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Answer :
After 8.00 hours, the concentration of ethanol is 0.45 m.
The rate of decomposition of ethanol follows a second order rate constant. This means that the rate of decomposition is proportional to the concentration of ethanol and the concentration of oxygen in the surrounding atmosphere. The rate constant for this reaction at 600 K is 4.00 x 10−5 m−1s−1.
If the initial concentration of ethanol is 0.50 m, then we can use the rate law to calculate the concentration after 8.00 hours. The equation for this is: [EtOH] = [EtOH]o × e−kt, where k is the rate constant (4.00 x 10−5 m−1s−1), t is the time in seconds (8.00 hours × 3600 seconds), and [EtOH]o is the initial concentration (0.50 m). Plugging in the values, we get [EtOH] = 0.50 × e−(4.00 x 10−5 m−1s−1 × 28800 seconds) = 0.45 m.
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