Chapter 13, Problem 13E.11P

Interpretation Introduction

Interpretation:

The standard molar entropy of ${\text{F}}_2^{-}$ at $\text{298}\text{K}$ has to be stated.

Concept introduction:

The randomness present in the system is known as its entropy.  It is an extensive property.  At equilibrium, the entropy of the system is zero.  It is denoted by $S$.  The entropy that is measured under the standard conditions of temperature is known as standard entropy.  The standard entropy change is denoted by $\Delta S$.

Answer to Problem 13E.11P

The standard molar entropy of ${\text{F}}_2^{-}$ at $\text{298}\text{K}$ is $\underset{\_}{199.4Jmo{l}^{-1}{K}^{-1}}$.

Explanation of Solution

The given value of fundamental vibrational wavenumber for the ground state of ${\text{F}}_2^{-}$, ${}^2{\displaystyle {\sum }_{\text{u}}^{\ast }}$, is $450{\text{cm}}^{-1}$.

The value of first two excited states of ${\text{F}}_2^{-}$ is $1.609\text{eV}$ and $1.702\text{eV}$.

The value of equilibrium internuclear distance is $190\text{pm}$ or $190\times {10}^{-12}\text{m}$.

The molar mass of ${\text{F}}_2^{-}$ is $\begin{array}{c}2\times 18.998\text{g/mol}=\text{37}\text{.996}\text{g/mol}\\ =\text{38}\text{.00}\text{g/mol}\end{array}$.

The value of Planck’s constant is $6.626068\times {10}^{-34}\text{J}\text{s}$.

The value of speed of light is $2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}$.

The value of Boltzmann’s constant is $1.3806503\times {10}^{-23}\text{J}{\text{K}}^{-1}$.

The expression to show the molar entropy, $S_{\text{m}}$  is given below.

    $S_{\text{m}}=\frac{U_{\text{m}}-U_{\text{m}}\left(0\right)}{T}+R\left(\ln \frac{q_{\text{m}}}{N_{\text{A}}}-1\right)$        (1)

Where,

• $U$ is the internal energy.
• $T$ is the temperature.
• $q_{\text{m}}$ is the molecular partition function.
• $N_{\text{A}}$ is the Avogadro’s number.
• $R$ is the universal gas constant.

The expression for the internal energy is given below.

    $\frac{U_{\text{m}}-U_{\text{m}}\left(0\right)}{T}=-N_{\text{A}}{\left(\frac{\partial \ln q_m}{\partial \beta }\right)}_V$        (2)

In the expression, $\frac{q_{\text{m}}}{N_{\text{A}}}$ is equal to $\frac{q_{\text{m}}^T}{N_{\text{A}}}=\text{Translational}\text{partition}\text{function}$, $q_{\text{m}}^R=\text{Rotational}\text{partition}\text{function}$ and $q_{\text{m}}^V=\text{Vibrational}\text{partition}\text{function}$.

So, the expression for $U_{\text{m}}-U_{\text{m}}\left(0\right)$ is written below.

    $U_{\text{m}}-U_{\text{m}}\left(0\right)=N_A\left[\langle E^T\rangle +\langle E^R\rangle +\langle E^V\rangle +\langle E^E\rangle \right]$        (3)

The expression for translational partition function, $\frac{q_{\text{m}}^T}{N_{\text{A}}}$ is given below.

    $\begin{array}{c}\frac{q_{\text{m}}^T}{N_{\text{A}}}=\frac{kT}{p^{\Theta }}{\left(\frac{\sqrt{2\pi mkT}}{h}\right)}^3\\ =\frac{{\left(\sqrt{2\pi k}\right)}^{3/2}}{p^{\Theta }{h}^3}{\left(T/K\right)}^{{}^{5/2}}{\left({\text{M/gmol}}^{-1}\right)}^{3/2}\end{array}$

Where,

• $h$ is the Planck’s constant.
• $\text{M}$ is the molar mass of ${\text{F}}_2^{-}$.
• $k$ is the Boltzmann constant.

Substitute $\frac{{\left(\sqrt{2\pi k}\right)}^{3/2}}{p^{\Theta }{h}^3}$ as $2.561\times {10}^{-2}$, $T/\text{K}$ as $298$ and $\text{M}$ as $38.00\text{g/mol}$ in the above equation.

    $\begin{array}{c}\frac{q_{\text{m}}^T}{N_{\text{A}}}=2.561\times {10}^{-2}\times {\left(298\right)}^{{}^{5/2}}{\left(38.00\right)}^{3/2}\\ =2.561\times {10}^{-2}\times \left(1.532995\times {10}^6\right)\times \left(234.25\right)\\ =9.20\times {10}^6\end{array}$

Thus, the value of translational partition function is $9.20\times {10}^6$.

