91Ó°ÊÓ

Calculate the hard-sphere collision theory rate constant for the reaction $$ \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \Longrightarrow \mathrm{NOCl}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) $$ at \(300 \mathrm{~K}\). The collision diameters of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) are \(370 \mathrm{pm}\) and \(540 \mathrm{pm}\), respectively. The Arrhenius parameters for the reaction are \(A=3.981 \times 10^{9} \mathrm{dm}^{3} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}\) and \(E_{\mathrm{a}}=\) \(84.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} .\) Calculate the ratio of the hard-sphere collision theory rate constant to the experimental rate constant at \(300 \mathrm{~K}\).

Step by step solution

01

- Understand the Problem

We need to calculate the ratio of the hard-sphere collision theory rate constant to the experimental rate constant at 300 K for a specific reaction. The given data includes collision diameters of the reactants, Arrhenius parameters, and temperature.

02

- Determine the Collision Rate Constant

The hard-sphere collision rate constant, \(k_c\), is calculated using the equation:\[k_c = Z_{AB} \cdot e^{-E_a/(RT)}\]where \(Z_{AB}\) is the collision frequency, \(R\) is the gas constant (8.314 J/(mol·K)), and \(T\) is the temperature in Kelvin (300 K).

03

- Calculate Collision Frequency

The collision frequency \(Z_{AB}\) is given by:\[Z_{AB} = N_A \sigma (8k_BT/\pi\mu)^{1/2}\]where \(N_A\) is Avogadro's number \(6.022 \times 10^{23} \space mol^{-1}\), \(\sigma\) is the collision cross-section \(= \pi (d_{NO} + d_{Cl_2})^2\), and \(\mu\) is the reduced mass.

04

- Calculate Collision Cross-Section

The collision cross-section \(\sigma\) is calculated as:\[\sigma = \pi (370 + 540)^2 \times 10^{-20} \space m^2\]This results in \(\sigma \approx 8.943 \times 10^{-19} \space m^2\).

05

- Calculate Reduced Mass

Calculate the reduced mass \(\mu\) using the molar masses of NO \(= 30 \space g/mol\) and \(Cl_2 = 70 \space g/mol\):\[\mu = \frac{30 \times 70}{30 + 70} \approx 21 \space g/mol \approx 21 \times 10^{-3} \space kg\]

06

- Calculate Z_{AB}

Substitute the known values into the \(Z_{AB}\) equation:\[Z_{AB} \approx 6.022 \times 10^{23} \times 8.943 \times 10^{-19} \times \left(\frac{8 \times 1.38\times 10^{-23} \times 300}{\pi \times 21\times 10^{-3}}\right)^{1/2}\]Compute \(Z_{AB}\) with numerical values to get a result in collisions per second.

07

- Calculate the Rate Constant k_c

Substitute \(Z_{AB}\), \(E_a = 84.9 \times 10^3\) J/mol, \(R = 8.314\) J/(mol·K), and \(T = 300\) K into the \(k_c\) expression:\[k_c = Z_{AB} \cdot e^{-84.9 \times 10^3/(8.314 \times 300)}\]This yields a value for \(k_c\).

08

- Calculate Experimental Rate Constant

The experimental rate constant \(k_{exp}\) can be derived from the Arrhenius equation:\[k_{exp} = A \cdot e^{-E_a/(RT)}\]Substitute \(A = 3.981 \times 10^9\) dm³/mol·s, \(E_a\), \(R\), and \(T\) to find \(k_{exp}\).

09

- Calculate the Ratio

The ratio is calculated by dividing the hard-sphere collision rate constant \(k_c\) by the experimental rate constant \(k_{exp}\). Compute this ratio using the values obtained in Steps 7 and 8.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Collision Frequency

Collision frequency, denoted as \(Z_{AB}\), is a key part of understanding how often reacting molecules collide during a chemical reaction. This is important because the likelihood of a reaction occurring increases with more frequent collisions.
To calculate \(Z_{AB}\), we use the formula:

• \[Z_{AB} = N_A \sigma \left(\frac{8k_BT}{\pi\mu}\right)^{1/2}\]
• \(N_A\) is Avogadro's number, representing the number of molecules in a mole.
• \(\sigma\) is the collision cross-section, effectively the target area for collisions.
• \(k_B\) is the Boltzmann constant, linking temperature to energy.
• \(T\) is the temperature in Kelvin; in the provided exercise, it's 300 K.
• \(\mu\) is the reduced mass of the system, affecting how masses interact during collision.

High collision frequency means more chances for reactions. The factors influencing \(Z_{AB}\) include temperature, which increases kinetic energy, and the collision cross-section, affected by the size of molecules.

Collision Cross-Section

The collision cross-section, \(\sigma\), plays a critical role in the calculation of collision frequency. It is a measure of the effective area where two molecules can collide, similar to the target size they present to each other. Larger cross-sections mean higher chances of collisions.
To calculate \(\sigma\), you can use the formula:

• \[\sigma = \pi (d_{NO} + d_{Cl_2})^2\]
• \(d_{NO}\) and \(d_{Cl_2}\) are the diameters of the reacting molecules NO and Cl\(_2\), converted into meters.

In the exercise, \(d_{NO}\) is 370 pm (picometers) and \(d_{Cl_2}\) is 540 pm. So, \(\sigma\approx 8.943 \times 10^{-19} \text{ m}^2\).
This calculation shows how big the molecular 'targets' are in a reaction, affecting how often they collide.

Reduced Mass

Reduced mass \(\mu\) is a concept utilized in the context of two-body systems in collisions, providing a simplified way to view mass for calculations like those in collision frequency. Imagine two objects joined by a hypothetical rod.
By using reduced mass, the calculations resemble those for a single point mass.

• The reduced mass \(\mu\) for two interacting particles is calculated by:
• \[\mu = \frac{m_1 m_2}{m_1 + m_2}\]
• Here, \(m_1\) and \(m_2\) are the masses of the two reacting species.

In the exercise, the masses are provided as molar masses: 30 g/mol for NO and 70 g/mol for Cl\(_2\).
This gives a \(\mu\approx 21 \times 10^{-3} \text{ kg/mol}\).
Using reduced mass simplifies mathematical work and provides a focused measure of the dynamics between the two interacting particles.

Arrhenius Equation

The Arrhenius equation is key for determining reactions' rate constants under different conditions. It ties the rate constant \(k_{exp}\) to temperature and activation energy \(E_a\), showing how temperature affects the speed of reactions.
The equation is:

• \[k_{exp} = A \cdot e^{-E_a/(RT)}\]
• \(A\) is the pre-exponential factor, representing the frequency of collisions resulting in a reaction.
• \(E_a\) is the activation energy required to start a reaction, provided here as 84.9 kJ/mol.
• \(R\) is the universal gas constant, \(8.314 \text{ J/mol·K}\).
• \(T\) is the temperature in Kelvin.

By substituting the given values, you can calculate the rate at which the reaction happens at 300 K.
Understanding this allows you to see why reactions speed up as it gets warmer, with higher temperature increasing \(e^{-E_a/(RT)}\), thus boosting \(k_{exp}\).