Question

If the rate laws are expressed using \(\mathrm{mol} \mathrm{L}^{-1}\) for concentrations and \(\mathrm{Pa}\) for pressure, what are the units of the second and third order rate constants, \(k_{2}\) and \(k_{3}\) ? Calculate the conversion factor for converting \(k_{2}\) values obtained in the units above to ones using molecules per \(\mathrm{cm}^{3}\) for concentration and atm for pressure.

Short answer

Units of \(k_2\) are \(\mathrm{L} \, \mathrm{mol}^{-1} \, \mathrm{s}^{-1}\) and \(k_3\) are \(\mathrm{L}^2 \, \mathrm{mol}^{-2} \, \mathrm{s}^{-1}\). Use conversion factors to change units for \(k_2\).

Step-by-step solution

Understand the Problem

We need to find the units of the second and third order rate constants \(k_2\) and \(k_3\). Additionally, we must convert \(k_2\) from units using \(\mathrm{mol}\, \mathrm{L}^{-1}\) and \(\mathrm{Pa}\) to units using molecules per \(\mathrm{cm}^{3}\) and \(\mathrm{atm}\). This involves dimensional analysis to express these rate constants in the appropriate units.

Second Order Rate Constant Units \(k_2\)

For a second order reaction, the rate law is \( \text{Rate} = k_2 [A][B] \). The rate, typically in \(\mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}\), involves two concentration terms \(\mathrm{mol} \, \mathrm{L}^{-1}\). Therefore, \(k_2\) must have units: \[ k_2 = \frac{\mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}}{(\mathrm{mol} \, \mathrm{L}^{-1})(\mathrm{mol} \, \mathrm{L}^{-1})} = \mathrm{L} \, \mathrm{mol}^{-1} \, \mathrm{s}^{-1} \]

Third Order Rate Constant Units \(k_3\)

For a third order reaction, the rate law is \( \text{Rate} = k_3 [A][B][C] \). This involves three concentration terms \(\mathrm{mol} \, \mathrm{L}^{-1}\). Therefore, \(k_3\) must have units: \[ k_3 = \frac{\mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}}{(\mathrm{mol} \, \mathrm{L}^{-1})^3} = \mathrm{L}^2 \, \mathrm{mol}^{-2} \, \mathrm{s}^{-1} \]

Convert Units for \(k_2\)

Now, we need to convert \(k_2\) from \(\mathrm{L} \, \mathrm{mol}^{-1} \, \mathrm{s}^{-1}\) to units using molecules per \(\mathrm{cm}^{3}\) and \(\mathrm{atm}\). Use the conversion factors: \(1 \text{ mol} = 6.022 \times 10^{23} \text{ molecules}\), \(1 \mathrm{L} = 1000 \mathrm{cm}^3\), and \(101325 \mathrm{Pa} = 1 \mathrm{atm}\). For \(k_2\), the conversion factor is: \[\left(\frac{1 \text{ mol}}{6.022 \times 10^{23} \text{ molecules}}\right)\left(\frac{1000 \mathrm{cm}^3}{1 \mathrm{L}}\right)\left(\frac{101325 \mathrm{Pa}}{1 \mathrm{atm}}\right)\]This converts the original units to: \( \text{cm}^{3} \text{ molecule}^{-1} \mathrm{s}^{-1}\), accounting for the change in pressure unit.

Key concepts

Second Order Reaction

In a second order reaction, the rate at which the reaction occurs is proportional to the concentration of two reactants. The rate law for such reactions is given by the expression \( \text{Rate} = k_2 [A][B] \). This equation indicates that the rate of the reaction depends on the product of the concentrations of reactants \([A]\) and \([B]\).

The rate constant, \(k_2\), for a second order reaction needs specific units to balance the equation. Since the rate is typically measured in \( \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1} \), and we have two concentration terms in \( \mathrm{mol} \, \mathrm{L}^{-1} \), the units for \(k_2\) are calculated by:

• Dividing the rate units by the product of the concentration units.
• Resulting in \(k_2 = \mathrm{L} \, \mathrm{mol}^{-1} \, \mathrm{s}^{-1} \).

These units indicate that \(k_2\) adjusts the rate of reaction to reflect the concentration of reactants involved.

Third Order Reaction

Third order reactions involve three reactant concentrations in their rate law: \( \text{Rate} = k_3 [A][B][C] \). For these reactions, the dependence on each reactant is such that the reaction rate increases three-fold when each concentration rises.

The units for the third order rate constant, \(k_3\), differ from second order reactions because there are three concentration terms in the equation. The unit calculation steps are as follows:

• Express the rate in \( \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1} \).
• Divide by the concentration units raised to the power of three \( \left(\mathrm{mol} \, \mathrm{L}^{-1}\right)^3 \).
• This results in units of \(k_3 = \mathrm{L}^2 \, \mathrm{mol}^{-2} \, \mathrm{s}^{-1} \).

Such units ensure the rate law for a third order reaction remains consistent across varied conditions.

Dimensional Analysis

Dimensional analysis is a crucial method used to ensure that equations or computations in chemistry maintain consistent units throughout the reaction or conversion process. It's essential for verifying the correctness of equations in terms of units.

For instance, when determining the units for rate constants \(k_2\) and \(k_3\), dimensional analysis involves breaking down the units of each component in the rate law:

• Identifying units for rate as \( \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1} \).
• Analyzing concentration units, which are \( \mathrm{mol} \, \mathrm{L}^{-1} \).
• Calculating the consistent units for \(k_2\) and \(k_3\) as shown in the previous sections.

Through dimensional analysis, conversion between different unit systems is performed accurately to ensure a seamless transition between measured and calculated results.

Unit Conversion

Unit conversion is fundamental in chemistry, especially when dealing with physical quantities like rate constants which can change depending on the measurement system.

When converting units for \(k_2\), the conversion from \( \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1} \) to \( \text{cm}^{3} \text{ molecule}^{-1} \mathrm{s}^{-1} \) involves:

• Using Avogadro's number to convert moles to molecules: \(1 \text{ mol} = 6.022 \times 10^{23} \text{ molecules}\).
• Converting liters to cubic centimeters: \(1 \mathrm{L} = 1000 \mathrm{cm}^3\).
• Switching from pressure in pascals to atmospheres: \(101325 \mathrm{Pa} = 1 \mathrm{atm}\).

These conversions require multiplying the initial units by the corresponding factors, ensuring accuracy of the rate constant's representation across different systems.

Pressure Units

Pressure is a key variable in reactions, often expressed in units like pascals (\( \mathrm{Pa} \)) or atmospheres (\( \mathrm{atm} \)). The choice of unit depends on the context of the pressure's application.

Each unit of pressure interacts differently in various formulas or conversion scenarios:

• Pascals are a standard SI unit, convenient for scientific computations.
• Atmospheres often make calculations and interpretations relatable to everyday conditions.

In the context of converting rate constants:

• Pa and atm conversions reflect the real-world implications of changes in pressure in chemical reactions.
• Maintaining unit consistency ensures that any applied mathematics remains accurate within the physical setting of reactant interaction.

Pressure units highlight the adaptability necessary when dealing with different physical environments in chemistry.