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 in (50%) 15 min thus half-life is\[=15\text{ }min={{T}_{50}}\] A change from 0.1 M to P.025 M is 75% and for first order reaction \[{{T}_{75}}=2\times 750=2\times 15=30\text{ }min\] \[{{T}_{50}}=15\text{ }min\] \[k=\frac{2.303\log 2}{{{T}_{50}}}=\frac{2.303\log 2}{15}\] \[a=0.1\,M\] \[(a-x)=0.025\,M\] 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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