\(2{{N}_{2}}{{O}_{5}}\,(g)\,\to \,4N{{O}_{2}}\,(g)\,+\,{{O}_{2}}(g)\)
\(\mathop {t = 0\,\,\min }\limits_{(\Pr essure\,\,\,in\,\,\,mm\,\,\,Hg)} \) \(50\) \(0\) \(0\)
\(\mathop {t =30\,\,\min }\limits_{(\Pr essure\,\,\,in\,\,\,mm\,\,\,Hg)} \) \(50-2p\) \(4p\) \(p\)
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 \(=50+3P=106.25\,\,mm\,Hg\)