How to treat stretch bond to calculate its enthalpy and entropy contribution?

Feb 3, 2018

For that you'll need to calculate the vibrational partition function, which allows you to get the energy, and then the enthalpy contribution from just vibrations. The entropy comes directly from the vibrational partition function and energy.

Since you apparently want this to be done using only frequencies and a temperature...

${\Theta}_{v i b , i} = t i l \mathrm{de} {\omega}_{i} / {k}_{B}$, ${k}_{B} = \text{0.695 cm"^(-1)"/K}$, $t i l \mathrm{de} {\omega}_{i} = {\text{frequency of that stretching mode in cm}}^{- 1}$.

${q}_{v i b , i} = {e}^{- {\Theta}_{v i b , i} / 2 T} / \left(1 - {e}^{- {\Theta}_{v i b , i} / T}\right)$

${q}_{v i b} = {\prod}_{i} {q}_{v i b , i} = {q}_{v i b , 1} {q}_{v i b , 2} \cdots$

${\left(E - {E}_{0}\right)}_{v i b} / N = {\left\langle\epsilon\right\rangle}_{v i b} = {k}_{B} \left[{\Theta}_{v i b} / \left(2\right) + {\Theta}_{v i b} / \left({e}^{{\Theta}_{v i b} / T} - 1\right)\right]$

${\left(H - {H}_{0}\right)}_{v i b} / N = {\left\langle\eta\right\rangle}_{v i b} = {\left\langle\epsilon\right\rangle}_{v i b} + {k}_{B} T$

${S}_{v i b} / N = {\left\langles\right\rangle}_{v i b} = {k}_{B} \ln {q}_{v i b} + \frac{{\left\langle\epsilon\right\rangle}_{v i b}}{T}$

Enthalpy units here are in $\text{cm"^(-1)"/molecule}$ and entropy units here are in $\text{cm"^(-1)"/molecule"cdot"K}$.

To check your work, you should get ${\left\langles\right\rangle}_{v i b} = \text{0.0129 cm"^(-1)"/molecule"cdot"K}$ for methane's four IR-active vibrational modes combined, each having (NIST) $t i l \mathrm{de} {\omega}_{i} = 2917 , 1534 , 3019 , {\text{1306 cm}}^{- 1}$, and degeneracies $1 , 2 , 3 , 3$.

(The degeneracy doesn't matter unless you want the heat capacities.)

Any partition function is written as

$q = {\sum}_{i} {g}_{i} {e}^{- \beta {\epsilon}_{i}}$

where ${g}_{i}$ is the degeneracy of state $i$, ${\epsilon}_{i}$ is the energy of state $i$, and $\beta = 1 / {k}_{B} T$ is a constant. ${k}_{B}$ is the Boltzmann constant and $T$ is temperature in $\text{K}$.

For vibrational energy states in a one-dimensional harmonic oscillator, ${g}_{\upsilon} = 1$, and epsilon_upsilon = ℏomega(upsilon+1/2), so we can write:

q_(vib) = sum_(upsilon) e^(-ℏomega(upsilon + 1/2)//k_BT)

Define the vibrational temperature as Theta_(vib) = (ℏomega)/(k_B), if ${k}_{B}$ is in $\text{J/K}$, or $t i l \mathrm{de} \frac{\omega}{k} _ B$ if ${k}_{B}$ is in $\text{cm"^(-1)"/K}$. Then:

${q}_{v i b} = {\sum}_{\upsilon} {e}^{- \left(\upsilon + \frac{1}{2}\right) {\Theta}_{v i b} / T}$

$= {e}^{- {\Theta}_{v i b} / 2 T} {\sum}_{\upsilon} {e}^{- \upsilon {\Theta}_{v i b} / T}$

$= {e}^{- {\Theta}_{v i b} / 2 T} {\sum}_{\upsilon} {\left({e}^{- {\Theta}_{v i b} / T}\right)}^{\upsilon}$

this has a nice power series solution of

$\textcolor{b l u e}{{q}_{v i b} = {e}^{- {\Theta}_{v i b} / 2 T} / \left(1 - {e}^{- {\Theta}_{v i b} / T}\right)}$

Next, the molecular energy is

$\textcolor{b l u e}{{\left\langle\epsilon\right\rangle}_{v i b}} = - {\left(\frac{\partial \ln {q}_{v i b}}{\partial \beta}\right)}_{V}$

$= {k}_{B} {T}^{2} {\left(\frac{\partial \ln {q}_{v i b}}{\partial T}\right)}_{V}$

$= {k}_{B} {T}^{2} {\left(\frac{\partial \ln \left[{e}^{- {\Theta}_{v i b} / 2 T} / \left(1 - {e}^{- {\Theta}_{v i b} / T}\right)\right]}{\partial T}\right)}_{V}$

$= {k}_{B} {T}^{2} {\left(\frac{\partial \left(- {\Theta}_{v i b} / \left(2 T\right) - \ln \left(1 - {e}^{- {\Theta}_{v i b} / T}\right)\right)}{\partial T}\right)}_{V}$

$= {k}_{B} {T}^{2} \left[{\Theta}_{v i b} / \left(2 {T}^{2}\right) - \frac{1}{1 - {e}^{- {\Theta}_{v i b} / T}} \cdot - {\Theta}_{v i b} / {T}^{2} \cdot {e}^{- {\Theta}_{v i b} / T}\right]$

$= {k}_{B} \left[{\Theta}_{v i b} / \left(2\right) + \frac{{\Theta}_{v i b} {e}^{- {\Theta}_{v i b} / T}}{1 - {e}^{- {\Theta}_{v i b} / T}}\right]$

$= \textcolor{b l u e}{{k}_{B} \left[{\Theta}_{v i b} / \left(2\right) + {\Theta}_{v i b} / \left({e}^{{\Theta}_{v i b} / T} - 1\right)\right]}$

If we want the enthalpy, $\left\langle\epsilon\right\rangle$ is related to ${C}_{V} / N$, and ${C}_{V} / N + {k}_{B} = {C}_{P} / N$, and $\left\langle\eta\right\rangle$ is related to ${C}_{P} / N$. Fortunately, since

${C}_{V} / N = {\left(\frac{\partial \left\langle\epsilon\right\rangle}{\partial T}\right)}_{V}$

then

${C}_{V} / N + {k}_{B} = {C}_{P} / N = {\left(\frac{\partial \left\langle\epsilon\right\rangle}{\partial T}\right)}_{V} + {k}_{B}$

and so, the vibrational molecular enthalpy is:

${\left(H - {H}_{0}\right)}_{v i b} / N = \textcolor{b l u e}{{\left\langle\eta\right\rangle}_{v i b}}$

$= \int {\left(\frac{\partial \left\langle\epsilon\right\rangle}{\partial T}\right)}_{V} + {k}_{B} \mathrm{dT}$

$= \textcolor{b l u e}{{\left\langle\epsilon\right\rangle}_{v i b} + {k}_{B} T}$

Finally, if we want the molecular entropy, then since I'm lazy, here is the TOTAL entropy from my homework. An equilibrium distribution of particles in everyday life allows this derivation. And the vibrational part only is for internal motions. Thus, the vibrational molecular entropy is:

${S}_{v i b} / N = \textcolor{b l u e}{{\left\langles\right\rangle}_{v i b} = {k}_{B} \ln {q}_{v i b} + \frac{{\left\langle\epsilon\right\rangle}_{v i b}}{T}}$