A) \[664.64\text{ }sec\]
B) \[646.6\text{ }sec\]
C) \[660.9\text{ }sec~\]
D) \[654.5\text{ }sec\]
Correct Answer: A
Solution :
Half of the reaction is completed in 100 second \[\therefore \] \[{{t}_{1/2}}=100\sec \] \[\therefore \] \[K=\frac{0.693}{100}{{\sec }^{-1}}\] For a first order reaction \[K=\frac{2.303}{t}\log \frac{a}{a-x}\] \[a=100,\] \[x=99,\] \[{{t}_{99%}}=?\] \[{{t}_{99%}}=\frac{2.303}{K}\log \frac{100}{100-99}\] \[=\frac{2.303\times 100}{0.693}\times \log 100\sec .\] \[=\frac{2.303\times 100\times 2}{0.693}\] \[=664.64\sec .\]
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