Question

When $NO$ and $N{O}_2$ are mixed, the following equilibria are readily obtained :
$2N{O}_2\rightleftharpoons N_2{O}_4{K}_p=6.8at{m}^{-1}$
and
$NO+N{O}_2\rightleftharpoons N_2{O}_3$
In an experiment when $NO$ and $N{O}_2$ are mixed in the ratio of $1:2$, the final total pressure was $5.05atm$ and the partial pressure of $N_2{O}_4$ was $1.7atm$. Calculate the equilibrium partial pressure of $NO$ in atmosphere :

• A. $1.05atm$
• B. $4atm$
• C. $6atm$
• D. $8atm$

Answer 1

The correct option is A $1.05atm$

For $I$ equilibrium,
$2N{O}_2\rightleftharpoons N_2{O}_4$
$K_p=\frac{P_{N_2{O}_4}}{P_{\left(N{O}_2\right)}^2}=\mathrm{6.8...}\left(i\right)$
$P_{N_2{O}_4}=1.7atm$
$\therefore$
By equation (i),
$P_{\left(N{O}_2\right)}=0.5atm$
The equilibria are maintained using $NO$ and $N{O}_2$ in the ratio $1:2$

$\text{For II equilibrium},NO+N{O}_2\rightleftharpoons N_2{O}_3\text{Initial pressure}P2P0\text{pressure at eqm}(p-x)(2p-x-3.4)x$
$3.4atm$ of $N{O}_2$ are used for $I$ equilibrium to have $P_{N_2{O}_4}=1.7atm$
$\text{at equilibrium}(p-x)0.5x$
$P_{N{O}_2}\text{is same for both the equlibria since both reactions are at equilibrium at a time}$
Total pressure at equilibrium ,(Given 5.05 atm)
$=P_{NO}+P_{N{O}_2}+P_{N_2{O}_3}+P_{N_2{O}_4}$
$=P-x+0.5+x+1.7$
$5.05=P+2.20$
$P=2.85atm$
$2P-x-3.4=0.5$
$2\times 2.85-x-3.4=0.5$
$x=1.80atm$
Hence, $P-x=2.85-1.80=1.05atm$