The translational energy, $\langle E^T\rangle$ is calculate by the expression given below.

    $\langle E^T\rangle =\frac{3}{2}kT$        (4)

The value of rotational constant is calculated by the expression given below.

    $\begin{array}{c}\tilde{B}=\frac{\hslash }{4\pi cI}\\ =\frac{\hslash }{4\pi c\mu R^2}\\ =\frac{\hslash }{4\pi c\times \frac{M}{2N_{\text{A}}}{R}^2}\end{array}$        (5)

Where,

• $\mu$ is the mass divided by the Avogadro’s number of fluorine in $\text{kg/mol}$.
• $R$ is the equilibrium internuclear distance.

Substitute the value of $M$, $N_{\text{A}}$, $c$, $\hslash$ as $1.0546\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}$ and $R$ as in equation (5).

    $\begin{array}{c}\tilde{B}=\frac{1.0546\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}}{4\times 3.14\times 2.998\times {10}^{10}\text{cm/s}\times \frac{19\times {10}^{-3}\text{kg}}{2\times 6.022\times {10}^{23}\text{mol}}\times {\left(190.\times {10}^{-12}\text{m}\right)}^2}\\ =\frac{1.0546\times {10}^{-34}\text{kg}{\text{m}}^2{\text{s}}^{-1}}{4\times 3.14\times 2.998\times {10}^{10}\text{cm/s}\times 1.57736\times {10}^{-26}\times {\left(190.\times {10}^{-12}\text{m}\right)}^2}\\ =\frac{1.0546\times {10}^{-34}}{2.1436\times {10}^{-34}\text{cm}}\\ =0.4918{\text{cm}}^{-1}\end{array}$

The expression for rotational partition function, $q_{\text{m}}^R$ is given below.

    $q_{\text{m}}^R=\frac{kT}{\sigma hc\tilde{B}}$        (6)

Where,

• $k$ is the Boltzmann constant.($1.3806503\times {10}^{-23}\text{J}{\text{K}}^{-1}$)
• $h$ is the Planck’s constant.
• $c$ is the speed of light.
• $\tilde{B}$ is the rotational constant.

Substitute the values of $k$, $h$, $c$, $T$ and $\tilde{B}$ in the above equation.

    $\begin{array}{c}{q}_{\text{m}}^R=\frac{1.3806503\times {10}^{-23}\text{J}{\text{K}}^{-1}}{6.62606957\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm/s}}\left(\frac{T}{\sigma \tilde{B}}\right)\\ =\frac{1.3806503\times {10}^{-23}\times {10}^{34}\times {10}^{-10}\text{J}{\text{K}}^{-1}}{19.86495}\left(\frac{T}{\sigma \tilde{B}}\right)\\ =0.6950\times \left(\frac{298}{2\times 0.4918}\right)\\ =210.6\end{array}$

The rotational energy, $\langle E^T\rangle$ is calculate by the expression given below.

    $\langle E^R\rangle =kT$        (4)

The value of vibrational partition function is given below.

    $q_{\text{m}}^V=\left(\frac{1}{1-e^{\frac{-hc\overline{\nu }}{kT}}}\right)$

Substitute the value of $\frac{-hc}{k}$ as $1.4388$, and temperature in the above equation.

    $\begin{array}{c}{q}_{\text{m}}^V=\left(\frac{1}{1-e^{\frac{1.4388\times \left(\overline{\nu }/c{m}^{-1}\right)}{298\text{K/K}}}}\right)\\ =\left(\frac{1}{1-e^{-217}}\right)\\ =\frac{1}{1-0.114}\\ =1.129\end{array}$

The value of vibrational partition function is $1.129$.

The expression for the vibrational energy is given below.

    $\langle E^V\rangle =\frac{hc\tilde{v}}{e^{\frac{hc}{kT}}}$

Substitute the value of $\frac{-hc}{k}$ as $1.4388$, $\tilde{v}$ as $450{\text{cm}}^{-1}$, $h$, $c$ and temperature in the above equation.

    $\begin{array}{c}\langle E^V\rangle =\frac{6.62606957\times {10}^{-34}\text{J}\text{s}\times 2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\times 450{\text{cm}}^{-1}}{e^{\left(\frac{1.4388\left(450\right)}{298}\right)-1}}\\ =\frac{8939.23045}{e^{21726}-1}\times {10}^{-34+10}\text{J}\\ =\frac{8939.23045}{7.78}\times {10}^{-24}\text{J}\\ =\text{1}\text{.149}\times {10}^{-21}\text{J}\end{array}$

The value of Boltzmann factor corresponding to the lowest excited electronic state is given below.

