
The equilibrium constant $K_p$ for the reaction $PC{l_{{5_{(g)}}}} \rightleftharpoons PC{l_{{3_{(g)}}}} + C{l_{{2_{(g)}}}}$ is 16. If the volume of the container is reduced to half of its original volume, the value of $K_p$ for the reaction at the same temperature will be:
A.32
B.64
C.16
D.4
Answer
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Hint: Equilibrium constant $K_p$ is equal to the ratio of the partial pressure of products and the partial pressure of reactants. The Partial pressures are raised to the powers that are equal to their numerical coefficients in the balanced chemical equation. The subscript ‘p’ in $K_p$ refers to the partial pressure.
Complete answer:
We know that a reaction is said to be at equilibrium when the Forward and the reverse reactions are the same. The concentration of both the reactant and the products remains the same, although the reaction still takes place. If one of the reactants is in a gaseous state, we can express the Equilibrium constant in partial pressure.
The given balanced equation is Phosphorus pentachloride is in equilibrium with phosphorous trichloride and Chlorine gas.
$PC{l_{{5_{(g)}}}} \rightleftharpoons PC{l_{{3_{(g)}}}} + C{l_{{2_{(g)}}}}$
$PC{{l}_{{{5}_{\left( g \right)}}}}$ is the reactant.
\[\text{PC}{{\text{l}}_{3\left( g \right)}}\]and \[\text{C}{{\text{l}}_{2\left( g \right)}}\]are the products.
So, $K_p$ will be given by:
$ \Rightarrow {K_p} = \dfrac{{\left[ {PC{l_{{5_{(g)}}}}} \right]}}{{\left[ {PC{l_{{3_{(g)}}}}} \right]\left[ {C{l_2}_{(g)}} \right]}}$
Now, the volume of the container is reduced to half. However, the volume change does not affect the $K_p$. The value of $K_p$ is independent of the volume of the container.
Only a change in temperature affects the value of $K_p$. But the temperature of the reaction is unchanged. So the value of $K_p$ will remain the same.
\[\Rightarrow {{K}_{p}}=\text{ }16.\]
Final answer: The correct answer is Option C: 16.
Note:
$K_p$ is expressed as the equilibrium constant only for the partial pressure of the homogeneous system that is gas. For the partial pressure of any heterogeneous system like solids or liquids, the equilibrium constant expression does not include their values.
Complete answer:
We know that a reaction is said to be at equilibrium when the Forward and the reverse reactions are the same. The concentration of both the reactant and the products remains the same, although the reaction still takes place. If one of the reactants is in a gaseous state, we can express the Equilibrium constant in partial pressure.
The given balanced equation is Phosphorus pentachloride is in equilibrium with phosphorous trichloride and Chlorine gas.
$PC{l_{{5_{(g)}}}} \rightleftharpoons PC{l_{{3_{(g)}}}} + C{l_{{2_{(g)}}}}$
$PC{{l}_{{{5}_{\left( g \right)}}}}$ is the reactant.
\[\text{PC}{{\text{l}}_{3\left( g \right)}}\]and \[\text{C}{{\text{l}}_{2\left( g \right)}}\]are the products.
So, $K_p$ will be given by:
$ \Rightarrow {K_p} = \dfrac{{\left[ {PC{l_{{5_{(g)}}}}} \right]}}{{\left[ {PC{l_{{3_{(g)}}}}} \right]\left[ {C{l_2}_{(g)}} \right]}}$
Now, the volume of the container is reduced to half. However, the volume change does not affect the $K_p$. The value of $K_p$ is independent of the volume of the container.
Only a change in temperature affects the value of $K_p$. But the temperature of the reaction is unchanged. So the value of $K_p$ will remain the same.
\[\Rightarrow {{K}_{p}}=\text{ }16.\]
Final answer: The correct answer is Option C: 16.
Note:
$K_p$ is expressed as the equilibrium constant only for the partial pressure of the homogeneous system that is gas. For the partial pressure of any heterogeneous system like solids or liquids, the equilibrium constant expression does not include their values.
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