A) 30 min
B) 15 min
C) 7.5 min
D) 60 min
Correct Answer: A
Solution :
Order\[=1\] Concentration changes from 0.8 M to 0.4 M (50%) in 15 min, thus half-life is \[15min={{T}_{50}}\] A change from 0.1 M to 0.025 M is 75% and for first order reaction \[{{T}_{75}}=2\times {{T}_{50}}=2\times 15=30\,min\]or\[{{T}_{50}}=15\min \] \[k=\frac{2.303\log 2}{{{T}_{50}}}=\frac{2.303\log 2}{15}\] \[a=0.1\text{ }M\] \[(a-x)=0.025\text{ }M\] For first order \[k=\frac{2.303}{t}\log \left( \frac{a}{a-x} \right)\] \[\frac{2.303\log 2}{15}=\frac{2.303}{t}\log \frac{0.1}{0.025}=\frac{2.303}{t}\log 4\] \[\therefore \]\[\frac{2.303\log 2}{15}=\frac{2\times 2.303\log 2}{t}\] \[\therefore \] \[t=30\text{ }\min \]
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