| Experi-ment | in \[\mathbf{mol}\text{ }{{\mathbf{L}}^{-\mathbf{1}}}\] | (in \[\mathbf{mol}\text{ }{{\mathbf{L}}^{-\mathbf{1}}}\] | Initial Rate of reaction (in \[\mathbf{mol}\text{ }{{\mathbf{L}}^{-\mathbf{1}}}\] \[\mathbf{mi}{{\mathbf{n}}^{\mathbf{-1}}}\]) |
| 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) 5
B) 1
C) 100
D) 10
Correct Answer: A
Solution :
\[6.93\,\,\times \,\,{{10}^{-\,3}}\,\,=\,\text{ }k\text{ }\times \text{ }{{\left( 0.1 \right)}^{x}}\text{ }{{\left( 0.2 \right)}^{y}}\] \[6.93\text{ }\times \text{ }{{10}^{-\,3}}=k\,\,\times \,\,{{(0.1)}^{x}}{{\left( 0.25 \right)}^{y}}\] So \[y=0\] and \[1.386\text{ }\times \text{ }{{10}^{-\,2}}=\text{ }k\text{ }{{\left( 0.2 \right)}^{x}}\,\,{{\left( 0.30 \right)}^{y}}\] Hit - So, \[r\,\,=\,\,k\,\,\times (0.1)\,\times \,{{(0.2)}^{0}}\] \[6.93\,\times \,{{10}^{-3}}\,=\,k\,\times \,0.1\,\,\times \,\,{{(0.2)}^{0}}\] \[k\,\,=\,6.93\,\,\times \,\,{{10}^{-2}}\,\] \[{{t}_{1/2}}\,=\,\frac{0.693}{2\,k}\,\,=\,\,\frac{0.693}{0.693\,\times \,{{10}^{-\,1}}\,\times \,2}\,\,=\,\,\frac{10}{2}\,\,=\,\,5\]
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