Question

A gas mixture containing \(\mathrm{CH}_{3}\) fragments, \(\mathrm{C}_{2} \mathrm{H}_{6}\) molecules, and an inert gas (He) was prepared at \(600 \mathrm{~K}\) with a total pressure of 5.42 atm. The elementary reaction $$ \mathrm{CH}_{3}+\mathrm{C}_{2} \mathrm{H}_{6} \longrightarrow \mathrm{CH}_{4}+\mathrm{C}_{2} \mathrm{H}_{5} $$ has a second-order rate constant of \(3.0 \times 10^{4} / M \cdot \mathrm{s} .\) Given that the mole fractions of \(\mathrm{CH}_{3}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) are 0.00093 and 0.00077 , respectively, calculate the initial rate of the reaction at this temperature.

Short answer

The initial reaction rate is approximately \(2.59 \times 10^{-4}\, \text{M/s}.\)

Step-by-step solution

Calculate partial pressures

First, calculate the partial pressures of \(\mathrm{CH}_{3}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) using the mole fractions and the total pressure. The partial pressure \(P_i\) for a gas component can be calculated using the formula: \[ P_i = X_i \times P_{\text{total}} \]where \(X_i\) is the mole fraction and \(P_{\text{total}}\) is the total pressure.For \(\mathrm{CH}_{3}\):\[ P_{\mathrm{CH}_{3}} = 0.00093 \times 5.42\, \text{atm} = 0.0050406\, \text{atm} \]For \(\mathrm{C}_{2} \mathrm{H}_{6}\):\[ P_{\mathrm{C}_{2} \mathrm{H}_{6}} = 0.00077 \times 5.42\, \text{atm} = 0.0041734\, \text{atm} \]

Calculate initial concentrations

Convert the partial pressures to concentrations using the ideal gas law, given by the formula:\[ C_i = \frac{P_i}{RT} \]where \(R\) is the gas constant \(0.0821\, \text{L atm / mol K}\) and \(T = 600\, \text{K}\).For \(\mathrm{CH}_{3}\): \[ C_{\mathrm{CH}_{3}} = \frac{0.0050406}{0.0821 \times 600} \approx 1.02 \times 10^{-4} \text{ M} \]For \(\mathrm{C}_{2} \mathrm{H}_{6}\): \[ C_{\mathrm{C}_{2} \mathrm{H}_{6}} = \frac{0.0041734}{0.0821 \times 600} \approx 8.47 \times 10^{-5} \text{ M} \]

Calculate initial reaction rate

The initial rate of the reaction can be calculated using the rate equation for a second-order reaction:\[ \text{Rate} = k \times [\mathrm{CH}_{3}] \times [\mathrm{C}_{2} \mathrm{H}_{6}] \]where \(k = 3.0 \times 10^{4} / M \cdot \text{s}\).Substitute the concentrations:\[ \text{Rate} = 3.0 \times 10^{4} \times 1.02 \times 10^{-4} \times 8.47 \times 10^{-5} \approx 2.59 \times 10^{-4} \text{ M/s} \]

Key concepts

Partial Pressure

Partial pressure is a key concept in understanding gas mixtures. When we have a mixture of various gases, each type of gas exerts its own pressure as if it were alone in the container. This is called the partial pressure. To calculate the partial pressure, you need to know the mole fraction, which represents the fraction of the total number of moles contributed by a particular gas in the mixture.

This can be computed using the formula:

• \[ P_i = X_i \times P_{\text{total}} \]

Where \( P_i \) is the partial pressure, \( X_i \) is the mole fraction, and \( P_{\text{total}} \) is the total pressure.

In our case, for

• \( \mathrm{CH}_3 \), we multiply its mole fraction, 0.00093, by the total pressure, 5.42 atm, to find its partial pressure \( P_{\mathrm{CH}_3} = 0.0050406 \) atm.
• Similarly, for \( \mathrm{C}_2 \mathrm{H}_6 \), its partial pressure is \( 0.0041734 \) atm.

Understanding partial pressure is essential for further calculations in gas reactions.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is usually expressed as:

• \[ PV = nRT \]

Where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the universal gas constant (0.0821 L atm / mol K)
• \( T \) is the temperature in Kelvin

In the context of this exercise, we use the ideal gas law to convert the partial pressures of gaseous species to concentrations needed for reaction rate calculations. The rearranged formula is:

• \[ C_i = \frac{P_i}{RT} \]

Using this, we calculate the concentrations of \( \mathrm{CH}_3 \) and \( \mathrm{C}_2 \mathrm{H}_6 \) based on their partial pressures, taking into account the temperature as specified in the exercise.

Second-Order Reaction

Reactions are classified into different orders based on how their reaction rates depend on the concentrations of reactants. A second-order reaction means that the rate is proportional to the square of the concentration of one reactant, or to the product of the concentrations of two different reactants. The rate law is written as:

• \[ \text{Rate} = k \times [A]^2 \]
• or for two reactants: \[ \text{Rate} = k \times [A] \times [B] \]

In our example, the reaction

• \( \mathrm{CH}_3 + \mathrm{C}_2 \mathrm{H}_6 \rightarrow \mathrm{CH}_4 + \mathrm{C}_2 \mathrm{H}_5 \)

is second-order, which means the rate depends on both \( \mathrm{CH}_3 \) and \( \mathrm{C}_2 \mathrm{H}_6 \) concentrations. Therefore, the initial rate is calculated by multiplying the rate constant \( k \) with the concentrations of each reactant.

Rate Constant

The rate constant, \( k \), is a crucial factor in the kinetics of a reaction. It provides information about the speed of the reaction under given conditions. The rate constant has different units depending on the order of the reaction. For a second-order reaction like the one described in the exercise, the units are \( M^{-1} \cdot s^{-1} \).

This different unit indicates that the rate constant accounts for two concentrations being multiplied together.

For the elementary reaction involving \( \mathrm{CH}_3 \) and \( \mathrm{C}_2 \mathrm{H}_6 \) with a given \( k = 3.0 \times 10^4 \) \( M^{-1} \cdot s^{-1} \), it quantifies how efficiently the reaction proceeds when both substances are present. This parameter helps us understand how fast reactants are transformed into products under specific conditions. A high \( k \) value reflects a fast reaction, while a lower \( k \) indicates a slower process.