A) 16 min
B) 8 min
C) 4 min
D) 32 min
Correct Answer: A
Solution :
Key Idea: \[t=\frac{2.303}{k}\log \frac{a}{a-x}\] Given t = 32 min a = 100 (initial amount) \[(a-x)=25\] (mass left after time 0 \[32=\frac{2.303}{k}\log \frac{100}{25}\] \[32k=2.303\log 4\] \[\therefore \] \[k=0.0432\] Now for 50% of the reaction \[k=0.0432,\,a=100,\,a-x=50\] \[t=\frac{2.303}{K}\log \frac{a}{a-x}\] \[=\frac{2.303}{0.0432}\log \frac{100}{50}\] \[=53.310\log 2\] \[=53.310\times 0.3010\] = 16 min
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