Question

For the dissociation of a gas $N_{2}O$ as:
$N_2{O}_5\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$
If '$D$' is the vapour density of equilibrium mixture and $P_0$ is the initial pressure of $N_2{O}_5\left(g\right)$, then its equilibrium pressure must be ($M$ is molecular mass of $N_2{O}_5$):

• A. $P_0\frac{(M-2D)}{3D}$
• B. $\frac{P_{0}M}{3D-M}$
• C. $\frac{P_{0}M}{2D}$
• D. $\frac{2P_{0}D}{M}$

Answer 1

The correct option is C $\frac{P_{0}M}{2D}$

$N_2{O}_5\rightleftharpoons 2N{O}_2+\frac{1}{2}{O}_2$

initially $P_0$ $0$ $0$

at eq $P_1-x$ $2x$ $\frac{1}{2}x$

$\therefore$ total equation pressure $=P_0+3\frac{3x}{2}$

We know that, $\frac{\left(VapourDensity\right)initially}{\left(VapourDensity\right)equation}=\frac{P_{equation}}{P_{initially}}=\frac{P_{equation}}{P_{initially}}$

Let $\alpha$ be the $V.D$ at equation ,also $VD=M/2\Rightarrow d=M/2$

$\therefore \frac{M}{2D}=\frac{P_0+3n/2}{P_0}$

$\frac{2}{3}P.(\frac{M}{2D}-1)=x$

$\therefore P_{eq}=P_0+\frac{3}{2}(\frac{2}{3}{P}_0\frac{M}{2D}-1)$

$=\frac{P.M}{2D}$