Question

Phosphorus pentachloride is dissociating $50\% $ 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) $P = 3{K_p}$
C) $P = \dfrac{{2{K_p}}}{3}$
D) ${K_p} = \dfrac{{2P}}{3}$

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,
${K_p} = \dfrac{{{{\left( {{P_C}} \right)}^c}{{\left( {{P_D}} \right)}^d}}}{{{{\left( {{P_A}} \right)}^a}{{\left( {{P_B}} \right)}^b}}}$
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.
${X_{PC{l_5}}} = \dfrac{{1 - 0.5}}{{3 \times 0.5}} = \dfrac{1}{3}$
${X_{PC{l_3}}} = \dfrac{{1 - 0.5}}{{3 \times 0.5}} = \dfrac{1}{3}$
${X_{C{l_2}}} = \dfrac{{1 - 0.5}}{{3 \times 0.5}} = \dfrac{1}{3}$
We know that the total pressure is P.
${K_p} = \dfrac{{\left( {{P_{PC{l_3}}}} \right)\left( {{P_{C{l_2}}}} \right)}}{{\left( {{P_{PC{l_2}}}} \right)}} = \dfrac{{\left( {{X_{PC{l_3}}} \times P} \right)\left( {{X_{C{l_2}}} \times P} \right)}}{{\left( {{X_{PC{l_2}}} \times P} \right)}}$
Let we substitute the values in the above equation we get,
$ \Rightarrow {K_p}\, = \dfrac{{\dfrac{1}{3} \times P \times \dfrac{1}{3} \times P}}{{\dfrac{1}{3} \times P}}$
On simplifying we get,
$ \Rightarrow P = 3{K_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,
${K_p} = \dfrac{{{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}{{\left[ {{A_x}{B_y}} \right]}}$
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.
$HA + {H_2}O\xrightarrow{{}}{H_3}{O^ + } + {A^ - }$
The constant ${K_a}$ of the solution is $4 \times {10^{ - 2}}$.
The dissociation constant of the reaction ${K_a}$ is written as,
${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}$
Let us imagine the concentration of \[\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]\] as x.
$4 \times {10^{ - 7}} = \dfrac{{{x^2}}}{{0.08 - x}}$
${x^2} = 4 \times {10^{ - 7}} \times 0.08$
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 $3.75$.