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# 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.

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 Ea values listed at https://kinetics.nist.gov/kinetics/ accessed 9/22/2021

30.7: Reactions Can Produce Vibrationally Excited Molecules is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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