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 ${27}^{\circ }C$ is $k_p=1atm$. In a closed container of volume $164L$ initially $10$ moles of $C{l}_2$ are present at ${27}^{\circ }C$.
What minimum mole of $B{r}_2\left(l\right)$ must be introduced into this container so that above equilibrium is maintained at total pressure of $2.25atm$. Vapour pressure of $B{r}_2\left(l\right)$ at ${27}^{\circ }C$ is $0.25atm$. Assume that volume occupied by liquid is negligible. $[R=0.082Latmmol{e}^{-1}{K}^{-1}$, Atomic mass of bromine $=80$].

Answer 1

${Br}_2\left(l\right)+{Cl}_2\left(g\right)\rightleftharpoons 2BrCl\left(g\right)$ at ${27}^{0}C$ $K_P=1$ Assuming sufficient quantity is there

At $t=0$ $C$ $10moles$ $0moles$ initially to maintain saturated vapour

$\downarrow$ pressure equilibrium

pressures $P^0{Br}_2$ $1.5atm$

at $t=0$

$\left(P\right)\left(164\right)=10\times 0.082\times 300$

$P=1.5atm$

At $\Rsh$ (doesn't change because of gas liquid physical equilibrium)

Eq'b $P_0{Br}_0$ $1.5(1-\alpha )$ $2\left(1.5\right)\left(\alpha \right)$

Total pressure $\Rightarrow \underset{}{{P_0}^{{Br}_2}}+1.5(1+\alpha )=2.25$

$0.25+(1+\alpha )1.5=2.25$

$1+\alpha =\frac{2}{1.5}=4/3\Rightarrow \alpha =1/3$

So $10/3$ moles of ${Cl}_2$ used to form ${BrCl}_2$

$\hookrightarrow$ So $10/3$ moles ${Br}_2$ also used.

Next $=$ Moles required to maintain $0.25$ atm saturated vapour pressure :

$\left(0.25\right)\left(164\right)=n\times \left(0.082\right)\times \left(300\right)$

$n=5/3moles/:$

So minimum initial requirement $\Rightarrow 10/3+5/3=5moles$