Question

Two solid compounds A and B dissociate into gaseous products at $20{}^\circ C$ as:
$\begin{align}
  & A(s)\rightleftharpoons {{A}^{'}}(g)+{{H}_{2}}S(g) \\
 & B(s)\rightleftharpoons {{B}^{'}}(g)+{{H}_{2}}S(g) \\
\end{align}$
At $20{}^\circ C$ pressure over excess solid A is 50mm and that over excess solid B is 68mm. Find:
a)relative number of moles of ${{A}^{'}}$ and $ {{B}^{'}}$ in the vapor phase over a mixture of the solids A and B.
b) what is the total pressure of gas over the solid mixture?
A. $\dfrac{{{n}_{{{A}^{'}}}}}{{{n}_{{{B}^{'}}}}}=0.2506,{{P}_{T}}=56.78mm$
B. $\dfrac{{{n}_{{{A}^{'}}}}}{{{n}_{{{B}^{'}}}}}=0.3520,{{P}_{T}}=73.45mm$
C. $\dfrac{{{n}_{{{A}^{'}}}}}{{{n}_{{{B}^{'}}}}}=0.5406,{{P}_{T}}=84.38mm$
D. $\dfrac{{{n}_{{{A}^{'}}}}}{{{n}_{{{B}^{'}}}}}=0.6250,{{P}_{T}}=92.58mm$

Answer 1

Hint: The total pressure of any mixture is the sum of partial pressure of its components. While the partial pressure of any substance is directly proportional to mole fraction of the solute which is the product of vapor pressure and mole fraction. ${{K}_{P}}$ is the product of the partial pressures of the components that is taken only for gases.

Complete answer:
a)As we know, the total pressure = ${{P}^{o}}_{A}+{{P}^{o}}_{B}$, so here we have dissociation of A as $A(s)\rightleftharpoons {{A}^{'}}(g)+{{H}_{2}}S(g)$
where the given pressure for A is 50mm that means${{P}^{o}}_{A}+{{P}^{o}}_{{{H}_{2}}S}=50mm$, so ${{P}^{o}}_{A}={{P}^{o}}_{{{H}_{2}}S}$ = 25 mm that is 1:1 ratio.
Now, ${{K}_{{{P}_{1}}}}={{P}^{o}}_{A}\times {{P}^{o}}_{{{H}_{2}}S}=25\times 25=625m{{m}^{2}}$
Similarly for the dissociation of B as, $B(s)\rightleftharpoons {{B}^{'}}(g)+{{H}_{2}}S(g)$
where the given pressure for A is 68mm that means${{P}^{o}}_{B}+{{P}^{o}}_{{{H}_{2}}S}=68mm$, so ${{P}^{o}}_{\mathbf{B}}={{P}^{o}}_{{{H}_{2}}S}$ = 34 mm that is 1:1 ratio.
Now, ${{K}_{{{P}_{2}}}}={{P}^{o}}_{B}\times {{P}^{o}}_{{{H}_{2}}S}=34\times 34=1156m{{m}^{2}}$
As both the reactions in equilibrium are happening at the same time, therefore,
$\begin{align}
  & A(s)\rightleftharpoons {{A}^{'}}(g)+{{H}_{2}}S(g) \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\
 & B(s)\rightleftharpoons {{B}^{'}}(g)+{{H}_{2}}S(g) \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,y \\
\end{align}$,
so the ${{K}_{{{P}_{1}}}}=x(x+y)$and${{K}_{{{P}_{2}}}}=y(x+y)$
Therefore, $\dfrac{{{K}_{{{p}_{1}}}}}{{{K}_{p}}_{2}}=\dfrac{x}{y}=\dfrac{625}{1156}$
The ratio of moles becomes similar to the ratio of pressures as the volume and temperature are constant.
So, $\dfrac{{{n}_{{{A}^{'}}}}}{{{n}_{{{B}^{'}}}}}=\dfrac{625}{1156}$ = 0.5406
So, the relative number of moles of ${{A}^{'}}$ and ${{B}^{'}}$ in the vapor phase over a mixture of the solids A and B is 0.506.
b) As from the above calculations we have $\dfrac{{{K}_{{{p}_{1}}}}}{{{K}_{p}}_{2}}=\dfrac{x}{y}=\dfrac{625}{1156}$, this implies that
x (x+y) = 625, therefore, x = 14.81 mm and y = 27.38 mm
Now, as we know, total pressure will be partial pressure of all components we have,
${{P}_{T}}={{P}^{o}}_{A}+{{P}^{o}}_{B}+{{P}^{o}}_{{{H}_{2}}S}$ = x + y + (x +y)
${{P}_{T}}$ = 2 (x+y)
${{P}_{T}}$ = 2 (14.81 + 27.38)
${{P}_{T}}$ = 84.38 mm
So, the total pressure of gas over the solid mixture is 84.38 mm

So, option C is correct.

Note:
The relative number of moles will be the number of moles of one component to the other. The number of moles are taken only in the vapor or gaseous phase. According to Henry’s law the partial pressure is directly proportional to the mole fraction of any component.