Question

When $NO$ and $N{O}_2$ are mixed, the following equilibrium are readily obtained, $2N{O}_2\rightleftharpoons N_2{O}_4,K_p=6.8{atm}^{-1}$ 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 (a) the equilibrium partial pressure of $NO$, (b) $K_p$ for: $NO+N{O}_2\rightleftharpoons N_2{O}_3$.

• A. $3.43{atm}^{-1}$
• B. $5.63{atm}^{-1}$
• C. $7.43{atm}^{-1}$
• D. None of these

Answer 1

Answer: (A)

$3.43{atm}^{-1}$

The equilibrium I and II are obtained by mixing $NO$ and $N{O}_2$ in the ratio $1:2$. Let pressure of $NO$ and $N{O}_2$ be $P$ and $2P$ respectively before reaction.

I equilibrium : $2N{O}_2\rightleftharpoons N_2{O}_4$
$K_p=\frac{P_{N_2{O}_4}^{\prime }}{{\left(P_{N{O}_2}^{\prime }\right)}^2}=6.8$ ...(i)
$\because P_{N_2{O}_4}^{\prime }=1.7atm$ (at equilibrium)
$\therefore$ By equation (i) $P_{N{O}_2}^{\prime }=0.5atm$ (at equilibrium) ...(ii)
Also, $P_{N{O}_2}^{\prime }$ used for attaining I equilibrium $=2\times P_{N_2{O}_4}^{\prime }$ formed $=2\times 1.7=3.4atm$ ...(iii)

II equilibrium : $NO+N{O}_2\rightleftharpoons N_2{O}_3$
Initial pressure $P2P0$
Pressure eqm $(P-x)(2P-x-3.4)x$

$\because 3.4atmN{O}_2$ are used for I equilibrium; also since both the equilibrium are attained simultaneously and thus $P_{N{O}_2}^{\prime }$ left at equilibrium in both should be same

i.e., $2P-x-3.4=0.5$ ...(iv)

The total pressure at equilibrium $=P_{NO}^{\prime }+P_{N{O}_2}^{\prime }+P_{N_2{O}_3}^{\prime }+P_{N_2{O}_4}^{\prime }$

or $5.05=(P-x)+(2P-x-3.4)+x+1.7$

$5.05=P-x+0.5+x+1.7$

$5.05=P+2.20$

$\therefore P=2.85atm$ ...(v)

By equation (iv) $2P-x-3.4=0.5$
$2\times 2.85-x-3.4=0.5$
$\therefore x=1.80atm$

Now, $K_p$ for II equilibrium

$K_p=\frac{P_{N_2{O}_3}^{\prime }}{P_{NO}^{\prime }\times P_{N{O}_2}^{\prime }}$

$=\frac{x}{(P-x)(2P-x-3.4)}$

$=\frac{1.8}{1.05\times 0.5}$

$=3.43{atm}^{-1}$