Question

\[\,\,\,\,\,\,X{{Y}_{2}}\]​dissociates as: \[\,X{{Y}_{2}}(g)\rightleftharpoons XY(g)+Y(g)\]
Initial pressure of \[\,\,\,\,\,\,X{{Y}_{2}}\]​ is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming volume of system to remain constant, the value of \[{{K}_{p}}\] ​is:
(A) 50
(B) 100
(C) 200
(D) 400

Answer 1

Hint: To find out the correct answer, we should note that Kp is the equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures.
Step by step answer:
First of all we should know that, \[{{K}_{p}}\] is the equilibrium constant calculated from the partial pressures of a reaction equation. We use it to find the relationship between product pressure and reactant pressure.
We should know about partial pressure. Dalton’s law states that all the partial pressures sum up to give total pressure.
In this question we have given a reaction in a gaseous phase.
First of all we will write the reaction, provided in question.
\[\begin{align}
  & ~\text{ }~\,\,\,\,\,\,\,\,\,\,\,\,\,\,X{{Y}_{2}}(g)\rightleftharpoons XY(g)+Y(g) \\
 & At, \\
 & t=0~\text{ }\to ~\,{{P}_{0}}~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\,\,0~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~~0 \\
 & t={{t}_{eq}}\to {{P}_{0}}~(1-\alpha )\,\,\,\,\,\,{{P}_{0}}\alpha \,\,\,\,\,\,\,\,\,\,{{P}_{0}}\alpha \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \, \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,400mm\,\,\,\,\,\,200mm\,\,\,\,\,200mm\,\,\,\,\,\,\,\,\,\, \\
\end{align}\]
It is given that \[~\,{{P}_{0}}~\text{ }\]= 600 mm
And from the question, we know that the total pressure of the system is 800mm. Or we can write it as follows:
\[{{P}_{0}}~(1+\alpha )\,\,\,\]= 800mm
After this by putting value of\[~\,{{P}_{0}}~\text{ }\], equation will be as:
\[\begin{align}
  & \,1+\alpha =\dfrac{800}{600}\,\,\,\,\,\,\,\, \\
 & \,\,1+\alpha =\dfrac{4}{3} \\
 & \alpha =\dfrac{1}{3} \\
 & {{K}_{p}}=\dfrac{200\times 200}{400}=100 \\
\end{align}\]

So, from this we can say that the value of equilibrium constant is 100. And the correct answer is option B.

Note: We should know that there are other equilibrium constants also. We should know that \[{{K}_{c}}\]\[{{K}_{c}}\text{ }and\text{ }{{K}_{p}}\] are the equilibrium constants of gaseous mixtures. However, the difference between the two constants is that \[{{K}_{c}}\] is defined by molar concentrations, whereas ${{K}_{p}}$ is defined by the partial pressures of the gasses inside a closed system.