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aliina [53]
3 years ago
11

The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 x 10-4 s -1 at

500 C. Cyclopropane (CH2CH2CH2) -- CH3-CH=CH2 (propene) If the initial concentration of the reactant was 0.25 M, what is the concentration after 8.8 min?
Chemistry
1 answer:
denpristay [2]3 years ago
8 0

Answer:

C_f=0.1755M

Explanation:

Hello,

Based on the information and the units of the given data, the integrated rate law turns out into:

\frac{dC}{dt}=-kC\\ C_f=C_0exp(-kt)\\C_f=0.25M*exp(-6.7x10^{-4}s^{-1}*8.8min*\frac{60s}{1min} )\\C_f=0.1755M

Best regards.

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Explanation:

Data

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As we can see, the reaction is balanced and the coefficients of all reactants and products are 1, but the number is not written in a balanced reaction.

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