A) \[0.0693\text{ }mol\text{ }mi{{n}^{-1}}\]
B) \[0.0693\times 2\text{ }mol\text{ }mi{{n}^{-1}}\]
C) \[0.0693\times 5\text{ }mol\text{ }mi{{n}^{-1}}\]
D) \[0.0693\times {{\left( 5 \right)}^{2}}mol\text{ }mi{{n}^{-1}}\]
Correct Answer: C
Solution :
[c] \[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{0.0693}=10\min \] Reactant after 10 min = 5 mol \[Rate\left( \frac{dx}{dt} \right)=k[A]=0.0693\times 5\,mol\,mi{{n}^{-1}}\]
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