30.7: Reactions Can Produce Vibrationally Excited Molecules

As shown at the end of section 30.5, using the center of mass reaction model allows us to state that

\[E_{R_{(int)}} + E_{R_{(trans)}} = E_{P_{(int)}} + E_{P_{(trans)}} \label{30.7.1} \]

where \(E_{R_{(int)}}\) and \(E_{P_{(int)}}\) represent the rotational, vibrational, and electronic energies collectively described as the internal energy of the reactants and the products, respectively. If we apply equation \(\ref{30.7.1}\) to the well-studied gas-phase reaction

\[\ce{F(g) + D_2(g) -> DF(g) + D(g)} \nonumber \]

we can discuss the distribution of a fixed total collision energy between \(E_{P_{(int)}}\) and \(E_{P_{(trans)}}\). Specifically, if we assume that \(\ce{F(g)}\) and \(\ce{D(g)}\) are in their ground electronic states, that \(\ce{D2(g)}\) is in its ground rotational, vibrational, and electronic states, and that \(\ce{DF(g)}\) is in its ground rotational and electronic states, we can focus on the possible vibrational states of the \(\ce{DF(g)}\) that can be populated. Figure \(\PageIndex{1}\) shows the potential energy curve of the process that is described below.

Figure \(\PageIndex{1}\): The potential energy diagram for the reaction \(\ce{F(g) + D_2(g) -> DF(g) + D(g)} \). ( CC BY-NC; Ümit Kaya via LibreTexts)

Expanding Equation \ref{30.7.1} to describe this assumed process, we get

\[E_{R_{(rot)}} + E_{R_{(vib)}} + E_{R_{(elec)}} + E_{R_{(trans)}} = E_{P_{(rot)}} + E_{P_{(vib)}} + E_{P_{(elec)}} + E_{P_{(trans)}} \label{30.7.2} \]

Let's assume that \(\ce{D2}\) and \(\ce{DF}\) are harmonic oscillators with \(\tilde{\nu}_{D_2}\) = 2990 cm^-1 and \(\tilde{\nu}_{DF}\) = 2907 cm^-1 .

If D₂ and DF are in their ground electronic states, then \(E_{R_{(elec)}} = -D_e(D_2)\) and \(E_{P_{(elec)}} = -D_e(DF)\). From experiments, we know that

\[-D_e(DF) - (-D_e(D_2)) = -140 kJ/mol \nonumber \]

We also know from experiments, that the activation energy for this reaction is about 6 kJ/mol¹, so if we assume that the relative translational energy of the reactants is 7.1 kJ/mol, they will have sufficient energy to overcome the activation energy barrier.

With these data, we can write Equation \(\ref{30.7.2}\) as

\[0 + E_{R_{(vib)}} - D_e(D_2) + 7.1 \dfrac{kJ}{mol} = 0 + E_{P_{(vib)}} - D_e(DF) + E_{P_{(trans)}} \nonumber \]

then

\[E_{R_{(vib)}} + 7.1 \dfrac{kJ}{mol} = E_{P_{(vib)}} + (-D_e(DF) - (-D_e(D_2)) + E_{P_{(trans)}} \nonumber \]

then

\[E_{R_{(vib)}} + 7.1 \dfrac{kJ}{mol} = E_{P_{(vib)}} - 140 \dfrac{kJ}{mol} + E_{P_{(trans)}}\nonumber \]

Because \(\ce{D2}\) is in its ground vibrational state, \(E_{R_{(vib)}} = \dfrac{1}{2}h\nu_{D_2} = 17.9 \dfrac{kJ}{mol}\)

Putting this altogether,

\[17.9 \dfrac{kJ}{mol} + 140 \dfrac{kJ}{mol} + 7.1 \dfrac{kJ}{mol} - E_{P_{(vib)}} = E_{P_{(trans)}} \nonumber \]

\[165 \dfrac{kJ}{mol} - E_{P_{(vib)}} = E_{P_{(trans)}} \label{30.7.3} \]

Equation \(\ref{30.7.3}\) tells us that the reaction will occur only if \(E_{P_{(vib)}}\) is less than 165 kJ/mol because \(E_{P_{(trans)}}\) must be a positive value.

If \(\ce{DF(g)}\) is a harmonic oscillator, then

\[\begin{align*} E_{P_{(vib)}} &= \left( v + \dfrac{1}{2} \right) h\nu_{DF} \\[4pt] &= \left( v + \dfrac{1}{2} \right)(34.8\, kJ/mol) < 165\, kJ/mol \end{align*} \nonumber \]

Vibrational states, \(v\) = 0, 1, 2, 3, and 4 will result in \(E_{P_{(vib)}}\) \(\leq\) 165 kJ/mol. Thus \(\ce{DF(g)}\) molecules in these five vibrational states will be produced by the reaction. Note that under these set of assumptions, the \(\ce{DF}\) molecules produced in different vibrational states will have correspondingly different \(E_{P_{(trans)}} \).

¹Average of experimental E_a values listed at https://kinetics.nist.gov/kinetics/ accessed 9/22/2021