# 30.8: The Velocity and Angular Distribution of the Products of a Reactive Collision


## Calculation of the speed distribution for DF molecules based upon the vibrational state of the molecule

This section describes how data from crossed molecular beam experiments allow us to describe the velocity and angular distribution of particles produced by the simple bi-molecular collision

$\ce{F(g) + D_2(g) -> DF(g) + D(g)} \label{react1}$

As noted in the previous section, the internal vibrational energy of the products will affect the velocity and distribution because if $$\ce{DF(g)}$$ is a harmonic oscillator, then

$E_{P_{(trans)}} + E_{P_{(vib)}} = 165 \dfrac{kJ}{mol} \nonumber$

and

$E_{P_{(vib)}} = \left( v + \dfrac{1}{2} \right)(34.8 \dfrac{kJ}{mol}) \nonumber$

Thus

\begin{align} E_{P_{(trans)}} + E_{P_{(vib)}} &= \dfrac{1}{2}\mu_Pv_{P_r} + \left( v + \dfrac{1}{2} \right)\left(34.8 \dfrac{kJ}{mol}\right) \nonumber \\[4pt] &= 165 \dfrac{kJ}{mol} \label{30.8.1} \end{align}

The reduced mass of the products $$\mu_P$$ is

$\dfrac{(21.01 \, g/mol)(2.014 \, g/mol)}{21.01 \, g/mol + 2.014 \, g/mol} = 1.84 \, g/mol = 0.00184 \, kg/mol \nonumber$

so that solving for the product velocity in Equation $$\ref{30.8.1}$$ for $$v_{P_r}$$ gives

$v_{P_r} = \sqrt{\left(\dfrac{2}{0.00184 \, kg/mol}\right) \left(1.65 \times 10^5 \, \dfrac{J}{mol} - \left(v + \dfrac{1}{2}\right)(3.48 \times 10^4) \dfrac{J}{mol} \right) } \nonumber$

It can be shown that the corresponding relative velocity of the $$\ce{DF}$$ molecule and the center of mass,$$|\vec v_{DF} - \vec v_{cm} |$$, is equal to $$\dfrac{m_D}{M}v_{P_r}$$ for this reaction, with $$M = m_{D_2} + m_F$$. Table $$\PageIndex{1}$$ shows the values of $$v_{P_r}$$ and $$|\vec v_{DF} - \vec v_{cm} |$$ for vibrational states $$v = 0-4$$.

Table $$\PageIndex{1}$$
Vibrational state, $$v$$ relative velocity of the products, $$v_{P_r}$$ in m/s $$|\vec v_{DF} - \vec v_{cm} |$$, in m/s
0 12,700 1111
1 11,100 971
2 9270 811
3 6930 606
4 3200 280

## Analysis of Crossed Molecular Beam Data

This link takes you to the 1986 Nobel Prize acceptance lecture of Yuan Tseh Lee, in which he describes the crossed molecular beam studies that his group carried out on the reaction \ref{react1}. On page 3 of the lecture, you will see a center-of-mass velocity flux contour map for the reaction. This map measures the speed distribution and the angular dispersion distribution as a result of collisions between one $$D_2$$ molecule approaching from the left and one F atom approaching from the right. Moving away from the center of the graph in any direction represents an increase in speed. The dashed line circles represent the maximum speed that a molecule in that given vibrational state can obtain. Notice that the highest vibrational state is closest to the nucleus. Thus, the map is consistent with theory, which claims that as the vibrational energy increases, the translational energy decreases.

The contour areas on the map represent a constant number of DF(g) molecules. The more closely spaced the contour lines, the greater the number of $$\ce{DF(g)}$$ molecules with that velocity in that vibrational state. An estimate of the population distribution is that roughly half the molecules are in the $$v = 3$$ state, roughly 25% of the molecules occupy the $$v = 2$$ state, and roughly 25% of the molecules occupy the $$v = 4$$ state. Very few molecules occupy the $$v = 0$$ or $$v = 1$$ states. The fact that there are molecules with velocities between the dashed lines shows that these $$\ce{DF}$$ molecules are in various rotational states within each vibrational state. If $$E_{rot}$$ = 0 and $$J = 0$$, the molecular velocity distribution would be centered on each dashed line. But because these molecules do exist in various rotational states, their $$E_{rot}$$ is greater than 0, and their translational energy is in between the energies of those molecules in rotational ground states. The distribution of vibrational states among the $$\ce{DF}$$ product molecules more closely resembles a normal distribution (i.e., Gaussian-like) than a Boltzmann distribution.

##### Example $$\PageIndex{1}$$

Determine the populations of the five lowest vibrational states of $$\ce{DF}(v)$$ relative to $$\ce{DF}(v=3)$$, assuming that the distribution is in thermal equilibrium at 298 K. You can assume that the $$\ce{DF}$$ molecules act as harmonic oscillators with $$\tilde{v_{DF}}$$ = 2907 cm-1.

###### Solution

If the $$\ce{DF(g)}$$ molecules are in thermal equilibrium, the population of vibrational states will follow a Boltzmann distribution. Therefore,

$N(v) = \Large{e}^{\small{-\dfrac{(v + 1/2)h\nu_{DF}}{k_BT}}}\nonumber$

and

$N(v =3) = \Large{e}^{\small{-\dfrac{(3 + 1/2)h\nu_{DF}}{k_BT}}}\nonumber$

so

$\dfrac{N(v)}{N(v=3)} = \Large{e}^{\small{-\dfrac{(v -3)h\nu_{DF}}{k_BT}}}\nonumber$

$$v$$ $$\dfrac{N(v)}{N(v=3)}$$
0 2.03 x 1018
1 1.60 x 1012
2 1.26 x 106
3 1
4 7.90 x 10-7

It is clear that when the molecules are at thermal equilibrium, the population of the vibrational states decreases with an increasing $$v$$, which is not the population pattern found from the experiment.

The map also shows that a majority of $$\ce{DF(g)}$$ molecules generally head back towards the direction from which the F atom originally approached the collision. Some even head directly back from whence they came. This type of collision is called a rebound reaction. We will see in the next section that not all reactions are rebound reactions.

A different style of contour map is shown in figure $$\PageIndex{1}$$. In this view, the density of the dots and the intensity of their color represent the number of particles being scattered at specific angles. This map does not show the vibrational state of the DF molecules.

30.8: The Velocity and Angular Distribution of the Products of a Reactive Collision is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.