Question

AB dissociates as $2A{{B}_{\left( g \right)}}\rightleftarrows 2{{A}_{\left( g \right)}}+{{B}_{2}}_{\left( g \right)}$
When the initial pressure of AB is 500 mm, the total pressure becomes 625 mm when the equilibrium is attained. Calculate ${{K}_{p}}$ for the reaction assuming volume remains constant.
(A) 500
(B) 125
(C) 750
(D) 375

Answer 1

Hint: ${{K}_{p}}$ is known as the equilibrium constant calculated from the partial pressures of a reaction equation. It is only necessary to know the partial pressures of reactants and products to solve given illustrations. Also, only one option can be correct out of all those given.

Complete answer:
Let us know about the equilibrium constant i.e. ${{K}_{p}}$ before moving towards the illustration;
${{K}_{p}}$ is the equilibrium constant used to express the relationship between product and reactant pressure. It is a unitless number even though it is related to the pressures.
As, ${{K}_{p}}$ is the ratio of partial pressure of products to the partial pressure of reactants which are raised to some powers which is exactly equal to the stoichiometric coefficients of them in a balanced equation.
For a given chemical reaction;
\[X+2Y\to {{Z}_{3}}\]
It is expressed as,
${{K}_{p}}=\dfrac{\left[ {{P}_{{{Z}_{3}}}} \right]}{\left[ {{P}_{X}} \right]{{\left[ {{P}_{Y}} \right]}^{2}}}$
Thus, for a given illustration we can solve for ${{K}_{p}}$ as;
Given data,
The initial pressure of AB = 500 mm
The total pressure of reaction when equilibrium is attained = 625 mm
The reaction can be stated as;
$2A{{B}_{\left( g \right)}}\rightleftarrows 2{{A}_{\left( g \right)}}+{{B}_{2}}_{\left( g \right)}$
Initial- 500 0 0
Final- 500-2P 2P P
At final conditions when equilibrium is attained;
Total pressure = 625 = 500 - 2P + 2P + P
Thus, P = 125 mm
${{P}_{AB}}=500-\left( 2\times 125 \right)=250mm$
${{P}_{A}}=2\times 125=250mm$
${{P}_{{{B}_{2}}}}=125mm$
Thus, ${{K}_{p}}$ is given as,
$\begin{align}
& {{K}_{p}}=\dfrac{\left[ {{P}_{{{B}_{2}}}} \right]{{\left[ {{P}_{A}} \right]}^{2}}}{{{\left[ {{P}_{AB}} \right]}^{2}}} \\
& {{K}_{p}}=\dfrac{125\times {{\left( 250 \right)}^{2}}}{{{\left( 250 \right)}^{2}}}=125mm \\
\end{align}$
Thus, the value of equilibrium constant i.e. ${{K}_{p}}$ is 125 mm.

Therefore, option (B) is correct.

Note: Do note that ${{K}_{p}}$ is a unitless quantity, but the units of pressures involved in the illustration must be consistent before we actually solve the problem.