Question

Two solid compounds $A$ and $B$ dissociate into gaseous products at ${20}^{o}C$ as :
(i) $A\left(s\right)\rightleftharpoons A^{\prime }\left(g\right)+H_{2}S\left(g\right)$
(ii) $B\left(s\right)\rightleftharpoons B^{\prime }\left(g\right)+H_{2}S\left(g\right)$
At ${20}^{o}C$ pressure over excess solid $A$ is $50mm$ and that over excess solid $B$ is $68mm$. Find the total pressure of gas over the solid mixture would be (in cm to the nearest integer)__________.

Answer 1

(i) For dissociation of $A\left(s\right)\rightleftharpoons A^{\prime }\left(g\right)+H_{2}S\left(g\right)$
$\because P_A^{\prime }+P_{H_{2}S}^{\prime }=50$
$\therefore P_A^{\prime }=25mm$ and $P_{H_{2}S}^{\prime }=25mm$ ($\because 1:1$ ratio)
$\therefore K_{p_1}=P_A^{\prime }\times P_{H_{2}S}^{\prime }=25\times 25=625{\left(mm\right)}^2$
Similarly for $B\left(s\right)\rightleftharpoons B^{\prime }\left(g\right)+H_{2}S\left(g\right)$
$K_{p_2}=P_B^{\prime }\times P_{H_{2}S}^{\prime }=34\times 34=1156{\left(mm\right)}^2$
(ii) For mixture showing an equilibrium simultaneously
$A\left(s\right)\rightleftharpoons A^{\prime }\left(g\right)+H_{2}S\left(g\right)$
$P$ at equilibrium $xx+y$
$B^{\prime }\left(s\right)\rightleftharpoons B^{\prime }\left(g\right)+H_{2}S\left(g\right)$
$P$ at equilibrium $yx+y$
$\therefore K_{p_1}=x(x+y)$ and $K_{p_2}=y(x+y)$
$\frac{K_{p_1}}{K_{p_2}}=\frac{x}{y}=\frac{625}{1156}$
$\because V$ and $T$ are same and thus, ratio of pressure is ratio of mole, i.e.,
$\frac{n_A^{\prime }}{n_B^{\prime }}=\frac{625}{1156}=0.5406$
(iii) $\frac{x}{y}=\frac{625}{1156}$ and $x(x+y)=625$
$\therefore x=14.81mm$ and $y=27.38mm$
$\therefore P_T=P_{A_1}^{\prime }+P_{B_1}^{\prime }+P_{H_{2}S}^{\prime }=x+y+(x+y)$
$=2(x+y)=2(14.81+27.38)=84.38mm\approx 8cm$