a
\((a)\) Rate law forr first order reaction \(=\,k[N_2O_5]\)

\(2{{N}_{2}}{{O}_{5}}\,(g)\,\to \,4N{{O}_{2}}\,(g)\,+\,{{O}_{2}}(g)\)

\(\mathop {t = 0\,\,\min }\limits_{(\Pr essure\,\,\,in\,\,\,mm\,\,\,Hg)} \)                 \(50\)                    \(0\)               \(0\)

\(\mathop {t =30\,\,\min }\limits_{(\Pr essure\,\,\,in\,\,\,mm\,\,\,Hg)} \)               \(50-2p\)               \(4p\)               \(p\)

Total pressure \(50-2p+4p+p=50+3p=87.5\,mm\,Hg\)

\(\therefore \,\,P\, = \,12.5\,mm\,Hg\)

\(\therefore \,\,{P_0}\, = 50\) and \(P\) \((t=30\,min)\)

          \(=\,25\) for \(N_2O_5\) reactant

\(\therefore \,\,k\, = \,\frac{{2.303}}{{30\,\,\min }}\, \times \,\log \,\left( {\frac{{50}}{{25}}} \right)\,\)  \( = \,\frac{{2.303}}{{60\,\,\min }}\, \times \,\log \,\left( {\frac{{50}}{x}} \right)\)

On solving \(x=12.5\,\,mm\,Hg\,=\,50-2p\)

\(\therefore P=18.75\,\,mm\,Hg\)

\(\therefore \) Total pressure \(=50+3P=106.25\,\,mm\,Hg\)