Chapter 13, Problem 13F.6P

Interpretation Introduction

Interpretation:

The values of $G_{\text{m}}^{\Theta }\left(10.0\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)$ and $G_{\text{m}}^{\Theta }\left(100.0\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)$ for ${\text{C}}_3$ have 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.6P

The values of $G_{\text{m}}^{\Theta }\left(10.0\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)$ and $G_{\text{m}}^{\Theta }\left(100.0\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)$ for ${\text{C}}_3$ are $\underset{\_}{756.43Jmo{l}^{-1}}$ and $\underset{\_}{2.10\times 10^{4}Jmo{l}^{-1}}$ respectively.

Explanation of Solution

The given moment of inertia for ${\text{C}}_3$ are $I=39.340m_u\stackrel{\ast }{{\text{A}}^2},39.032m_u\stackrel{\ast }{{\text{A}}^2}$ and $0.3082m_u\stackrel{\ast }{{\text{A}}^2}$

The value of $m_u$ is $1.6605{\text{4×10}}^{\text{-27}}\text{kg}$.

The given vibrational wavenumbers for ${\text{C}}_3$ are $63.4,1224.5,$ and $2040{\text{cm}}^{-1}$.

The molar mass of ${\text{C}}_3$ is $36.033\text{g/mol}$.

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.

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

Where,

$q_{\text{m}}^0$ is the molecular partition function.

$R$ is the universal gas constant.

$T$ is the temperature.

$\frac{q_{\text{m}}^0}{N_{\text{A}}}=\frac{q_{\text{m}}^T}{N_{\text{A}}}{q}^R{q}^V{q}^R{q}^E$

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

The expression for translational partition function, $\frac{q_{\text{m}}^T}{N_{\text{A}}}$ for ${\text{C}}_3$ at $1\text{0}\text{K}$ 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 ${\text{C}}_3$.

$k$ 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 $\text{10}\text{K}$ and $\text{M}$ as $36.033\text{g/mol}$ in the above equation.

    $\begin{array}{c}\frac{q_{\text{m}}^T}{N_{\text{A}}}=2.561\times {10}^{-2}\times {\left(10\right)}^{{}^{5/2}}{\left(36.033\text{g/mol}\right)}^{3/2}\\ =2.561\times {10}^{-2}\times \left(316.2278\right)\times \left(216.2971\right)\\ =1752\end{array}$

Thus the value of translational partition function for ${\text{C}}_3$ at $\text{10}\text{K}$ is $1752$.

The expression for rotational partition function, $q^R$ for ${\text{C}}_3$ which is a nonlinear molecule given below.

    $\begin{array}{c}{q}^R=\frac{1}{\sigma }\frac{kT}{hc}{\left(\frac{\pi }{ABC}\right)}^{1/2}\\ =\frac{1.0270}{\sigma }\times {\left(\frac{T/K}{ABC/c{m}^{-3}}\right)}^{3/2}\mathrm{...............}\left(2\right)\end{array}$

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.

The value of $ABC$ is calculated by the expression given below.

    $ABC={\left(\frac{\hslash }{4\pi c}\right)}^3\frac{1}{I_A{I}_B{I}_C}$

Substitute the values of $\hslash$ as $1.055\times {10}^{-34}\text{J}\text{s}$, $c$ $2.998\times {10}^{10}\text{cm/s}$ and moment of inertia for ${\text{C}}_3$ are $I_A=39.340m_u\stackrel{\ast }{{\text{A}}^2},I_B=39.032m_u\stackrel{\ast }{{\text{A}}^2}$ and $I_C=0.3082m_u\stackrel{\ast }{{\text{A}}^2}$ in the above equation.

