Chapter 13, Problem 13F.5P

Interpretation Introduction

Interpretation:

The value of $G_{\text{m}}^{\Theta }\left(200\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)$ for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ has to be calculated.

Concept introduction:

A thermodynamic potential that is utilized in the calculation of the highest reversible function taking place in the thermodynamic system at the constant value pressure and temperature is known as Gibbs-free energy.  It is denoted by $\Delta G°$.

Answer to Problem 13F.5P

The value of $G_{\text{m}}^{\Theta }\left(200\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)$ for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is $\underset{\_}{4.576\times 10^{4}Jmo{l}^{-1}}$.

Explanation of Solution

The given rotational constants for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ are $B=13109.4,2409.8$ and $2139.7\text{MHz}$.

The given rotational symmetry number is $2$.

The given vibtational wavenumbers for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ are $753,542,310,127,646$ and $419{\text{cm}}^{-1}$ which are all degenerate.

The molar mass of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is $102.9\text{g/mol}$.

The given temperature is $\text{200}\text{K}$.

The value of Planck’s constant is $6.626\times {10}^{-34}\text{J}\cdot \text{s}$.

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

The expression that is used to calculate the standard molar Gibbs energy, $G_{\text{m}}^{\Theta }$, is mentioned as follows.

    $G_{\text{m}}^{\Theta }\left(2000\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)=RT\ln \frac{q_{\text{m}}^0}{N_{\text{A}}}$        (1)

Where,

• $q_{\text{m}}^0$ is the molecular partition function.
• $R$ is the universal gas constant.
• $T$ is the temperature.

In the above expression, $\frac{q_{\text{m}}^0}{N_{\text{A}}}$ is equal to $\frac{q_{\text{m}}^T}{N_{\text{A}}}=\text{Translational}\text{partition}\text{function}$, $\frac{q_{\text{m}}^R}{N_{\text{A}}}=\text{Rotational}\text{partition}\text{function}$ and $\frac{q_{\text{m}}^V}{N_{\text{A}}}=\text{Vibrational}\text{partition}\text{function}$.

The expression for translational partition function, $\frac{q_{\text{m}}^T}{N_{\text{A}}}$ for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ 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 ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$.
• $k$ is the Boltzmann constant.($1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}$)

Substitute $\frac{{\left(\sqrt{2\pi k}\right)}^{3/2}}{p^{\Theta }{h}^3}$ as $2.561\times {10}^{-2}$, $T$ as $\text{200}\text{K}$ and $\text{M}$ as $102.9\text{g/mol}$ in the above equation.

    $\begin{array}{c}\frac{q_{\text{m}}^T}{N_{\text{A}}}=2.561\times {10}^{-2}\times {\left(200\right)}^{{}^{5/2}}{\left(102.9\right)}^{3/2}\\ =2.561\times {10}^{-2}\times \left(565685.425\right)\times \left(1043.8139\right)\\ =1.512\times {10}^7\end{array}$

Thus the value of translational partition function for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is $1.512\times {10}^7$.

The expression for rotational partition function, $\frac{q_{\text{m}}^R}{N_{\text{A}}}$ for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ which is a non linear molecule given below.

    $q_{\text{m}}^R=\frac{1}{\sigma }\frac{kT}{hc}{\left(\frac{\pi }{ABC}\right)}^{1/2}$        (2)

Where,

• $k$ is the Boltzmann constant.($1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}$)
• $h$ is the Planck’s constant.
• $c$ is the speed of light.
• $ABC$ are the rotational constants of the molecules along three directions.
• $\sigma$ is the symmetry number equals to $2$.

Substitute the values of $k$, $h$, $c$ and $\sigma$ in the above equation.

    $\begin{array}{c}{q}_{\text{m}}^R=\frac{1}{2}{\left(\frac{\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)T}{6.626\times {10}^{-34}\text{J s}\times 2.998\times {10}^{10}\text{cm/s}}\right)}^{1.5}{\left(\frac{3.14}{ABC}\right)}^{0.5}\\ =\frac{1}{2}\left(\frac{{\left(1.38\times {10}^{-23}{\text{K}}^{-1}\right)}^{1.5}{\left(3.14\right)}^{0.5}}{{\left(6.626\times {10}^{-34}\text{J s}\times 2.998\times {10}^{10}\text{cm/s}\right)}^{{}^{1.5}}}\right)\left(\frac{{\left(T\right)}^{1.5}}{{\left(ABC\right)}^{0.5}}\right)\\ =\frac{1.0270}{2}\left(\frac{{\left(T/K\right)}^{1.5}{\left(c\right)}^{1.5}}{{\left(ABC/{\text{cm}}^{-3}\right)}^{0.5}}\right)\end{array}$        (3)

Substitute the value of $T=\text{200}\text{K}$ and $c$ as $2.998\times {10}^{10}\text{cm/s}$, $A$ as $13109.4\text{MHz}$ $B$ as $2409.8\text{MHz}$ and $C$ as $2139.7\text{MHz}$ in the above equation.

