Question

Phosphorus pentachloride is dissociating $50\mathrm{\%}$ at ${250}^{\circ }C$ at total pressure of $Patm$. If the equilibrium constant is$K_p$, then which of the following reactions is numerically correct?
A) $K_p=3P$
B) $P=3K_p$
C) $P={\displaystyle \frac{2K_p}{3}}$
D) $K_p={\displaystyle \frac{2P}{3}}$

Answer 1

Hint: We know that the equilibrium constant $K_p$ describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.
The expression can be given as,
$K_p={\displaystyle \frac{{\left(P_C\right)}^c{\left(P_D\right)}^d}{{\left(P_A\right)}^a{\left(P_B\right)}^b}}$
The equilibrium constant $K_p$ is related to the equilibrium constants in terms of molar concentration.
$P_A$, $P_B$, $P_C$ and $P_D$ are partial pressures of A, B, C and D respectively.

Complete step by step solution:
The dissociation equation of $PC{l}_5$ can be written as,
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
It is given the value of $X=0.5$
Let us calculate the mole fractions of the components.
$X_{PC{l}_5}={\displaystyle \frac{1-0.5}{3\times 0.5}}={\displaystyle \frac{1}{3}}$
$X_{PC{l}_3}={\displaystyle \frac{1-0.5}{3\times 0.5}}={\displaystyle \frac{1}{3}}$
$X_{C{l}_2}={\displaystyle \frac{1-0.5}{3\times 0.5}}={\displaystyle \frac{1}{3}}$
We know that the total pressure is P.
$K_p={\displaystyle \frac{\left(P_{PC{l}_3}\right)\left(P_{C{l}_2}\right)}{\left(P_{PC{l}_2}\right)}}={\displaystyle \frac{\left(X_{PC{l}_3}\times P\right)\left(X_{C{l}_2}\times P\right)}{\left(X_{PC{l}_2}\times P\right)}}$
Let we substitute the values in the above equation we get,
$\Rightarrow K_p={\displaystyle \frac{{\displaystyle \frac{1}{3}}\times P\times {\displaystyle \frac{1}{3}}\times P}{{\displaystyle \frac{1}{3}}\times P}}$
On simplifying we get,
$\Rightarrow P=3K_p$
Therefore,option B is correct.

Additional information:
We know that dissociation constant is an equilibrium constant which measures the tendency of a bigger object to separate reversibly into smaller components; the equilibrium constant also can be called ionization constant.
For a general reaction:
$A_x{B}_y\rightleftharpoons xA+yB$
The dissociation constant of the reaction is,
$K_p={\displaystyle \frac{{\left[A\right]}^x{\left[B\right]}^y}{\left[A_x{B}_y\right]}}$
The equilibrium concentrations of A, B are $$\left[A\right]$$,$$\left[B\right]$$, and $$\left[A_x{B}_y\right]$$.

Note:
The idea of the stability constant is applied in various fields of chemistry and pharmacology. In protein-ligand binding the stability constant describes the attraction among a protein and a ligand. A small stability constant indicates an addition strongly among the ligand. Within the case of antibody-antigen requisite the upturned equilibrium constant is engaged and is named affinity constant.
Using the equilibrium constant we can calculate the $pH$ of the reaction,
Example:
Write the dissociation equation of the reaction.
$HA+H_{2}O\stackrel{}{\to }{H}_3{O}^{+}+A^{-}$
The constant $K_a$ of the solution is $4\times {10}^{-2}$.
The dissociation constant of the reaction $K_a$ is written as,
$K_a={\displaystyle \frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]}}$
Let us imagine the concentration of $$\left[H_3{O}^{+}\right]\left[A^{-}\right]$$ as x.
$4\times {10}^{-7}={\displaystyle \frac{x^2}{0.08-x}}$
$x^2=4\times {10}^{-7}\times 0.08$
On simplifying we get,
$\Rightarrow x=1.78\times {10}^{-4}$
The concentration of Hydrogen is $1.78\times {10}^{-4}$
We can calculate the $pH$ of the solution is,
$pH=-\log \left[H^{+}\right]=3.75$
The $pH$ of the solution is $3.75$.