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 \(\mathrm{L} ?\)

Short answer

The mass of propene produced in 10.0 minutes is approximately 12.02 grams.

Step-by-step solution

Write the first-order rate law equation for the reaction

Since the reaction is first-order with respect to cyclopropane, we can write the rate law equation as: \[\frac{-d[\ce{C3H6}]}{dt} = k[\ce{C3H6}]\] Where: \( [\ce{C3H6}] \): concentration of cyclopropane \(t \): time (in seconds) \(k\): the rate constant (in \(s^{-1}\))

Integrate the rate law equation and solve for final concentration of cyclopropane

Integrating the rate law equation and rearranging it to find the final concentration of cyclopropane, we get: \[[\ce{C3H6}]_t = [\ce{C3H6}]_0 e^{-kt}\] Where: \([\ce{C3H6}]_0 \): initial concentration of cyclopropane (\(0.0300 M\)) \([\ce{C3H6}]_t \): concentration of cyclopropane at time \(t\) \(t = 10.0 \ min × 60 \ \frac{sec}{min} = 600 \ sec \) Plug in the given values and solve for \([\ce{C3H6}]_t\): \[[\ce{C3H6}]_t = 0.0300 \times e^{-6.22 \times 10^{-4} \times 600} \approx 0.0186 \ \text{M}\]

Determine the concentration of formed propene using stoichiometry

The stoichiometry of the reaction is 1:1; thus the increase in the concentration of propene is equal to the decrease in the concentration of cyclopropane: \[[\ce{CH2CHCH3}] = [\ce{C3H6}]_0 - [\ce{C3H6}]_t = 0.0300 - 0.0186 = 0.0114 \ \text{M} \]

Calculate the mass of propene produced

Now that we have the concentration of propene, we can find the mass using the molecular weight of propene (\(\text{MW}= 42.08 \frac{g}{mol} \)) and the given volume: mass of propene = concentration of propene × molecular weight × volume mass of propene \(= 0.0114 \ M \times 42.08 \ \frac{g}{mol} \times 2.50 \ L \) \( \approx \)\(12.02 \ g \) So, the mass of propene produced in 10.0 minutes is approximately 12.02 grams.

Key concepts

Understanding Rate Law

The rate law is a fundamental concept in chemical kinetics that defines the relationship between the concentration of reactants and the rate of a chemical reaction. It is usually expressed in the form of an equation that illustrates how the rate of reaction depends on the concentration of each reactant raised to a certain power, known as the reaction order.

In the rate law equation, \[\text{rate} = k[\text{Reactant}]^n\], 'k' represents the rate constant, a factor that is determined experimentally and is specific to a particular reaction at a given temperature. The exponent 'n' represents the order of the reaction with respect to the reactant concentration.

• For a first-order reaction, the exponent 'n' is 1, indicating that the reaction rate is directly proportional to the reactant's concentration.
• In a second-order reaction, 'n' is 2, suggesting that the rate is proportional to the square of the reactant's concentration.
• Zero-order reactions have 'n' equal to 0, signifying that the rate is independent of the reactant's concentration.

Having a precise equation for the rate law allows chemists to predict how changes in concentration will affect the reaction rate, which is key for controlling processes in industries and understanding natural phenomena.

First-Order Reaction Kinetics

First-order reactions are a type of chemical reaction where the rate is directly proportional to the concentration of a single reactant. These reactions have a characteristic half-life that is independent of the initial concentration of the reactants. The integrated rate law for a first-order reaction is given by the formula:\[\ln\left([\text{Reactant}]\right)= -kt + \ln\left([\text{Reactant}]_0\right)\]In practice, this means that in a first-order reaction, each time interval of a fixed length leads to a reduction in the reactant's concentration by the same factor, regardless of the actual concentration at the start of the interval. In our cyclopropane example, the decrease in concentration over 10 minutes is predictable once we know the rate constant, k, and the initial concentration.Understanding the kinetics of first-order reactions is crucial because they are commonplace in nature, such as in radioactive decay and many enzymatic processes. In industrial applications, such as pharmaceutical synthesis, first-order kinetics helps predict the conversion of reactants to products, which is essential for manufacturing efficiency and cost-effectiveness.

Stoichiometry in Reaction Mass Calculations

Stoichiometry is the aspect of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction, based on the balanced chemical equation. It allows us to predict how much product will be formed from a given amount of reactant and vice versa. Stoichiometry is rooted in the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
When it comes to stoichiometry in a chemical reaction like the one discussed, you start by looking at the coefficients in the balanced equation, which provides the mole ratio between reactants and products. In our example, cyclopropane converts to propene in a 1:1 mole ratio.

• To calculate the amount of product formed, you need to note changes in the concentration of the reactants.
• These changes, multiplied by the reaction volume and the molar mass of the product, give you the mass of product formed.

In chemical manufacturing and laboratory settings, stoichiometry is crucial for determining the optimal amounts of starting materials and for scaling reactions up or down. It prevents waste, saves cost by ensuring precise measurements, and helps in impacting environmental conservation positively by reducing unnecessary chemical usage.