A) 0
B) 2
C) 1
D) 4
Correct Answer: B
Solution :
We know that, \[\frac{{{({{t}_{1/2}})}_{1}}}{{{({{t}_{1/2}})}_{2}}}={{\left[ \frac{{{a}_{2}}}{{{a}_{1}}} \right]}^{n-1}}\] where, n = order of the reaction Given, \[({{t}_{1/2}})=0.1\,s,\]\[\,{{a}_{1}}=400\] \[{{({{t}_{1/2}})}_{2}}=0.8\,s,\]\[\,{{a}_{2}}=50\] On substituting the values, \[\frac{0.1}{0.8}={{\left[ \frac{50}{400} \right]}^{n-1}}\] On taking log both sides \[\log \frac{0.1}{0.8}=(n-1)\log \frac{50}{400}\] \[\log \frac{1}{8}=(n-1)\log \frac{1}{8}\] \[0.90=(n-1)\,0.90\] \[n-1=1,\]\[n=2\]
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