Q.A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?

Step 1. 40% complete means 60% of A remains, so $[A]_0/[A]=100/60$.

Step 2. $k=\dfrac{2.303}{50}\log\dfrac{100}{60}=0.04606\times\log(1.6667)=0.04606\times0.2218=0.01022\ \text{min}^{-1}$.

Step 3. For 80% completion, 20% of A remains, so $t_{80\%}=\dfrac{2.303}{k}\log\dfrac{100}{20}=\dfrac{2.303}{0.01022}\times\log5$. …

Compute k from the 40%-completion data using the first order rate law, then use the same k (or the direct ratio o …

• Using $[A]_0/[A]=100/40$ (mistaking the PERCENTAGE COMPLETED for the fraction REMAINING) instead of the correc …