1) A first-order reaction is 25 % completed in 40 minutes. What is the rate constant k for the reaction? A) $\frac{2.303\times \log1.33}{40}$ B) $2.303\times \log\frac{4}{3}$ C) $\frac{2.303\times \log4}{40\times3}$ D) $\frac{2.303 }{40}\times\log \frac{1}{4}$ Answer: Option AExplanation: Given, Time (t)= 40 min a=100 (a-x)=100-25=75 First order reaction, $Rate(k)=\frac{2.303}{t}\log\left(\frac{a}{a-x}\right)=\frac{2.303\times\log 1.33}{40}$