    $\begin{array}{c}\exp \left(\frac{-\left(1.609\text{eV}\right)\left(1.602\times {10}^{-19}{\text{eV}}^{-1}\right)}{1.381\times {10}^{-23}T{\text{K}}^{-1}\times 298\text{K}}\right)=\exp \left(\frac{-\left(2.577618\times {10}^4{\text{eV}}^{-1}\right)}{411.538}\right)\\ =\exp \left(6.263\times {10}^{-3}\times {10}^4\right)\\ =6.3\times {10}^{-28}\end{array}$

The value of $q_{\text{m}}^{\text{E}}$ equals to the degeneracy of the ground state as calculated below.

    $\begin{array}{c}{q}_{\text{m}}^{\text{E}}=2\times \left(1+\left(6.3\times {10}^{-28}\right)\right)\\ =2\end{array}$

Thus, the value of electronic partition function is equal to $2$.

The expression to calculate the expression, $R\left(\ln \frac{q_{\text{m}}}{N_{\text{A}}}-1\right)$ is given below.

    $R\left(\ln \frac{q_{\text{m}}}{N_{\text{A}}}-1\right)=R\left(\ln \frac{q_{\text{m}}^{\text{T}}}{N_{\text{A}}}{q}_{{}^{\text{m}}}^R{q}_{\text{m}}^{\text{V}}{q}_{\text{m}}^{\text{E}}-1\right)$

Substitute the calculated values of $\frac{q_{\text{m}}^{\text{T}}}{N_{\text{A}}}$, $q_{{}^{\text{m}}}^R$, $q_{\text{m}}^{\text{V}}$ and $q_{\text{m}}^{\text{E}}$ in the above expression.

    $\begin{array}{c}R\left(\ln \frac{q_{\text{m}}}{N_{\text{A}}}-1\right)=8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\left(\ln \left(9.20\times {10}^6\times 210.6\times 1.129\times 2\right)-1\right)\\ =8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\left(\ln \left(4374.9\times {10}^6\right)-1\right)\\ =8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times \left(22.19-1\right)\\ =176.3\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\end{array}$

Thus, the value for the expression $R\left(\ln \frac{q_{\text{m}}}{N_{\text{A}}}-1\right)$ is $176.3\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$.

The expression to calculate the internal energy is given below.

    $\begin{array}{l}\frac{U_{\text{m}}-U_{\text{m}}\left(0\right)}{T}=\frac{N_A}{T}\left[\langle E^T\rangle +\langle E^R\rangle +\langle E^V\rangle +\langle E^E\rangle \right]\\ \frac{U_{\text{m}}-U_{\text{m}}\left(0\right)}{T}=\frac{N_A}{T}\left[\frac{3}{2}kT+kT+1.149\times {10}^{-21}\text{J}\right]\\ \frac{U_{\text{m}}-U_{\text{m}}\left(0\right)}{T}=\frac{5}{2}R\times \frac{N_A}{T}\left[1.149\times {10}^{-21}\text{J}\right]\end{array}$

Substitute the value of $R$, $N_{\text{A}}$ and $T$ in the above expression.

    $\begin{array}{c}\frac{U_{\text{m}}-U_{\text{m}}\left(0\right)}{T}=\frac{5}{2}\times 8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times \frac{6.022\times {10}^{23}\text{mol}}{298\text{K}}\left[1.149\times {10}^{-21}\text{J}\right]\\ =20.7862\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}+0.0232\times {10}^2\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\\ =23.1062\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\end{array}$

Thus, the value of $\frac{U_{\text{m}}-U_{\text{m}}\left(0\right)}{T}=23.1062\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$.

Substitute the values of $\frac{U_{\text{m}}-U_{\text{m}}\left(0\right)}{T}=23.1062\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$ and $R\left(\ln \frac{q_{\text{m}}}{N_{\text{A}}}-1\right)$ as $176.3\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$ in equation (1).

    $\begin{array}{c}{S}_{\text{m}}=23.1062\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}+176.3\text{J}{\text{mol}}^{-1}{\text{ K}}^{-1}\\ =\underset{\_}{199.4Jmo{l}^{-1}{K}^{-1}}\end{array}$

Therefore, the value of molar entropy is $\underset{\_}{199.4Jmo{l}^{-1}{K}^{-1}}$.