    $\begin{array}{c}{q}_{\text{m}}^R=\frac{1.0270}{2}\left(\frac{{\left(200\right)}^{1.5}{\left(2.998\times {10}^{10}\text{cm/s}\right)}^{1.5}}{{\left(\left(13109.4\times 2409.8\times 2139.7\right)\frac{{10}^6/\text{s}}{\text{MHz}}/{\text{cm}}^{-3}\right)}^{0.5}}\right)\\ =\frac{1.0270}{2}\left(\frac{1.46822\times {10}^{19}}{{\left(\left(6.7595\times {10}^{10}\right)\frac{{10}^6/\text{s}}{\text{MHz}}\right)}^{0.5}}\right)\\ =2.9\times {10}^4\end{array}$

Thus, the value of rotational partition function of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is $2.9\times {10}^4$.

As, ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ possesses six vibrational degrees of freedom and all are having partition function, so, the value of vibrational partition function for the vibrational degrees of freedom of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$  is given below.

    $q_{\text{m}}^V=\left(\frac{1}{1-e^{\frac{-1.4388\left(\overline{\nu {\text{cm}}^{-1}}\right)}{T/K}}}\right)$        (4)

Where,

• $\overline{\nu }$ is the vibrational wave number.

The six vibrational modes of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ are labeled as (1), (2), (3), (4), (5) and (6).

Substitute the value of $\overline{\nu }$ as $753{\text{cm}}^{-1}$ and temperature in equation (4).

    $\begin{array}{c}{q}_{\text{m}}^{V(1)}=\left(\frac{1}{1-e^{\frac{-1.4388\left(\text{753}\text{}{\text{cm}}^{-1}\right)}{200}}}\right)\\ =1.004\end{array}$

Substitute the value of $\overline{\nu }$ as $542{\text{cm}}^{-1}$ and temperature in equation (4).

    $\begin{array}{c}{q}_{\text{m}}^{V(2)}=\left(\frac{1}{1-e^{\frac{-1.4388\left(\text{542}{\text{cm}}^{-1}\right)}{200}}}\right)\\ =1.021\end{array}$

Substitute the value of $\overline{\nu }$ as $310{\text{cm}}^{-1}$ and temperature in equation (4).

    $\begin{array}{c}{q}_{\text{m}}^{V(3)}=\left(\frac{1}{1-e^{\frac{-1.4388\left(\text{310}{\text{cm}}^{-1}\right)}{200}}}\right)\\ =1.120\end{array}$

Substitute the value of $\overline{\nu }$ as $127{\text{cm}}^{-1}$ and temperature in equation (4).

    $\begin{array}{c}{q}_{\text{m}}^{V(4)}=\left(\frac{1}{1-e^{\frac{-1.4388\left(\text{127}{\text{cm}}^{-1}\right)}{200}}}\right)\\ =1.670\end{array}$

Substitute the value of $\overline{\nu }$ as $646{\text{cm}}^{-1}$ and temperature in equation (4).

    $\begin{array}{c}{q}_{\text{m}}^{V(5)}=\left(\frac{1}{1-e^{\frac{-1.4388\left(646{\text{cm}}^{-1}\right)}{200}}}\right)\\ =1.010\end{array}$

Substitute the value of $\overline{\nu }$ as $419{\text{cm}}^{-1}$ and temperature in equation (4).

    $\begin{array}{c}{q}_{\text{m}}^{V(6)}=\left(\frac{1}{1-e^{\frac{-1.4388\left(419{\text{cm}}^{-1}\right)}{200}}}\right)\\ =1.052\end{array}$

So, the total vibrational partition function is calculated below.

    $\begin{array}{c}{q}_{\text{m}}^V={\displaystyle \prod _{i=1}^6{q}_{\text{m}}^{V(i)}}\\ =q_{\text{m}}^{V(1)}\times q_{\text{m}}^{V(2)}\times q_{\text{m}}^{V(3)}\times q_{\text{m}}^{V(4)}\times q_{\text{m}}^{V(5)}\times q_{\text{m}}^{V(6)}\end{array}$

Substitute the value of $q_{\text{m}}^{V(1)}\times q_{\text{m}}^{V(2)}\times q_{\text{m}}^{V(3)}\times q_{\text{m}}^{V(4)}\times q_{\text{m}}^{V(5)}\times q_{\text{m}}^{V(6)}$ in the above equation.

    $\begin{array}{c}{q}_{\text{m}}^V=1.004\times 1.021\times 1.120\times 1.670\times 1.010\times 1.052\\ =2.037\end{array}$

Substitute the value of $q_{\text{m}}^E$, $q_{\text{m}}^T$, $q_{\text{m}}^R$, $R$ as $8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$, $T$ and $q_{\text{m}}^V$ for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ in equation (1).

    $\begin{array}{c}{G}_{\text{m}}^{\Theta }\left(200\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)=8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 200\text{K}\ln \left[\left(1.512\times {10}^7\right)\left(2.9\times {10}^4\right)\left(2.037\right)\left(1\right)\right]\\ =1662.9\times \ln \left(8.9318\times {10}^{11}\right)\\ =\underset{\_}{4.576\times 10^{4}Jmo{l}^{-1}}\end{array}$

Therefore, the value of $G_{\text{m}}^{\Theta }\left(200\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)$ for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is $\underset{\_}{4.576\times 10^{4}Jmo{l}^{-1}}$.