Question

Two solid compounds $A$ and $B$ dissociate into gaseous products at 20${}^o$C as;
$A\left(s\right)\rightleftharpoons A^{\prime }\left(g\right)+H_{2}S\left(g\right)$
$B\left(s\right)\rightleftharpoons B^{\prime }\left(g\right)+H_{2}S\left(g\right)$

At ${20}^{0}C$ pressure over excess solid $A$ is $50$ mm and that over excess solid $B$ is $68$ mm. Find:
a) relative number of moles of $A^{\prime }$ and $B^{\prime }$ in the vapour phase over a mixture of the solids $A$ and $B$.
b) what is the total pressure of gas over the solid mixture ?.

• A. $\frac{n_{A^{\prime }}}{n_{B^{\prime }}}=0.2506$, $P_T$ $=56.78$ mm
• B. $\frac{n_{A^{\prime }}}{n_{B^{\prime }}}=0.3520$, $P_T$ $=73.45$ mm
• C. $\frac{n_{A^{\prime }}}{n_{B^{\prime }}}=0.5406$, $P_T$ $=84.38$ mm
• D. $\frac{n_{A^{\prime }}}{n_{B^{\prime }}}=0.6250$, $P_T$ $=92.58$ mm

Answer 1

The correct option is C $\frac{n_{A^{\prime }}}{n_{B^{\prime }}}=0.5406$, $P_T$ $=84.38$ mm

(A) For dissociation of

$A\left(s\right)\rightleftharpoons A^{\prime }\left(g\right)+H_{2}S\left(g\right)$

$\because P_A^{\prime }+P_{H_{2}S\left(g\right)}^{\prime }=50$

$\therefore P_A^{\prime }=25mmand{P}_{H_{2}S\left(g\right)}=25mm[\because (Given1:1ratio\left)\right]$

$\therefore K_{P_1}=P_A^{\prime }\times P_{H_{2}S}^{\prime }=25\times 25=625({mm)}^2$

Similarly, for

$B\left(s\right)\rightleftharpoons B^{\prime }\left(g\right)+H_{2}S\left(g\right)$

$\therefore K_{P_2}=P_B^{\prime }\times P_{H_{2}S}^{\prime }=34\times 34=1156({mm)}^2$

(B) For mixture showing an equilibrium simultaneously,

$A\left(s\right)\rightleftharpoons A^{\prime }\left(g\right)+H_{2}S\left(g\right)$;

$x$ $x$

$B\left(s\right)\rightleftharpoons B^{\prime }\left(g\right)+H_{2}S\left(g\right)$

$y$ $y$

$\therefore K_{P_1}=x(x+y)and{K}_{P_2}=y(x+y)$

$\therefore \frac{K_{P_1}}{K_{P_2}}=\frac{x}{y}$

$\therefore \frac{x}{y}=\frac{625}{1156}$

$\because VandT$ are same and thus, ratio of pressure is ratio of moles i.e.,

$\frac{n_{A^{\prime }}}{n_{B^{\prime }}}=\frac{625}{1156}=0.5406$

(C) $\because \frac{x}{y}=\frac{625}{1156}$ and $x(x+y)=625$

$\therefore x=14.81mmandy=27.38mm$

$\therefore P_T=P_A^{\prime }+P_B^{\prime }+P_{H_{2}S}^{\prime }=x+y+(x+y)$

$P_T=2(x+y)=2(14.81+27.38)$

$P_T=84.38mm$