Question

The rate constant for a first order reaction is $6.909{min}^{-1}$. Therefore, the time required in minutes for the participation of $75$% of the initial reactant is:

• A. $\frac{2}{3}\log 2$
• B. $\frac{2}{3}\log 4$
• C. $\frac{3}{2}\log 2$
• D. $\frac{3}{2}\log 4$

Answer 1

The correct option is A $\frac{2}{3}\log 2$

$\ln \left(\frac{A_0}{A}\right)=kt\ln \left(\frac{A_0}{1/4A_0}\right)=kt\frac{\ln \left(4\right)}{6.909}=t\frac{2\log 2}{3}=t$