A) 0.5
B) 1
C) 2
D) 0
Correct Answer: C
Solution :
\[{{t}_{1/2}}\propto \frac{1}{{{[{{A}_{0}}]}^{n\,-\,1}}}\] \[\therefore \,\,\,\,\,\,\,\frac{{{({{t}_{1/2}})}_{1}}}{{{({{t}_{1/2}})}_{2}}}=\frac{[{{A}_{0}}]_{2}^{n-1}}{[{{A}_{0}}]_{1}^{n-1}}\,\,=\,\,{{\left[ \frac{{{[{{A}_{0}}]}_{2}}}{{{[{{A}_{0}}]}_{1}}} \right]}^{n\,-\,1}}\] \[\frac{t}{t/2}={{\left( \frac{2a}{a} \right)}^{n\,-\,1}}or\,\,2\,\,=\,\,{{2}^{n\,-\,1}}\] or \[(n-1)=1\,\,or\,\,n=2\]
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