Question

The equilibrium constant for the reaction
$B{r}_2\left(l\right)+C{l}_2\left(g\right)\rightleftharpoons 2BrCl\left(g\right)$ at $27{}^{o}C$ is $K_p=1atm$. In a closed container of volume 164 L initially 10 moles of $C{l}_2$ are present at $27{}^{o}C$. What minimum moles of $B{r}_2\left(l\right)$ must be introduced into this container so that the above equilibrium is maintained at a total pressure of 2.25 atm. Vapour pressure of $B{r}_2\left(l\right)$ at $27{}^{o}C$ is 0.25 atm. Assume that volume occupied by liquid is neglegible.
(Atomic mass of $B{r}_2\left(l\right)=80$.

Answer 1

To maintain the mentioned equilibrium there must be some mass of $B{r}_2\left(l\right)$ at equilibrium.
$B{r}_2\left(l\right)\rightleftharpoons B{r}_2\left(g\right);K_p=0.25atm$
So some $B{r}_2\left(l\right)$ will be required for this conversion. Moles of $B{r}_2\left(l\right)$ required for above equilibrium,
$n=\frac{PV}{RT}$
$n=\frac{0.25\times 164L}{0.082Latmmo{l}^{-1}{K}^{-1}\times 300K}=\frac{5}{3}$

Now for equilibrium
$B{r}_2\left(l\right)+C{l}_2\left(g\right)\rightleftharpoons 2BrCl\left(g\right)$
t = 0 x 10
t =$t_{eq}$ 0 10 - x 2x
(where x be the moles of $B{r}_2\left(l\right)$ required just to maintain above equilibrium)
Total gaseous moles = ( 10 + x)
$P_{C{l}_2}+P_{BrCl}=2atm$
so
$P_{C{l}_2}=\frac{10-x}{10+x}\times 2atm$

$P_{BrCl}=\frac{2x}{10+x}\times 2atm$

$K_p=\frac{P_{BrCl}^2}{P_{C{l}_2}}$

$K_p=\frac{4x^2\times (10+x)}{(10+x{)}^2(10-x)}\times 2$

on solving
$x=\frac{10}{3}$
Hence total moles of $B{r}_2\left(l\right)$ required to maintain both of above equlibiria = $\frac{5}{3}+\frac{10}{3}=5mol$