Question

Given, \[X{Y_2}\]dissociates as:
\[X{Y_2}(g) \rightleftharpoons XY(g) + Y(g)\]
Initial pressure \[X{Y_2}\] is 600mm Hg. The total pressure at equilibrium is 800mm 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: In order to calculate the value of \[{K_p}\], we must first know what a \[{K_p}\]is. \[{K_p}\] is the equilibrium constant of the reaction which is expressed in terms of the partial pressure of the reactants and products.

Complete answer:
- \[X{Y_2}\]dissociates as XY and Y. The reaction is given as
\[X{Y_2}(g) \rightleftharpoons XY(g) + Y(g)\]
- Initial pressure of \[X{Y_2}\] is given as 600 mm Hg and the total pressure at equilibrium is given as 800 mm Hg. x mm Hg of \[X{Y_2}\] will dissociate in order to reach the equilibrium. x mm Hg of XY and Y each will be formed at the equilibrium.
Total pressure is found to be
600 – x + x + x = 600 + x mm Hg
In the question, the total pressure at equilibrium is given as 800 mm Hg.
Therefore,
600 + x = 800
x = 200 mm Hg
The partial pressure of \[X{Y_2} = {P_{X{Y_2}}}\]= 600 – x = 600 – 200 = 400 mm Hg
The partial pressure of XY = \[{P_{XY}}\]= x = 200 mm Hg
The partial pressure of Y = \[{P_Y}\]= x = 200 mm Hg
The value of \[{K_p}\]is found to be
\[{K_p} = \dfrac{{{P_{XY}} \times {P_Y}}}{{{P_{X{Y_2}}}}}\]
  \[{K_p} = \dfrac{{200 \times 200}}{{400}} = 100\]
\[{K_p}\]= 100 mm Hg
The value of \[{K_p}\]is found to be 100 mm Hg.

Therefore, the correct answer is option (B) 100.

Note:
We have to remember that certain factors can affect the \[{K_p}\] (equilibrium constant in terms of the partial pressure), they are:
- Change in concentration
- Addition of the catalyst
- Addition of the inert gas
- Change in the temperature
- Change in the pressure.