Question

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

• A. 3 atm
• B. 2 atm
• C. 4 atm
• D. 1 atm

Answer 1

The correct option is D 1 atm

For I equilibrium $2N{O}_2\rightleftharpoons N_2{O}_4$
$K_P=\frac{P_{N_2{O}_4}}{(P_{N{O}_2}{)}^2}=6.8$ ....(i)
$P_{N_2{O}_4}=1.7atm\therefore \text{By eq.}\left(i\right)P_{N{O}_2}^{\prime }=0.5atm$
The equilibria are maintained using $NO$ and $N{O}_2$ in the ratio 1 : 2
$\begin{array}{cccccc}\text{For II equilibrium}& NO& +& N{O}_2\rightleftharpoons N_2{O}_3& & \\ \text{Initial pressure}& P& & 2P& & 0\\ \text{Pressure at equilibrium}& (p-x)& & (2P-x-3.4)& & x\end{array}$
3.4 atm of $N{O}_2$ are used for I equilibrium to have $P_{N_2{O}_4}=1.7atm$
At equilibrium (P - x) 0.5 x

($\because P_{N{O}_2}$ is same for both the equilibria since both reactions are at equilibrium at a time)
Total pressure at equilibrium (Given 5.05 atm)
$=P_{NO}+P_{N{O}_2}+P_{N_2{O}_3}+P_{N_2{O}_4}$
$=P-x+0.5+x+1.7$
$5.05=P+2.20$
$5.05=P+2.20$
$P=2.85atm$
$2\times 2.85-x-3.4=0.5$
$x=1.80atm$
$P_{NO}=P-x$
$P_{NO}=2.85-1.80=1.05atm\approx 1atm$