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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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Answer:

The half-life would be 1372 seconds.

The concentration remains after 2 hours is 0.000177 M.

Explanation:

Given the initial concentration [tex](A_0)[/tex] of was [tex]0.00670\ M[/tex].

And the rate constant [tex](k)[/tex] is [tex]5.05\times 10^{-4}\ s^{-1}[/tex]

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

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

[tex]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[/tex]

So, half-life would be 1372 seconds.

Now, we will find its concentration after 2 hours. That means [tex]t=2\times60\times60=7200\ seconds[/tex]

Use the equation

[tex]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[/tex]

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

ajeigbeibraheem

The half-life of cyclopropane at this temperature [tex]\mathbf{1.372 \times 10^3\ s}[/tex].The concentration of cyclopropane that remains after 2.00 hours is [tex]\mathbf{1.765 \times 10^{-4} \ M}[/tex]

Considering the first-order reaction described by the equation, the half-life for a first-order reaction can be computed as:

[tex]\mathbf{t_{1/2} = \dfrac{0.693}{k} }[/tex]

where;

  • rate constant K = 5.05 × 10⁻⁴ s⁻¹

∴

[tex]\mathbf{t_{1/2} = \dfrac{0.693}{5.05 \times 10^{-4} \ s^{-1}} }[/tex]

[tex]\mathbf{t_{1/2} =0.1372 \times 10^4\ s}[/tex]

[tex]\mathbf{t_{1/2} =1.372 \times 10^3\ s}[/tex]

For a first-order reaction, the integrated rate law can be used to determine the concentration of cyclopropane that remains after 2.00 hours:

[tex]\mathbf{In [A]_t = -kt + In[A]_o}[/tex]

where;

  • [tex]\mathbf{[A]_t}[/tex] = concentration at time (t)
  • [tex]\mathbf{[A]_o}[/tex] = initial concentration

When the time (t) = 2.00 hours

=( 2 × 60 × 60) seconds

= 7200 s

∴

[tex]\mathbf{In [A]_t = -(5.05 \times 10^{-4} \ s^{-1} \times 7200 s) + In(0.00670 \ M)}[/tex]

[tex]\mathbf{In [A]_t = -3.636 -5.006 }[/tex]

[tex]\mathbf{In [A]_t =-8.642 \ M }[/tex]

[tex]\mathbf{[A]_t =1.765 \times 10^{-4} \ M}[/tex]

Therefore, we can conclude that the concentration of cyclopropane that remains after 2.00 hours is [tex]\mathbf{1.765 \times 10^{-4} \ M}[/tex]

Learn more about the first-order reaction here:

https://brainly.com/question/2877303?referrer=searchResults

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