• # question_answer $6{{I}^{-}}(aq)+Br{{O}_{3}}^{-}(aq)+6{{H}^{+}}(aq)\xrightarrow{{}}3{{I}_{2}}(aq)+$ $B{{r}^{-}}(aq)+3{{H}_{2}}O(\ell )$ these data were obtained when this reaction was studies. $[{{I}^{-}}],M$ $[Br{{O}_{3}}^{-}],M$ $[{{H}^{+}}],M$ Reaction rate $(mol\,{{L}^{-1}}{{s}^{-1}})$ 0.0010 0.0020 0.010 $8.0\times {{10}^{-5}}$ 0.0020 0.0020 0.010 $1.6\times {{10}^{-4}}$ 0.0020 0.0040 0.010 $1.6\times {{10}^{-4}}$ 0.0010 0.0040 0.020 $1.6\times {{10}^{-4}}$ What is the units of the rate constant for this reaction? A) ${{s}^{-1}}$                          B) $mol\,\,{{L}^{-1}}{{s}^{-1}}$ C) $L\,\,mo{{l}^{-1}}{{s}^{-1}}$ D) ${{L}^{2}}\,\,mo{{l}^{-2}}{{s}^{-1}}$

 $R=K{{[{{I}^{-}}]}^{a}}{{[BrO_{3}^{-}]}^{b}}{{[{{H}^{+}}]}^{c}}$ $8\times {{10}^{-5}}=K{{(0.001)}^{a}}{{(0.002)}^{b}}{{(0.01)}^{c}}$                  .....(i) $1.6\times {{10}^{-4}}=K{{(0.002)}^{a}}{{(0.002)}^{b}}{{(0.01)}^{c}}$               .....(ii) $1.6\times {{10}^{-4}}=K{{(0.002)}^{a}}{{(0.004)}^{b}}{{(0.01)}^{c}}$   .....(iii) $1.6\times {{10}^{-4}}=K{{(0.001)}^{a}}{{(0.004)}^{b}}{{(0.02)}^{c}}$   ....(iv) (ii)-(i)                 $2=2a\Rightarrow a=1$ (iii)-(ii)               $1=2b\Rightarrow b=0$ (iv)-(i)                $2=2c\Rightarrow c=1$ $R=K[{{I}^{-}}][{{H}^{+}}]$ unit of K  $mo{{l}^{-1}}$ litre ${{\sec }^{-1}}$.