Question

For the reaction $AB(g)\rightleftarrows A(g)+B(g)$ , AB is 33% dissociated at a total pressure of P, therefore is related to ${{K}_{P}}$ by one of the following options-
(A) $P={{K}_{P}}$
(B) $P=3{{K}_{P}}$
(C) $P=4{{K}_{P}}$
(D) $P=8{{K}_{P}}$

Answer 1

Hint: First find no. of moles at equilibrium to calculate partial pressures of gases AB, A and B. Then use formula \[{{K}_{p}}=\dfrac{{{P}_{A}}\times {{P}_{B}}}{{{P}_{AB}}}\] to calculate equilibrium constant at constant pressure.

Complete answer:
33% dissociation of AB means out of one mole of AB $\dfrac{1}{3}$​ mole is dissociated.

	$AB(g)$	$\rightleftarrows $	$A(g)$	$B(g)$
No. of moles initially	1		0	0
No. of moles at equilibrium	$1-\dfrac{1}{3}$		$\dfrac{1}{3}$	$\dfrac{1}{3}$

Total no. of moles $=(1-\dfrac{1}{3})+\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{4}{3}$
As we know that partial pressure of a gas in a mixture of gases is given by the formula-
     \[{{P}_{A}}={{X}_{A}}{{P}_{Total}}\]
Where ${{P}_{A}}=$ partial pressure of gas A
${{X}_{A}}=$ Mole fraction of gas A$=\dfrac{no\text{. of moles of gas A}}{\text{Total no}\text{. of moles of gases}}$
${{P}_{Total}}=$ total pressure of mixture.
partial pressure of gas AB$={{P}_{AB}}=\dfrac{2/3}{4/3}{{P}_{Total}}=\dfrac{1}{2}{{P}_{Total}}$
partial pressure of gas A$={{P}_{A}}=\dfrac{1/3}{4/3}{{P}_{Total}}=\dfrac{1}{4}{{P}_{Total}}$
partial pressure of gas B$={{P}_{B}}=\dfrac{1/3}{4/3}{{P}_{Total}}=\dfrac{1}{4}{{P}_{Total}}$
As we know that
     \[{{K}_{p}}=\dfrac{{{P}_{A}}\times {{P}_{B}}}{{{P}_{AB}}}\] (i)
Where ${{K}_{p}}=$ equilibrium constant at constant pressure.
Putting all the values in equation (i), we get
     \[\begin{align}
  & {{K}_{p}}=\dfrac{\dfrac{1}{4}{{P}_{Total}}\times \dfrac{1}{4}{{P}_{Total}}}{\dfrac{1}{2}{{P}_{Total}}}=\dfrac{1}{8}{{P}_{Total}} \\
 & {{P}_{Total}}=8{{K}_{p}} \\
\end{align}\]

Hence the correct option is (D) $P=8{{K}_{P}}$.

Note:
Equilibrium constants are defined as the ratio of concentrations at equilibrium for a reaction at a specific temperature. When we use the symbol ${{K}_{C}}$the subscript c means that all concentrations are being expressed in terms of molar concentration. When the reactants and products are gases, we can determine the equilibrium constant in terms of partial pressures where ${{K}_{p}}=$ equilibrium constant at constant pressure.