Q.Describe the graphical representation of first order reaction.

Using $t=\dfrac{2.303}{k}\log\dfrac{a}{a-x}$, 75% completion needs $\log 4$ and 50% completion needs $\log 2$; since $\log 4=2\log 2$, the 75%-completion time is exactly twice the 50%-completion time, giving $t_{50\%}=30$ min.

For a first-order reaction, $t=\dfrac{2.303}{k}\log\dfrac{a}{a-x}$. At 75% completion, $x=0.75a$, so $a-x=0.25a$ and $\dfrac{a}{a-x}=4$; thus $60=\dfrac{2.303}{k}\log 4$. At 50% completion (this is simply $t_{1/2}$), $x=0.5a$, so $a-x=0.5a$ and $\dfrac{a}{a-x}=2$; thus $t_{50\%}=\dfrac{2.303}{k}\log 2$. Since $\log 4=\log(2^2)=2\log 2$, the 75%-completion time is exactly double the 50%-completion time: $60=2\times t_{50\%}$, so $t_{50\%}=30$ …

Decomposition of $N_2O_5$ obeys $-\dfrac{d[N_2O_5]}{dt}=k[N_2O_5]$ intrinsically, making it a classic true first-order reaction rather than a pseudo first-order one.

A reaction is called 'pseudo first order' when its true molecularity/order is higher (e.g. second order) but it experimentally behaves as first order because one reactant (often the solvent, like water in ester hydrolysis or sucrose inversion) is present in such large excess that its concentration stays effectively constant. The decomposition of $N_2O_5$ (whether in the gas phase or dissolved in an inert solvent like $CCl_4$) is different — its rate genuinely depends only on the first power of $[N_2O_5]$ itself, with no other species' concentration involved in the rate law, so it is one of the standard …