Question

Two equilibria are simultaneously exist in a vessel at $25{}^{\circ }C$
$NO\left(g\right)+N{O}_2\left(g\right)\stackrel{K_{p1}}{\rightleftharpoons }{N}_2{O}_3\left(g\right)$
$2N{O}_2\left(g\right)\stackrel{K_{p2}=8atm}{\rightleftharpoons }{N}_2{O}_4\left(g\right)$
If initially only NO and $N{O}_2$ are present in a 3:5 mole ratio and the total pressure at equilibrium is 5.5 atm with the pressure of $N{O}_2$ at 0.5 atm, calculate $K_{p1}$.

• A. $0.5$
• B. $4$
• C. $0.4$
• D. $2.5$

Answer 1

The correct option is C $0.4$

If initially only $NO$ and $N{O}_2$ are present in a 3:5 mole ratio then initial pressure ratio will also be 3:5, so let initial pressure be 3p and 5p.
So,
$NO\left(g\right)+N{O}_2\left(g\right)\stackrel{K_{p1}}{\rightleftharpoons }{N}_2{O}_3\left(g\right)$
$\begin{array}{cccc}3p& 5p& 0& \text{initial pressure}\\ 3p-x& 5p-x-2y& x& \text{at equilibrium}\end{array}$
And
$2N{O}_2\left(g\right)\stackrel{K_{p2}=8atm}{\rightleftharpoons }{N}_2{O}_4$
$5p-x-2yy$
As pressure of $N{O}_2=0.5$ atm as given so 5p - x - 2y = 0.5
So, from 2nd equilibrium $K_{p2}=8=\frac{p_{N_2{O}_4}}{(p_{N{O}_2}{)}^2}$
$\Rightarrow \frac{p_{N_2{O}_4}}{(0.5{)}^2}=8\Rightarrow p_{N_2{O}_4}=2atm=y$
As total pressure is given = 5.5 atm, so
$(3p-x)+(5p-x-2y)+x+y=5.5$
$\Rightarrow (3p-x)+0.5+x+2=5.5$
$\Rightarrow p=1atm$
Now from, $5p-x-2y=0.5,\Rightarrow x=0.5atm$
So, $K_{p1}=\frac{p_{N_2{O}_3}}{p_{NO}\times p_{N{O}_2}}$
$=\frac{x}{(3p-x)\times 0.5}$
$=\frac{0.5}{(3-0.5)\times 0.5}=\frac{1}{2.5}=0.4$