| \[[{{I}^{-}}],M\] | \[[Br{{O}_{3}}^{-}],M\] | \[[{{H}^{+}}],M\] | 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 | \[1.6\times {{10}^{-4}}\] |
A) \[{{s}^{-1}}\]
B) \[mol\,\,{{L}^{-1}}{{s}^{-1}}\]
C) \[L\,\,mo{{l}^{-1}}{{s}^{-1}}\]
D) \[{{L}^{2}}\,\,mo{{l}^{-2}}{{s}^{-1}}\]
Correct Answer: C
Solution :
| \[R=K{{[{{I}^{-}}]}^{a}}{{[BrO_{3}^{-}]}^{b}}{{[{{H}^{+}}]}^{c}}\] |
| \[8\times {{10}^{-5}}=K{{(0.001)}^{a}}{{(0.002)}^{b}}{{(0.01)}^{c}}\] .....(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 \[{{\sec }^{-1}}\]. |
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