In the reaction \[{{N}_{2}}{{O}_{4}}(g)\,\,\,\,2N{{O}_{2}}(g)\]. If D & d are the vapour densities at initial stage & at equilibrium then what will be the value of \[\frac{D}{d}\] at point A in the
A)0
B)1.5
C)1
D)0.5
Correct Answer:
C
Solution :
\[\alpha =\frac{D-d}{(n-1)d}\]
For given reaction n=2 & \[\alpha =0\] point A so \[\frac{D}{d}=1\]