Question

In a first-order decomposition reaction. 58.6% of a compound decomposes in 11.7 min. How long (in min) does it take for 80.2% of the compound to decompose

Answer 1

Answer:

t = 21.5 min.

Explanation:

Hello!

In this case, since the kinetics of a first-order reaction is:

$\frac{[A]}{[A]_0}=exp(-kt)$

Thus, since we are given the 11.7 min for a 58.6-% consumption, we can compute the rate constant, k:

$ln(1-0.586)=-kt\\\\k=\frac{ln(0.414)}{-t}=\frac{-0.882}{11.7min}=0.0754min^{-1}$

Now, for the second problem, as the new consumption is 80.2%, we can compute the required time as shown below:

$ln(1-0.802)=-kt\\\\t=\frac{ln(198)}{k} \\\\t=\frac{-1.62}{0.0754min^{-1}}\\\\t=21.5min$

Best regards!