Question

When sulphur in the form of $\text{ }{{\text{S}}_{\text{2}}}\text{(g) }$ is heated at $\text{ 900 K }$, the initial partial pressure of $\text{ }{{\text{S}}_{8}}\text{(g) }$ which was 1 atm falls by $\text{ 29}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ at equilibrium. This is because of the conversion of some $\text{ }{{\text{S}}_{8}}\text{(g) }$to $\text{ }{{\text{S}}_{\text{2}}}\text{(g) }$. Find the $\text{ }{{\text{K}}_{\text{P}}}\text{ }$ for reaction, $\text{ }{{\text{S}}_{8}}\text{(g) }\rightleftharpoons \text{ 4}{{\text{S}}_{\text{2}}}\text{(g) }$

Answer 1

Hint: For an ideal gaseous mixture, each component follows Dalton’s law of partial pressure i.e.$\text{ }{{\text{P}}_{\text{i}}}\text{ = }{{\text{X}}_{\text{i}}}\text{P }$ where P is the total pressure and $\text{ }{{\text{p}}_{\text{i}}}\text{ }$ is the partial pressure of the ith component with mole fraction $\text{ }{{\text{X}}_{\text{i}}}\text{ }$ in the mixture. Then the equilibrium constant for a reaction is,
$\begin{align}
& \text{ }a\text{A+}b\text{B }\rightleftharpoons m\text{M +}n\text{N } \\
& \text{ }\Rightarrow {{\text{K}}_{\text{p}}}\text{ = }\dfrac{\text{P}_{\text{M}}^{\text{m}}\text{P}_{\text{N}}^{\text{n}}}{\text{P}_{\text{A}}^{\text{a}}\text{P}_{\text{B}}^{\text{b}}}\text{ } \\
\end{align}$

Complete step by step solution:
We have given that sulphur in the form of $\text{ }{{\text{S}}_{\text{2}}}\text{(g) }$ is heated at $\text{ 900 K }$ .the partial pressure of $\text{ }{{\text{S}}_{8}}\text{(g) }$ sulphur form is 1 atm.at equilibrium, the partial pressure of the $\text{ }{{\text{S}}_{8}}\text{(g) }$ forms drops by $\text{ 29}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ .
We are interested to determine the equilibrium constant $\text{ }{{\text{K}}_{\text{P}}}\text{ }$ of the reaction. The reaction of $\text{ }{{\text{S}}_{8}}\text{(g) }$from to $\text{ }{{\text{S}}_{\text{2}}}\text{(g) }$ form is given as follows,
$\text{ }{{\text{S}}_{8}}\text{(g) }\rightleftharpoons \text{ 4}{{\text{S}}_{\text{2}}}\text{(g) }$
For an ideal gaseous mixture, each component follows Dalton’s law of partial pressure i.e.$\text{ }{{\text{P}}_{\text{i}}}\text{ = }{{\text{X}}_{\text{i}}}\text{P }$ where P is the total pressure and $\text{ }{{\text{p}}_{\text{i}}}\text{ }$ is the partial pressure of the ith component with mole fraction $\text{ }{{\text{X}}_{\text{i}}}\text{ }$ in the mixture. Then the equilibrium constant for a reaction is,
$\begin{align}
& \text{ }a\text{A+}b\text{B }\rightleftharpoons m\text{M +}n\text{N } \\
& \text{ }\Rightarrow {{\text{K}}_{\text{p}}}\text{ = }\dfrac{\text{P}_{\text{M}}^{\text{m}}\text{P}_{\text{N}}^{\text{n}}}{\text{P}_{\text{A}}^{\text{a}}\text{P}_{\text{B}}^{\text{b}}}\text{ } \\
\end{align}$
We have given the following reaction between the $\text{ }{{\text{S}}_{8}}\text{(g) }$ to give $\text{ }{{\text{S}}_{\text{2}}}\text{(g) }$.Initially the $\text{ }{{\text{S}}_{8}}\text{(g) }$ has a pressure equal to 1 atm. At equilibrium the $\text{ 29}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ of the pressure is dropped .Thus reaction is given as follows,
$\text{ }\begin{matrix}
{} & {{\text{S}}_{\text{8}}}\text{(g)} & \rightleftharpoons & \text{4}{{\text{S}}_{\text{2}}}\text{(g)} \\
\text{Initial P} & 1 & {} & 0 \\
\text{At e}{{\text{q}}^{\text{m}}} & 1-\alpha & {} & 4\alpha \\
\end{matrix}\text{ }$
Where \[\alpha \] is a degree of dissociation of sulphur.
Here $\text{ 29}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ of the pressure is dropped. Thus the value of the degree of dissociation is equal to $\text{ }\alpha \text{ = }\dfrac{29}{100}\text{ = 0}\text{.29 }$
At equilibrium, the equilibrium constant for the reaction in terms of partial pressure is written as follows,
$\text{ }{{\text{K}}_{\text{p}}}\text{ = }\dfrac{{{\left( {{\text{P}}_{{{\text{S}}_{\text{2}}}}} \right)}^{4}}}{{{\text{P}}_{{{\text{S}}_{\text{8}}}}}}\text{ = }\dfrac{{{\left( 4\times 0.29 \right)}^{4}}}{\left( 1-0.29 \right)}=\dfrac{{{\left( 1.16 \right)}^{4}}}{0.71}=2.55\text{ }$

Thus the value of the equilibrium constant for the reaction is $\text{ }{{\text{K}}_{\text{P}}}\text{ = }2.55$.

Note: The attainment of equilibrium can be recognized by considering the observable properties such as pressure, concentration, density, or colour . The equilibrium constant can be written in terms of concentration as follows,
$\text{ }{{\text{K}}_{\text{C}}}\text{ = }\dfrac{\text{C}_{\text{M}}^{\text{m}}\text{C}_{\text{N}}^{\text{n}}}{\text{C}_{\text{A}}^{\text{a}}\text{C}_{\text{B}}^{\text{b}}}\text{ }$
Where C is the concertation, A and B are reactants and M and N are the product.