A) \[6.932\] min
B) 100 min
C) \[0.6932\times {{10}^{-3}}\] min
D) \[0.6932\times {{10}^{-2}}\] min
Correct Answer: B
Solution :
\[r=k{{[\text{reactant}]}^{-1}}\]\[\therefore k=\frac{0.693\times {{10}^{-2}}}{1}\]also \[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{0.693\times {{10}^{-2}}}=100\,\min \].You need to login to perform this action.
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