• # question_answer For the$2A(g)\xrightarrow{\,}\,3B(g),\,{{t}_{1/2}}=12\,\min$. Initial pressure exerted by A is 640 mm of Hg. The pressure of the reaction mixture after the time period of 36 min will be A) 560 mm Hg      B) 680 mm Hg       C) 920 mm Hg       D) 600 mm Hg

 For first order reaction $\underset{640-2x}{\mathop{2A(g)}}\,\,\xrightarrow{\,}\,\underset{3x}{\mathop{3B\,(g)}}\,$ $\therefore$      ${{p}_{Total}}=640+x$ $k=\frac{2.303}{t}\,\log \,\frac{a}{a-x}$ $\frac{0.693}{{{t}_{1/2}}}=\frac{2.303}{36}\log \,\frac{640}{640-2x}$ $\frac{2.303\,\log \,2}{12}=\frac{2.303}{36}\,\log \,\frac{640}{640-2x}$ $\log {{2}^{2}}=\log \,\frac{640}{640-2x}$ $x=280$ $\therefore$      ${{p}_{Total}}\,=640+280\,=920\,mm$