Question

At\[\text{ T}\left( \text{K} \right)\] , the partial pressures of$\text{ S}{{\text{O}}_{\text{2}}}$, $\text{ }{{\text{O}}_{\text{2}}}$ and $\text{ S}{{\text{O}}_{3}}$ are\[0.662\] , \[0.100\] and \[0.331\] atm, respectively for the reaction $\begin{matrix}
\text{ 2S}{{\text{O}}_{\text{2}}}(g) & \text{+} & {{\text{O}}_{\text{2}}}(g) & \rightleftharpoons & \text{2S}{{\text{O}}_{\text{3}}}(g) \\
\end{matrix}$at equilibrium. What is the partial pressure in atm, of $\text{ }{{\text{O}}_{\text{2}}}$when the equilibrium partial pressures of $\text{ S}{{\text{O}}_{\text{2}}}$and $\text{ S}{{\text{O}}_{3}}$ are equal at the same temperature?
A) \[0.4\]
B) \[0.8\]
C) \[0.25\]
D) \[2.5\]

Answer 1

Hint: For gaseous state, the equilibrium constant ${{\text{K}}_{\text{p}}}$ is written in terms of the partial pressure, the reactant and pressure. Here, The ${{\text{K}}_{\text{p}}}$ can be calculated by considering the partial pressure of $\text{ S}{{\text{O}}_{\text{2}}}$,$\text{ S}{{\text{O}}_{3}}$ and $\text{ }{{\text{O}}_{\text{2}}}$. Since, we are interested to find out the partial pressure of $\text{ }{{\text{O}}_{\text{2}}}$at a condition $\text{p}_{\text{S}{{\text{O}}_{\text{2}}}}^{{}}\text{ = p}_{\text{S}{{\text{O}}_{3}}}^{{}}$, the equilibrium constant will remains the same.

