A) First order
B) Zero order
C) Second order
D) Third order
Correct Answer: C
Solution :
The half-life period of a reaction of nth order is related to initial concentration of reactant as \[{{t}_{1/2}}\propto \frac{1}{{{[{{A}_{0}}]}^{n-1}}}\] For two different half-life periods, their concentration are \[\frac{{{({{t}_{1/2}})}_{1}}}{{{({{t}_{1/2}})}_{2}}}=\frac{[{{A}_{0}}]_{2}^{n-1}}{{{[{{A}_{0}}]}^{n-1}}}={{\left\{ \frac{{{[{{A}_{0}}]}_{2}}}{{{[{{A}_{0}}]}_{1}}} \right\}}^{n-1}}\] or \[\frac{1}{1/2}={{\left( \frac{2}{1} \right)}^{n-1}}\] or \[2={{2}^{n-1}}\] or \[n-1=1\] or \[n=2\] \[\therefore \]Order of reaction\[=2\]
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