Problem 90 - Chemistry Matter and Change

Question

Hydrocarbons Heating cyclopropane \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\) converts it to propene \(\left(\mathrm{CH}_{2}=\mathrm{CHCH}_{3}\right) .\) The rate law is first order in cyclopropane. If the rate constant at a particular temperature is \(6.22 \times 10^{-4} \mathrm{s}^{1}\) and the concentration of cyclopropane is held at \(0.0300 \mathrm{mol} / \mathrm{L},\) what mass of propene is produced in 10.0 \(\mathrm{min}\) in a volume of 2.50 \(\mathrm{L}\) ?

Step-by-Step Solution

Answer

1.16 grams of propene is produced.

Step 1: Identify the Reaction Order

The problem states that the reaction is first order in cyclopropane. This implies that the rate of the reaction depends on the concentration of cyclopropane raised to the first power. The rate law for a first-order reaction is given by: \[ \text{Rate} = k [\text{C}_{3}\text{H}_{6}] \] where \( k \) is the rate constant and \([\text{C}_{3}\text{H}_{6}]\) is the concentration of cyclopropane.

Step 2: Use the First-Order Kinetics Equation

The integrated rate law for a first-order reaction is: \[ [A]_t = [A]_0 e^{-kt} \] where \([A]_t\) is the concentration of \(A\) at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant. In this case, \([A]_0 = 0.0300 \, \text{mol/L}\), \(k = 6.22 \times 10^{-4} \, \text{s}^{-1}\), and \(t = 10.0 \, \text{min}\).

Step 3: Convert Time to Seconds

To use the rate constant \(k\) in \(\text{s}^{-1}\), convert the time from minutes to seconds. \[ 10.0 \, \text{min} = 10.0 \, \times \, 60 \text{s/min} = 600 \, \text{s} \]

Step 4: Calculate Final Cyclopropane Concentration

Substitute the known values into the integrated rate law to find the final concentration of cyclopropane: \[ [\text{C}_{3}\text{H}_{6}]_t = 0.0300 \, \text{mol/L} \times e^{-6.22 \times 10^{-4} \, \text{s}^{-1} \times 600 \, \text{s}} \approx 0.0190 \, \text{mol/L} \]

Step 5: Determine the Amount of Cyclopropane Consumed

Find the amount of cyclopropane consumed by subtracting the final concentration from the initial concentration: \[ 0.0300 \, \text{mol/L} - 0.0190 \, \text{mol/L} = 0.0110 \, \text{mol/L} \] Multiply by the volume in liters to find the moles consumed: \[ 0.0110 \, \text{mol/L} \times 2.50 \, \text{L} = 0.0275 \, \text{mol} \]

Step 6: Convert Moles of Propene to Mass

Since the amount of cyclopropane converted is equal to the amount of propene produced, you have 0.0275 mol of propene. Calculate the mass of propene using its molar mass: \[ \text{Molar mass of propene} = (12.01 \times 3 + 1.01 \times 6) \, \text{g/mol} = 42.08 \, \text{g/mol} \] \[ \text{Mass of propene} = 0.0275 \, \text{mol} \times 42.08 \, \text{g/mol} = 1.16 \, \text{g} \]

Step 7: Final Answer

The mass of propene produced in 10.0 minutes in a volume of 2.50 L is approximately 1.16 grams.

Key Concepts

Reaction Kinetics

Reaction kinetics is the study of the rates at which chemical reactions occur and the factors that affect these rates. To understand reaction kinetics, one must first be aware of the concept of reaction order. In the given problem, the reaction is first order, which means the rate of reaction is directly proportional to the concentration of one reactant. This is expressed in the rate law as:

• \[ \text{Rate} = k [\text{C}_{3}\text{H}_{6}] \] , where \( k \) represents the rate constant.

The unit of \( k \) for a first-order reaction is typically \( \text{s}^{-1} \), meaning per second. The integrated rate law for a first-order reaction is essential in determining how the concentration of a reactant changes over time:

• \[ [A]_t = [A]_0 e^{-kt} \]

Understanding these kinetics helps us to predict how quickly a reaction will proceed and allows calculation of remaining concentrations at any given time. This is crucial in industries and research where precise reaction control is necessary.

Cyclopropane

Cyclopropane is a type of hydrocarbon known for its distinctive ring structure consisting of three carbon atoms. This triangle-shaped molecule is highly strained due to its small bond angles (60 degrees), which are much less than the ideal 109.5 degrees found in more stable aliphatic compounds. As a result, cyclopropane is highly reactive. When heated, it readily undergoes transformations, such as converting into propene. Cyclopropane's instability makes it an interesting subject in reaction studies. Because it can easily convert to more stable compounds, it is used to understand the mechanics of ring-opening reactions, a type of rearrangement process. Such insights are valuable in synthetic chemistry and the development of pharmaceuticals, where similar transformations are often desired to create specific molecular structures. Moreover, cyclopropane has a historical significance as it was once used as an anesthetic. However, due to its flammability and explosive nature when mixed with air, it has been largely replaced by safer alternatives in medical use.

Propene Production

Propene, also known as propylene, is a colorless gas vital in the chemical industry. It is a key monomer used in the production of polypropylene, one of the most widely used plastics in the world. The transition from cyclopropane to propene is not only an interesting chemical transformation but also a practical demonstration of chemical reactivity that leads to industrially valuable products. In the conversion of cyclopropane to propene, the breaking and reforming of bonds lead to a more stable compound, demonstrating the concept of stability driving reactions. This reaction also illustrates the importance of understanding reaction mechanisms. Knowing how and why cyclopropane rearranges into propene can inform broader chemical synthesis techniques and pathways.

• Propene's importance extends beyond plastics. It's also a precursor in the synthesis of important chemicals and polymers, such as acrylonitrile, which is used in synthetic rubber and rigid plastics.
• Additionally, the production of propene highlights a sustainable angle due to the emphasis on converting existing, less useful chemicals into more valuable ones. This kind of transformation is at the heart of green chemistry, which seeks to minimize waste and make more efficient use of resources.