• # question_answer A hypothetical reaction: ${{A}_{2}}+{{B}_{2}}\xrightarrow{{}}2AB$ follows mechanism as given below: ${{A}_{2}}A+A$............ (fast)         (${{k}_{c}}$ is equilibrium constant) ${{A}_{2}}+{{B}_{2}}\xrightarrow{{{k}_{1}}}AB+B$ ..............(slow)  (${{k}_{1}}$ rate constant) $A+BAB$ .................. (fast)      (${{k}_{2}},{{k}_{3}}$ are rate constant) Give the rate law. A) $r={{k}_{1}}\sqrt{{{k}_{c}}}\,{{[{{A}_{2}}]}^{1/2}}[{{B}_{2}}]$   B) $r=\frac{{{k}_{1}}}{{{k}_{c}}}\,{{[{{A}_{2}}]}^{1/2}}[{{B}_{2}}]$ C) $r=\sqrt{{{k}_{1}}{{k}_{c}}}\,\,{{[{{A}_{2}}]}^{1/2}}[{{B}_{2}}]$ D) $r=\frac{{{k}_{1}}}{\sqrt{{{k}_{c}}}}\,{{[{{A}_{2}}]}^{1/2}}[{{B}_{2}}]$

 [a] Rate is governed by slowest step $A+{{B}_{2}}\xrightarrow{{{k}_{1}}}AB+B$ $r={{k}_{1}}\,[A]\,\,[{{B}_{2}}]$ (i) From     ${{A}_{2}}A+A$
 ${{k}_{c}}=\frac{{{[A]}^{2}}}{[{{A}_{2}}]}$ (ii) $[A]=\sqrt{{{k}_{c}}}\,\,{{[{{A}_{2}}]}^{1/2}}$ $r={{k}_{1}}\sqrt{{{k}_{c}}}\,\,{{[{{A}_{2}}]}^{1/2}}\,[{{B}_{2}}]$ order is $=\frac{1}{2}+1=\frac{3}{2}$.