Question

For the reaction $N_2{O}_4\left(g\right)\leftrightharpoons 2N{O}_2\left(g\right),$ if percentage dissociation of $N_2{O}_4$ are 20% 45%, 65 & 80% , then the sequence of vapor densities,

• A. $d_{20}>d_{45}>d_{60}>d_{80}$
• B. $d_{80}>d_{65}>d_{45}>d_{20}$
• C. $d_{20}=d_{45}=d_{60}=d_{80}$
• D. $(d_{2}0=d_{45})>(d_{65}=d_{80})$

Answer 1

The correct option is A $d_{20}>d_{45}>d_{60}>d_{80}$

$(VD{)}_{mix}=\frac{M_{mix}}{2}=\frac{M}{2(1+\alpha )}$
So, as $\alpha$ (percentage dissociation) increases vapour density $(VD{)}_{mix}$ decreases.
Hence the sequence of observed vapour densities will be:
$d_{20}>d_{45}>d_{60}>d_{80}$