Complete step by step answer:
In an ideal gaseous mixture, each component obeys Dalton’s law of partial pressures that is the total pressure exerted by the component on the mixture is equal to the partial pressure of the component. It is represented as,
$\text{ }{{\text{p}}_{\text{i}}}={{x}_{\text{i}}}\text{P }$,
Where P is the total pressure and ${{\text{p}}_{\text{i}}}$ is equal to the partial pressure of the ‘i’ component.
For a gaseous state reaction in which the A and B in gaseous state reacts and produces M and N in the gaseous state. The reaction is as shown below,
$\text{aA(g) + bB(g) }\rightleftharpoons \text{ mM(g) + nN(g)}$
We have, \[{{\text{p}}_{\text{A}}}\text{=}{{\text{X}}_{\text{A}}}\text{P}\] ,\[{{\text{p}}_{B}}\text{=}{{\text{X}}_{B}}\text{P}\], \[{{\text{p}}_{M}}\text{=}{{\text{X}}_{M}}\text{P}\] and \[{{\text{p}}_{N}}\text{=}{{\text{X}}_{N}}\text{P}\]
The equilibrium constant for the reaction at the equilibrium condition is given as,
${{\text{K}}_{\text{p}}}\text{ = }\dfrac{\text{p}_{\text{M}}^{\text{m}}\text{ p}_{\text{N}}^{\text{n}}}{\text{p}_{\text{A}}^{\text{a}}\text{ p}_{\text{B}}^{\text{b}}}$
Where A and B are reactant and M and N are the product of the reaction.
Here, we are given,
Partial pressure of $\text{ S}{{\text{O}}_{\text{2}}}$ $\text{p}_{\text{S}{{\text{O}}_{\text{2}}}}^{{}}\text{ = 0}\text{.662 atm}$
Partial pressure of $\text{ S}{{\text{O}}_{3}}$ $\text{p}_{\text{S}{{\text{O}}_{3}}}^{{}}\text{ = 0}\text{.331 atm}$
Partial pressure of $\text{ }{{\text{O}}_{\text{2}}}$ $\text{p}_{\text{ }{{\text{O}}_{\text{2}}}}^{{}}\text{ = 0}\text{.100 atm}$
We are interested to find out the partial pressure of oxygen when the $\text{p}_{\text{S}{{\text{O}}_{\text{2}}}}^{{}}\text{ = p}_{\text{S}{{\text{O}}_{3}}}^{{}}$
The sulfur dioxide reacts in presence of oxygen to produce sulphur trioxide. The reaction is reversible. It is depicted as follows,
$\begin{matrix}
   \text{ 2S}{{\text{O}}_{\text{2}}} & \text{+} & {{\text{O}}_{\text{2}}} & \rightleftharpoons & \text{2S}{{\text{O}}_{\text{3}}} \\
\end{matrix}$
Let’s first find out the equilibrium constant${{\text{K}}_{\text{p}}}$. Using the law of mass action for product and reactant we get,
${{\text{K}}_{\text{p}}}\text{ = }\dfrac{\text{p}_{\text{S}{{\text{O}}_{\text{3}}}}^{\text{2}}\text{ }}{\text{p}_{\text{S}{{\text{O}}_{\text{2}}}}^{\text{2}}\text{ p}_{{{\text{O}}_{\text{2}}}}^{\text{1}}}$
Substitute the values in the equilibrium constant, we have
$\begin{align}
 & \text{ }{{\text{K}}_{\text{p}}}\text{ = }\dfrac{{{\left( 0.331 \right)}^{\text{2}}}\text{ }}{{{\left( 0.662 \right)}^{\text{2}}}\text{ }\left( 0.100 \right)} \\
& \Rightarrow {{\text{K}}_{\text{p}}}\text{ = }\dfrac{\left( 0.109 \right)\text{ }}{\left( 0.438 \right)\text{ }\left( 0.100 \right)} \\
& \Rightarrow {{\text{K}}_{\text{p}}}\text{ = }\dfrac{\left( 0.109 \right)\text{ }}{\left( 0.0438 \right)} \\
& \Rightarrow {{\text{K}}_{\text{p}}}\text{ = }\dfrac{\left( 0.109 \right)\text{ }}{\left( 0.0438 \right)} \\
 & \Rightarrow {{\text{K}}_{\text{p}}}\text{ = 2}\text{.48 }\simeq \text{ 2}\text{.5 } \\
\end{align}$
Therefore, the equilibrium constant ${{\text{K}}_{\text{p}}}\text{ = 2}\text{.5 }$.
We want to find the $\text{p}_{\text{ }{{\text{O}}_{\text{2}}}}^{'}\text{ }$, when $\text{p}_{\text{S}{{\text{O}}_{\text{2}}}}^{{}}\text{ = p}_{\text{S}{{\text{O}}_{3}}}^{{}}$
Substitute values in the equation of equilibrium constant ${{\text{K}}_{\text{p}}}$.we have,
$\begin{align}
& \text{ }{{\text{K}}_{\text{p}}}\text{ = }\dfrac{\text{p}_{\text{S}{{\text{O}}_{\text{3}}}}^{\text{2}}\text{ }}{\text{p}_{\text{S}{{\text{O}}_{\text{2}}}}^{\text{2}}\text{ p}_{{{\text{O}}_{\text{2}}}}^{'}} \\
& \Rightarrow 2.5\text{ = }\dfrac{\left( \text{p}_{\text{S}{{\text{O}}_{\text{3}}}}^{\text{2}} \right)\text{ }}{\left( \text{p}_{\text{S}{{\text{O}}_{3}}}^{\text{2}} \right)\text{ p}_{{{\text{O}}_{\text{2}}}}^{'}}\text{ }\because \text{p}_{\text{S}{{\text{O}}_{\text{2}}}}^{{}}\text{ = p}_{\text{S}{{\text{O}}_{3}}}^{{}}\text{ , }{{\text{K}}_{\text{p}}}=2.5 \\
 & \Rightarrow 2.5\text{ = }\dfrac{\text{1 }}{\text{ p}_{{{\text{O}}_{\text{2}}}}^{'}} \\
 & \Rightarrow \text{p}_{{{\text{O}}_{\text{2}}}}^{'}\text{ = }\dfrac{1}{2.5} \\
& \therefore \text{p}_{{{\text{O}}_{\text{2}}}}^{'}\text{ = 0}\text{.4} \\
\end{align}$
Therefore, the partial pressure of oxygen $\text{p}_{{{\text{O}}_{\text{2}}}}^{'}\text{ }$when $\text{p}_{\text{S}{{\text{O}}_{\text{2}}}}^{{}}\text{ = p}_{\text{S}{{\text{O}}_{3}}}^{{}}$is equalled to$\text{0}\text{.4}$.

Hence, (A) is the correct option.


Note: The equilibrium constant can be written in the terms of activities$\text{ (}{{\text{a}}_{\text{i}}}\text{) }$, molar concentration$\text{ (}{{\text{C}}_{\text{i}}}\text{) }$, or mole fractions $\text{ (}{{\text{X}}_{\text{i}}}\text{) }$of the species involved in the reaction. The equilibrium constant in terms of concentration is written as,
$\begin{align}
& \text{for reaction,} \\
& \text{ aA + bB }\rightleftharpoons \text{ cC + dD} \\
& {{\text{K}}_{\text{p}}}\text{ = }\dfrac{{{\left[ \text{C} \right]}^{\text{c}}}{{\left[ \text{D} \right]}^{\text{d}}}\text{ }}{{{\left[ \text{A} \right]}^{\text{a}}}{{\left[ \text{B} \right]}^{\text{b}}}} \\
\end{align}$