Question

Hydrocarbons Heating cyclopropane (C₃H₆) converts it to propene (CH₂ =CHCH₃). The rate law is first order in cyclopropane. If the rate constant at a particular temperature is 6.22 x 10⁻⁴⁵s⁻¹ and the concentration of cyclopropane is held at 0.0300 moll, what mass of propene is produced in 10.0 min in a volume of 2.50 L?

Answer 1

To calculate the mass of propene produced in 10.0 minutes, we need to use the rate law equation. The concentration of cyclopropane is given as 0.0300 mol/L and the rate constant is 6.22 x 10⁻⁴⁵s⁻¹ at the given temperature. Using the rate law, we can calculate the concentration of cyclopropane and then use it to calculate the number of moles of propene produced in 10.0 minutes. Finally, we can calculate the mass of propene produced using the molar mass of propene.

Step-by-step explanation:

In a first-order reaction, the rate of the reaction depends on the concentration of only one reactant. In this case, the rate of the reaction is first order in cyclopropane. The rate law for the reaction is given as Rate = k[C₃H₆], where k is the rate constant. The rate constant at the given temperature is 6.22 x 10⁻⁴⁵s⁻¹. The concentration of cyclopropane is 0.0300 mol/L and the volume of the reaction is 2.50 L.

To calculate the mass of propene produced in 10.0 minutes, we need to use the equation: Mass of propene = moles of propene x molar mass of propene. First, we need to calculate the number of moles of propene produced. The reaction is 1:1, so the number of moles of propene produced is equal to the number of moles of cyclopropane consumed. We can use the rate law to calculate the rate of the reaction: Rate = k[C₃H₆].

From the rate law, we know that the rate of the reaction is equal to the rate constant multiplied by the concentration of cyclopropane. We can rearrange the equation to solve for the concentration of cyclopropane: [C₃H₆] = Rate / k. Plugging in the values, we get [C₃H₆] = (6.22 x 10⁻⁴⁵s⁻¹)(0.0300 mol/L).

Now we can calculate the moles of propene produced in 10.0 minutes. Since the reaction is first order in cyclopropane, the rate of consumption of cyclopropane is equal to the rate of production of propene. The rate of consumption of cyclopropane is given by the equation: Rate = -d[C₃H₆] / dt. Plugging in the values, we get -d[C₃H₆] / dt = (6.22 x 10⁻⁴⁵s⁻¹)(0.0300 mol/L).

Now we can rearrange the equation to solve for -d[C₃H₆]: -d[C₃H₆] = (6.22 x 10⁻⁴⁵s⁻¹)(0.0300 mol/L)(10.0 min). To convert minutes to seconds, we multiply by 60: -d[C₃H₆] = (6.22 x 10⁻⁴⁵s⁻¹)(0.0300 mol/L)(10.0 min)(60 s/min).

Next, we can convert -d[C₃H₆] to d[CH₂ =CHCH₃]. Since the reaction is 1:1, the moles of cyclopropane consumed is equal to the moles of propene produced. Therefore, d[CH₂ =CHCH₃] = -d[C₃H₆]. Plugging in the values, we get d[CH₂ =CHCH₃] = (6.22 x 10⁻⁴⁵s⁻¹)(0.0300 mol/L)(10.0 min)(60 s/min).

Finally, we can calculate the mass of propene produced. The molar mass of propene (CH₂ =CHCH₃) is 42.08 g/mol. Therefore, the mass of propene produced in 10.0 minutes is: Mass of propene = (6.22 x 10⁻⁴⁵s⁻¹)(0.0300 mol/L)(10.0 min)(60 s/min)(42.08 g/mol).

Answer 2

The half-life of the isomerization of cyclopropane to propene at 499 °C is approximately 19.37 minutes. Approximately 35.7% of cyclopropane remains after 0.75 hours.

Step-by-step explanation:

The rate constant for the isomerization of cyclopropane to propene at 499 °C is 5.95 x 10-4 s ¹. To find the half-life of the reaction, we can use the integrated rate law for a first-order reaction. The half-life is calculated using the formula: t_(1/2) = (ln 2) / k, where k is the rate constant. Plugging in the values, we find that the half-life of this reaction is approximately 1162 seconds (or 19.37 minutes).

To determine the fraction of cyclopropane that remains after 0.75 hours, we need to calculate the amount of time in seconds. Since there are 60 minutes in an hour and 60 seconds in a minute, 0.75 hours is equal to 45 minutes or 2700 seconds. The fraction of cyclopropane that remains after this time is given by the equation: remaining_fraction = e^(-kt), where k is the rate constant and t is the time. Plugging in the values, we find that approximately 35.7% of cyclopropane remains after 0.75 hours.