The time it takes for the partial pressure of cyclopropane to decrease from 1 atm to 0.125 atm at 500°C is 57 minutes.
A first-order reaction is a unimolecular reaction and the rate of a first-order reaction process is determined by the concentration of only one reactant.
The half-life for first-order reaction can be represented as:
[tex]\mathbf{t_{1/2} = \dfrac{0.693}{k} }[/tex]
Given that;
[tex]\mathbf{19 = \dfrac{0.693}{k} }[/tex]
[tex]\mathbf{k = \dfrac{0.693}{19} }[/tex]
k = 0.0365 / min
Similarly, for a first-order reaction, the order of reaction can be computed as:
[tex]\mathbf{K = \Big(\dfrac{2.303}{t} \Big) log \dfrac{P_o}{P}}[/tex]
[tex]\mathbf{0.0365 = \Big(\dfrac{2.303}{t} \Big) log \Big(\dfrac{1.0}{0.125}\Big)}[/tex]
[tex]\mathbf{0.0365t = 2.303 \times 0.903}[/tex]
[tex]\mathbf{t = \dfrac{2.303 \times 0.903}{0.0365}}[/tex]
t = 56.97 minutes
t ≅ 57 minutes
Therefore, we can conclude that the time it takes for the partial pressure of cyclopropane to decrease from 1 atm to 0.125 atm at 500°C is 57 minutes.
Learn more about first-order of reaction here:
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