Question

At \[87^\circ C\], the following equilibrium is established \[{H_2}\left( g \right) + S\left( s \right) \rightleftharpoons {H_2}S\left( g \right)\;\;\;\]\[\;{K_p} = 7 \times {10^{ - 2}}\]
If 0.50 mole of hydrogen and 1.0 mole of sulphur are heated to \[87^\circ C\] in 1.0 L vessels, what will be the partial pressure of \[{H_2}S\] at equilibrium?
A) 0.966 atm
B) 1.38 atm
C) 0.0327 atm
D) 9.66 atm

Answer 1

Hint: Raoult’s law states that the partial vapour pressure of a component of an ideal mixture is the vapour pressure of the solvent multiplied by its mole fraction of solvent. In an ideal mixture it is presumed that there are equal intermolecular interactions between all molecules inside the mixture. It can be expressed as follows:
$p = {p^0}x$
Where,
${p^0} = $ vapour pressure of solvent
$x = $ mole fraction of the solvent
In this question to calculate the partial pressure of the compound hydrogen sulphide, we can use the equation of Raoult’s law and partial pressure and
Relation of partial pressure can be expressed as follows:
$PV = nRT$, here n is the number of moles, R is gas constant, T is temperature, P is pressure and V is volume.

Complete step by step answer:
Let us write down the equilibrium reaction given in question:
${H_2}(g) + S(s) \rightleftharpoons {H_2}S(g)$
It is given in question that initially 0.50 moles of hydrogen is present and 1.0 mole of sulphur.
Let us assume that x mole of this hydrogen got utilized during the reaction and produced x mole of sulphide.
Therefore, at equilibrium, $0.50 - x$ moles of hydrogen and x moles of hydrogen sulphide are present.
The expression for the equilibrium constant is ${K_p} = \dfrac{{{P_{{H_2}S}}}}{{{P_{{H_2}}}}}$…… (i)
Where, $P = $ Pressure
From Raoult’s law we know that,
${P_{{H_2}S}} = {X_{{H_2}S}} \times P$
${P_{{H_2}}} = {X_{{H_2}}} \times P$
Where,
$X = $ Mole fraction
${X_{{H_2}S}} = \dfrac{x}{{0.5}}$
\[{X_{{H_2}}} = \dfrac{{0.5 - x}}{{0.5}}\]
Therefore,
${P_{{H_2}S}} = \dfrac{x}{{0.5}} \times P$
${P_{{H_2}}} = \dfrac{{0.5 - x}}{{0.5}} \times P$
Substituting the above values in equation (i)
$7 \times {10^{ - 2}} = \dfrac{x}{{0.5 - x}}$
$ \Rightarrow x = 0.0327$
Therefore, the number of moles of hydrogen sulphide, i.e. $x = 0.0327$.
Now the total pressure can be calculated from Ideal gas equation,
$PV = nRT(R = 0.0821)$
Substituting the above calculated values in this we get,
$P = \dfrac{{nRT}}{V} = \dfrac{{0.5 \times 0.0821 \times 360}}{1} = 14.77atm$
Therefore, Partial pressure of ${H_2}S$,
${P_{{H_2}S}} = {X_{{H_2}S}} \times P$
\[ \Rightarrow {P_{{H_2}S}} = \dfrac{x}{{0.5}} \times P\]
\[ \Rightarrow {P_{{H_2}S}} = \dfrac{{0.0327}}{{0.5}} \times 14.77\]
\[ \Rightarrow {P_{{H_2}S}} = 0.966atm\]
Partial pressure of \[{H_2}S = 0.966atm\].

Therefore, option A is correct.

Note: Vapour pressure is defined as measure of the ability of a substance or material to change its state into the vapour or the gaseous state, and it is directly dependent on the temperature, with increases in temperature it increases.