| Experiment | [A] (in mol \[{{\text{L}}^{-1}}\]) | [B] (in mol \[{{\text{L}}^{-1}}\]) | Initial Rate of Reaction (in mol \[{{\text{L}}^{-1}}\] \[\text{mi}{{\text{n}}^{-1}}\]) |
| (I) | 0.10 | 0.20 | \[6.93\times {{10}^{-3}}\] |
| (II) | 0.10 | 0.25 | \[6.93\times {{10}^{-3}}\] |
| (III) | 0.20 | 0.30 | \[1.386\times {{10}^{-2}}\] |
A) 10
B) 5
C) 100
D) 1
Correct Answer: B
Solution :
| \[6.93\times 10{{\,}^{-\,3}}\,=\text{K}\,\times \,{{(0.1)}^{x}}{{(0.2)}^{y}}\] |
| \[6.93\times 10{{\,}^{-\,3}}\,=\,\text{K}\,\times \,{{(0.1)}^{x\,}}\,{{(0.25)}^{y}}\] |
| So, \[y=0\] |
| and \[1.386\times {{10}^{\,-\,2}}\,=\,\text{K}\,\times \,{{(0.2)}^{x}}\,{{(0.30)}^{y}}\] |
| \[\frac{1}{2}\,=\,{{\left( \frac{1}{2} \right)}^{x}}\,x\,=\,1\] |
| So, \[r\,=\,\text{K}\,\times \,(0.1)\,\times \,{{(0.2)}^{0}}\] |
| \[6.93\times 10{{\,}^{-\,3}}\,=\,\text{K}\,\times 0.1\,\times \,{{(0.2)}^{0}}\] |
| \[\text{K}\,=\,6.93\times {{10}^{-\,2}}\] |
| \[{{t}_{\frac{1}{2}}}\,=\,\frac{0.693}{2\text{K}}\,=\,\frac{0.693}{0.69\times {{10}^{\,-\,1}}\times 2}\,=\,\frac{10}{2}\,=\,5.\] |
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