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If a 2.3 L reaction vessel initially contains 721 torr of cyclopropane at 485 °C, how long in minutes will it take for the partial pressure of cyclopropane to drop to below 104 torr?

a. 22 minutes
b. 32 minutes
c. 42 minutes
d. 52 minutes

1 Answer

4 votes

Final answer:

To calculate the time required for the partial pressure of cyclopropane to drop below 104 torr, we need the rate constant at the specified temperature (485 °C) which is not provided; thus, we cannot accurately determine the answer without making assumptions.

Step-by-step explanation:

The question asks to determine how long it will take for the partial pressure of cyclopropane to drop below 104 torr in a 2.3 L reaction vessel starting from 721 torr at 485 °C. Although the information provided involves reactions and conditions at 499 °C rather than 485 °C, it usually implies a first-order rate law where the rate of reaction is proportional to the concentration of the reactant. However, without additional details such as the rate constant at the specified temperature (485 °C) or a relationship between temperature and reaction rate for cyclopropane, an accurate determination of the time required for the pressure to drop to the specified level cannot be made provided the conditions of the reaction; temperature and pressure, half-life data, nor the rate constants are not directly applicable to the given initial state of the reaction. Therefore, it is not possible to calculate the time needed for the given reaction to reach the specified pressure without the correct corresponding information or making unjustified assumptions.

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