3}}}}=\dfrac{\text{moles of PC}{{\text{l}}_{3}}}{\begin{align}
& total\text{ number of moles at equilibrium} \\
& \text{=}\dfrac{x}{a-x+x+x} \\
\end{align}}\]
= $\dfrac{x}{a+x}$
Mole fraction of $PC{{l}_{3}}$ is 0.25 so mole fraction of chlorine gas at equilibrium will be also 0.25
Partial pressure of any component is a product of total pressure and mole fraction of that component.
The partial pressure of chlorine at equilibrium is:
Total pressure and mole fraction of chlorine =2 x 0.25=0.5
So, the correct answer is “Option C”.
Note: mole fraction of $C{{l}_{2}}$ can be directly calculated as 0.25 as at equilibrium, moles of $C{{l}_{2}}$ is equal to moles of $PC{{l}_{3}}$.Partial pressure of required component can be calculated by multiplying total pressure and mole fraction of that component .Mole fraction is ratio of number of moles of that component to total number of moles of all component.
Complete step by step answer:
-Mole fraction of any component is ratio of number of moles of that component to total number of moles of all components present in the system.
-As given \[PC{{l}_{5}}\leftrightarrows\ PC{{l}_{3}}+C{{l}_{2}}\]
As given in data initially moles of $PC{{l}_{5}}$ are $a$ , Initial moles of products will be zero.
At equilibrium, $x$ moles of products are formed so moles of $PC{{l}_{5}}$ at equilibrium will be a-$x $and moles of $PC{{l}_{3}}$ will be $x$ and moles of $C{{l}_{2}}$ will be $x$ at equilibrium.
Mole fraction of $PC{{l}_{3}}$ and \[C{{l}_{2}}\] at equilibrium are same as number of moles of both are equal at equilibrium.
\[{{X}_{C{{l}_{2}}}}={{X}_{PC{{l}_{3}}}}=\dfrac{\text{moles of PC}{{\text{l}}_{3}}}{\begin{align}
& total\text{ number of moles at equilibrium} \\
& \text{=}\dfrac{x}{a-x+x+x} \\
\end{align}}\]
= $\dfrac{x}{a+x}$
Mole fraction of $PC{{l}_{3}}$ is 0.25 so mole fraction of chlorine gas at equilibrium will be also 0.25
Partial pressure of any component is a product of total pressure and mole fraction of that component.
The partial pressure of chlorine at equilibrium is:
Total pressure and mole fraction of chlorine =2 x 0.25=0.5
So, the correct answer is “Option C”.
Note: mole fraction of $C{{l}_{2}}$ can be directly calculated as 0.25 as at equilibrium, moles of $C{{l}_{2}}$ is equal to moles of $PC{{l}_{3}}$.Partial pressure of required component can be calculated by multiplying total pressure and mole fraction of that component .Mole fraction is ratio of number of moles of that component to total number of moles of all component.
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