Chemistry: Consider the first-order reaction described by the equation Cyclopropane gas isomerizes to propene gas. At a certain temperature, the rate constant for this reaction is 5.05 × 10 − 4 s − 1 . Calculate the half-life of cyclopropane at this temperature. t 1 / 2 = s Given an initial cyclopropane concentration of 0.00670 M , calculate the concentration of cyclopropane that remains after 2.00 hours. concentration.

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Consider the first-order reaction described by

kennedy3076

19.01.2022

The half-life would be 1372 seconds.

The concentration remains after 2 hours is 0.000177 M.

Explanation:

Given the initial concentration (A_0) of was $0.00670\ M$ .

And the rate constant (k) is $5.05\times 10^{-4}\ s^{-1}$

We need to calculate the half-life and the concentration remains (A) after 2 hours.

First, we will use the formula of half-life of the first order reaction.

$t_{\frac{1}{2} }=\frac{0.693}{K}\\\\t_{\frac{1}{2} }=\frac{0.693}{5.05\times 10^{-4}}\\\\t_{\frac{1}{2} }=1372\ seconds$

So, half-life would be 1372 seconds.

Now, we will find its concentration after 2 hours. That means $t=2\times60\times60=7200\ seconds$

Use the equation

$A=A_oe^{-kt}\\A=0.00670\times e^{-5.05\times10^{-4}\times7200}\\A=0.00670\times e^{-3.636}\\A=0.00670\times0.026\\A=0.000177\ M$

So, the concentration remains after 2 hours is 0.000177 M.

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