Question

$X{Y}_{2}$ dissociates as:
$X{ Y }_{ 2 }(g)\quad \rightleftharpoons \quad XY(g)\quad +\quad Y(g)$
When the initial pressure of $X{Y}_{2}$ is 600mm Hg, the total equilibrium pressure is 800mm Hg. Calculate K for the reaction assuming that the volume of the system remains unchanged.
a.) 50
b.) 100
c.) 166.6
d.) 400.0

Answer 1

Hint: ${K}_{p}$ or K is the equilibrium constant calculated in terms of partial pressures of the reactants and the products in a reaction. It is defined for the gases in the reaction equation.

Complete step by step answer:
It is given that the initial pressure of $X{Y}_{2}$ is 600mm Hg but the total pressure of the system is 800mm Hg. We need to find out the equilibrium constant for the following reaction.

$X{ Y }_{ 2 }(g)\quad \rightleftharpoons \quad XY(g)\quad +\quad Y(g)$

Let 'z' mm Hg of $X{Y}_{2}$ dissociate to reach equilibrium and form z mm Hg of $XY$ and z mm Hg of $Y$ at equilibrium. Now, at equilibrium (600 - z) mm Hg of $X{Y}_{2}$ will be present.

	$X{Y}_{2}$	XY	Y
Pressure at t=0	600	0	0
Pressure at $t={t}_{eq}$	600-z	z	z

Therefore, total pressure = (600-z) + z + z = (600 + z)mm Hg
But it is given that total pressure is 800mm Hg
$\implies 600 +z = 800$
or, z = 200mm Hg

Thus, the partial pressure at equilibrium are:
${P}_{X{Y}_{2}}$ = 600 - z = 600 - 200 = 400mm Hg
${P}_{XY}$ = z = 200mm Hg
${P}_{Y}$ = z =200mm Hg

Now, the equilibrium constant at constant volume is given as:
$K\quad =\quad \cfrac { { P }_{ XY }{ P }_{ Y } }{ { P }_{ X{ Y }_{ 2 } } }$
Substituting the above found values in the equation, we get,
$K\quad =\quad \cfrac { 200\quad \times \quad 200 }{ 400 }$
$\implies K\quad =\quad 100$

Therefore, the value of K is 100. So, the correct answer is “Option B”.

Note: Students tend to write brackets around the individual partial pressure terms. This should not be done as even the round parenthesis around the partial pressures can be confused with the square brackets. This may confuse students between ${K}_{p}$ and ${K}_{c}$.