Question

At a certain temperature, the following equilibrium is established:

$C{O}_{\left(g\right)}+NO{{}_2}_{\left(g\right)}\rightleftharpoons CO{}_2\left(g\right)+N{O}_{\left(g\right)}$

One mole of each of the four gases is mixed in a one litre container and the reaction is allowed to reach equilibrium. When excess of baryta water is added to the equilibrium mixture, the weight of white precipitate obtained is 236.4 g. The equilibrium constant $K_c$ of the reaction is: (molar mass of $Ba$ is 137 g/mol):

• A. 1.2
• B. 2.25
• C. 3.6
• D. 4.8

Answer 1

The correct option is B 2.25

$CO\left(g\right)+N{O}_2\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+NO\left(g\right)$
initially,
1 mol 1 mol 1 mol 1 mol
at equillibrium,
1-x 1-x 1+x 1+x

To find x we can use the following:

$Ba(OH{)}_2+C{O}_2\to BaC{O}_3+H_{2}O$
white ppt here is $BaC{O}_3$ which is 236.4 gm = 1.2 mol
which means $C{O}_2$ available will be 1.2 mols
$\Rightarrow$ x= 0.2
So,
$K_c=\frac{{1.2}^2}{{0.8}^2}=2.25$