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IRINA_888 [86]
3 years ago
10

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
Chemistry
1 answer:
gavmur [86]3 years ago
6 0

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!

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