Q.
$18.4 \,g \,N _{2} O _{4}$ was placed in $1 \,L$ vessel at $400\, K$
and allowed to attain the following equilibrium $N _{2} O _{4}(g) \rightleftharpoons 2 NO _{2}(g)$
If the total pressure at equilibrium was $10.64$ bar, approximate $K_{p}$ is $(R=0.083 \,L$ bar $K ^{-1} \,mol ^{-1} ) $ (Assume $N _{2} O _{4}, NO _{2}$ as ideal gases)
AP EAMCETAP EAMCET 2019
Solution:
From ideal gas equation,
$p V =n R T$
$p_{ N _{2} O _{4}} \times 1 =\frac{18.4}{92} \times 0.083 \times 400 $
$p_{ N _{2} O _{4}} =6.64$ bar
For the equilibrium reaction,
$N_2O_4(g) $
$\ce{<=>}$
$2NO_2(g)$
Initial pressure
P
0
At equilibrium
$P-p_{i}$
$2p_{i}$
$p_{T} =p-p_{i}+2 p_{i} $
$p_{T} =p+p_{i} $
$\Rightarrow \,p_{i}=p_{T}-p $
$p_{i} =10.64-6.64 $
$=4.00 $
At equilibrium,
$\therefore \,p_{ N _{2} O _{4}} =p-p_{i}=6.64-4 $
$=2.64$ bar
$p_{ NO _{2}} =2 p_{i}=2 \times 4$
$=8 $ bar
$\therefore \, K_{p} =\frac{\left[p_{ NO _{2}}\right]^{2}}{\left[p_{ N _{2} O _{4}}\right]} $
$=\frac{[8]^{2}}{2.64}=24.29$
| $N_2O_4(g) $ | $\ce{<=>}$ | $2NO_2(g)$ | |
|---|---|---|---|
| Initial pressure | P | 0 | |
| At equilibrium | $P-p_{i}$ | $2p_{i}$ |