A) zero
B) first
C) second
D) more than zero but less than first
Correct Answer: A
Solution :
| [a] For a zero order reaction \[{{t}_{1/2}}\] is directly proportional to the initial concentration of the reactant \[{{[R]}_{0}}\] |
| \[{{t}_{1/2}}\propto {{[R]}_{0}}\] |
| For a first order reaction |
| \[k=\frac{2.303}{t}\log \frac{{{[R]}_{0}}}{[R]}\,\,at\,\,\,{{t}_{1/2}},[R]=\frac{{{[R]}_{0}}}{2}\] |
| So, the above equation becomes |
| \[K=\frac{2.303}{{{t}_{1/2}}}\log \frac{{{[R]}_{0}}}{{{[R]}_{0}}/2}\] |
| \[{{t}_{1/2}}=\frac{2.303}{K}=\log 2=\frac{2.303}{K}\times .3010\] |
| \[{{t}_{1/2}}=\frac{.693}{k}\] |
| i.e., half-life period is independent of initial concentration of a reactant. |
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