Question

40% of a first order reaction is completed in 50 minutes. How much time will it take for the completion of 80% of this reaction?

Answer 1

Rate equation for First order reaction:

$=\frac{2.303}{t}\log \left(\frac{R_0}{R_t}\right)Att=50min,R_{50}=(1-0.4)R_0=0.6R_{0}Att=xmin,R_x=(1-0.8)R_0=0.2R_0$

Substituting, the values we get,

$\frac{2.303}{50}\log \left(\frac{R_0}{0.6R_0}\right)=\frac{2.303}{x}\log \left(\frac{R_0}{0.2R_0}\right)x=50\left[\frac{1-\log \left(2\right)}{1-\log \left(6\right)}\right]=50\times \frac{1-0.414}{1-0.778}=131.98\approx 132min$