Question

The following reaction is first order: \(\mathrm{C}_{2} \mathrm{H}_{6} \rightarrow 2 \mathrm{CH}_{3}\). If the rate constant is equal to \(5.5\) \(\times 10^{-4} \mathrm{~s}^{-1}\) at \(1000 \mathrm{~K}\), how long will it take for \(0.35\) mol of \(\mathrm{C}_{2} \mathrm{H}_{6}\) in a \(1.00 \mathrm{~L}\) container to decrease to \(0.10 \mathrm{~mol}\) in the same container? a. \(38 \mathrm{~min}\) b. \(26 \mathrm{~min}\) c. \(19 \mathrm{~min}\) d. \(68 \mathrm{~min}\)

Short answer

It takes approximately 38 minutes for the reaction.

Step-by-step solution

Identify the formula

For a first-order reaction, the formula to calculate the time required is: \[t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]} \right)\] where \(t\) is the time, \(k\) is the rate constant, \([A]_0\) is the initial concentration of the reactant, and \([A]\) is the concentration of the reactant at time \(t\).

Convert amounts to concentrations

Since the reaction occurs in a 1.00 L container, the initial concentration \([A]_0\) is 0.35 mol/L and the concentration \([A]\) at time \(t\) is 0.10 mol/L.

Substitute the known values into the equation

Using the formula from Step 1, substitute the given values:\[t = \frac{1}{5.5 \times 10^{-4} \, s^{-1}} \ln \left( \frac{0.35}{0.10} \right)\]Solve for \(t\): first, calculate the natural log, then divide by the rate constant.

Calculate the natural logarithm

Calculate \(\ln \left( \frac{0.35}{0.10} \right)\) to find the factor by which the initial concentration has changed:\[\ln \left( \frac{0.35}{0.10} \right) = \ln (3.5) = 1.25276\]

Calculate time

Now calculate \(t\):\[t = \frac{1}{5.5 \times 10^{-4}} \times 1.25276 = 2277.75 \, \text{seconds}\]Convert this time into minutes by dividing by 60:\[\frac{2277.75}{60} \approx 37.96 \, \text{minutes} \approx 38 \, \text{minutes}\]

Choose the correct option

The calculated time is approximately 38 minutes, which corresponds to option (a).

Key concepts

Rate Constant

In the world of reaction kinetics, the rate constant, often denoted as \( k \), is a vital parameter. It defines the speed at which a chemical reaction proceeds. For first-order reactions, like the decomposition of ethane, the rate constant has the unit \( \mathrm{s^{-1}} \). This indicates that the reaction rate depends linearly on the concentration of one reactant.

The smaller the rate constant, the slower the reaction. Conversely, a higher rate constant leads to a faster reaction. In the given exercise, the rate constant is \( 5.5 \times 10^{-4} \mathrm{~s}^{-1} \). This tells us how the concentration of the reactant decreases over time in a first-order reaction.

• The rate constant remains constant only under steady conditions, like constant temperature.
• It's unique to every reaction and can vary with changes in temperature or the presence of a catalyst.

Understanding the rate constant helps in predicting how long a reaction will take to complete, making it a cornerstone of chemical kinetics.

Natural Logarithm

The natural logarithm (\( \ln \)) is a mathematical function that is frequently used in kinetics, particularly for first-order reactions. It helps in simplifying the exponential decrease of reactant concentration over time.

In the formula \( t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]} \right) \), the natural logarithm converts the ratio of concentrations into a linear scale, which allows us to calculate the time needed for the reaction.

• Natural logarithms are expressed as powers of Euler's number \( e \), which is approximately 2.718.
• They play a crucial role in dealing with exponential growth or decay models, like the decrease of reactants in a chemical reaction.

In our context, \( \ln(3.5) \approx 1.25276 \), helps us determine how much the reactant concentration has dropped, allowing us to calculate the specific time elapsed.

Time Calculation for Reactions

One of the main goals in the study of kinetics is to forecast the time it takes for a reaction to reach a certain completion point. For first-order reactions, you can do this using the equation: \[ t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]} \right) \] Here, \( t \) is the time variable we need to determine. By plugging in the rate constant and the concentration values into this formula, we can find the duration needed for a concentration shift.

For instance, substituting the given values \( k = 5.5 \times 10^{-4} \mathrm{~s}^{-1} \), \( [A]_0 = 0.35 \) mol/L, and \( [A] = 0.10 \) mol/L into the equation gives us the reaction time in seconds.

• This calculation is significant because it allows scientists to control and optimize reactions in various industries.
• Finding the precise time for reactions helps in managing resources and predicting the outcomes in production settings.

In this problem, you will find that the time calculated amounts to approximately 38 minutes.

Concentration Conversions

Concentration is a key factor in chemical reactions and is typically measured in moles per liter (mol/L), emphasizing the number of moles of a substance in a given volume. In first-order reaction calculations, getting these conversions right is crucial.

For the exercise, we consider the ethanol molecule concentration change, from an initial \[ [A]_0 = 0.35 \text{ mol/L} \] to \[ [A] = 0.10 \text{ mol/L} \] within a 1.00 L container. Recognizing these concentrations correctly allows us to apply the kinetics equations accurately.

• Moles are a standard unit in chemistry expressing the amount of a chemical substance.
• Converting quantities to concentrations helps in directly applying them to kinetic formulas, especially when assessing reaction rates over a period.

This attention to concentration details ensures precise calculations, impacting the understanding of how reactants convert in reactions.