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

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.