26. Solving a chemistry problem
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5. The kinetics of high-temperature decomposition of acetic acid corresponds to the kinetics of a first-order parallel reaction. The reaction scheme is as follows:
CH3COOH=CH4CO2 (k1)
CH3COOH=CH2CO H2O (k2)
The half-life of the acid is 0.09 s, and the ratio of the number of moles of methane formed in the first stage to the number of moles of ketene formed in the second stage at the same time is 0.83. Calculate the value of the rate constants for each of the reaction steps and the total rate constant of the process.
CH3COOH=CH4CO2 (k1)
CH3COOH=CH2CO H2O (k2)
The half-life of the acid is 0.09 s, and the ratio of the number of moles of methane formed in the first stage to the number of moles of ketene formed in the second stage at the same time is 0.83. Calculate the value of the rate constants for each of the reaction steps and the total rate constant of the process.
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- Added to the site 29.04.2023
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If you do not open the file - install the RAR archiver. Inside the archive you will find a solution in Word format.
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26. Solving a chemistry problem
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