At a certain temperature the following equilibrium is established,…
$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g}) ightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{NO}(\mathrm{g})$
One mole of each of the four gases is mixed in one litre container and the reaction is allowed to reach equilibrium state. When excess of baryta water is added to the equilibrium mixture, the weight of white precipitate obtained is $236.4 \mathrm{~g} .$ The equilibrium constant, $\mathrm{K}_{\mathrm{C}}$ of the reaction is
- $1.2$
- $2.25$
- $2.1$
- $3.6$
Solution
Moles initially $1 \quad 1$
Moles at equib. $1-x 1-x \quad 1+x \quad 1+x$
$\mathrm{CO}_{2}+\mathrm{Ba}(\mathrm{OH})_{2} \longrightarrow \mathrm{BaCO}_{3}+\mathrm{H}_{2} \mathrm{O}$
Moles of $\mathrm{BaCO}_{3}=\frac{236.4}{197}=1.2$
Moles of $\mathrm{CO}_{2}$ at equilibrium $=1.2$ or,
$1+\mathrm{x}=1.2 ; \mathrm{x}=0.2 \quad \therefore \quad \mathrm{K}_{\mathrm{C}}=\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}ight)^{2}=\left(\frac{1.2}{0.8}ight)^{2}=2.25$
Asked in: JEE-TOPICTESTS-CHEMISTRY