A) 10 times that required for half of the reaction
B) 100 times that required for two-third of the reaction
C) 10 times that required for one-fourth of the reaction
D) 20 times that required for half of the reaction
Correct Answer: A
Solution :
Time required for 99.9% of the reaction, \[{{T}_{1}}=\frac{2.303}{k}\log \frac{100}{100-99.9}\] \[=\frac{2.303}{k}\log \frac{100}{0.1}=\frac{6.909}{k}\] Again, time required for half of the reaction, \[{{T}_{2}}=\frac{0.7693}{k}\] ?. (i) From Eqs. (i) and (ii), we have \[{{T}_{1}}=10\times {{T}_{2}}\]
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