$\begin{array}{c}ABC={\left(\frac{1.055\times {10}^{-34}\text{J}\text{s}}{4\times 3.14\times 2.998\times {10}^{10}\text{cm/s}}\right)}^3\frac{1}{m_u\left(39.340\stackrel{\ast }{{\text{A}}^2}\times 39.032\stackrel{\ast }{{\text{A}}^2}\times 0.3082\stackrel{\ast }{{\text{A}}^2}\right)}\\ ={\left(\frac{1.055\times {10}^{-34}\text{J}\text{s}}{4\times 3.14\times 2.998\times {10}^{10}\text{cm/s}}\right)}^3\times \frac{{10}^{60}{\text{m}}^{-6}}{{\left(1.66054\times {10}^{-27}\text{kg}\right)}^3\times \stackrel{\ast }{{\text{A}}^6}\left(39.340\stackrel{\ast }{{\text{A}}^2}\times 39.032\stackrel{\ast }{{\text{A}}^2}\times 0.3082\stackrel{\ast }{{\text{A}}^2}\right)}\\ =2.1993\times {10}^{-5-102-30}\frac{{\text{kg}}^3{\text{m}}^6{\text{s}}^{-3}}{{\text{cm}}^3{\text{s}}^{-3}}\times \frac{{10}^{60}{\text{m}}^{-6}}{{\left(1.66054\times {10}^{-27}\text{kg}\right)}^3\times \stackrel{\ast }{{\text{A}}^6}\left(39.340\times 39.032\times 0.3082\right)}\\ =10.134{\text{cm}}^{-3}\end{array}$

Substitute the values of $ABC$ as $10.134{\text{cm}}^{-3}$, $T=10\text{K}$ and $\sigma$ as $2$ in equation (2).

    $\begin{array}{c}{q}^R=\frac{1.0270}{2}\times \times \left(\frac{{\left(10\right)}^{3/2}}{{\left(10.134\right)}^{1/2}}\right)\\ =\frac{1.0270}{2}\times \left(\frac{31.62}{3.18}\right)\\ =\mathrm{5.1......................}\left(3\right)\end{array}$

Thus, the value of rotational partition function of ${\text{C}}_3$ at $\text{10}\text{K}$ is $5.1$.

The vibrational degrees of freedom of ${\text{C}}_3$ is given below.

    $q_{}^V=\left(\frac{1}{1-e^{\frac{-hc\tilde{v}}{kT}}}\right)\mathrm{..........................}\left(4\right)$

Where,

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

Substitute the value of $h$, $c$, $k$ T and $\overline{\nu }$ as $63.4{\text{cm}}^{-1}$ in equation (4).

    $\begin{array}{c}{q}_{}^V={\left(1-\exp \left(\frac{6.626\times {10}^{-34}\text{J}\text{s}\times 2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\times 63.4{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\times 10\text{K}}\right)\right)}^{-1}\\ ={\left(1-\exp \left(-91.2\times {10}^{-34+33}\right)\right)}^{-1}\\ ={\left(1-1.0945\times {10}^{-4}\right)}^{-1}\\ =1.0001\end{array}$

Thus, the vibrational number for ${\text{C}}_{\text{3}}$ at $\text{10}\text{K}$ is $1.0001$.

The degeneracy of electronic ground sate is $q^E=1$ because the lowest-mode of partition function is also one.

Substitute the value of $q^E$, $q^T$, $q^R$, $R$ as $8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$, $T$ and $q^V$ for ${\text{C}}_3$ in equation (1).

    $\begin{array}{c}{G}_{\text{m}}^{\Theta }\left(10\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)=8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 10\text{K}\ln \left[\left(152\right)\left(5.1\right)\left(1\right)\left(1\right)\right]\\ =\underset{\_}{756.43Jmo{l}^{-1}}\end{array}$

Therefore, the value of $G_{\text{m}}^{\Theta }\left(10\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)$ for ${\text{C}}_3$ is $\underset{\_}{756.43Jmo{l}^{-1}}$.

For $\text{100}\text{K}$, substitute $\frac{{\left(\sqrt{2\pi k}\right)}^{3/2}}{p^{\Theta }{h}^3}$ as $2.561\times {10}^{-2}$, $T$ as $\text{100}\text{K}$ and $\text{M}$ as $36.033\text{g/mol}$ in the above equation.

    $\begin{array}{c}\frac{q_{\text{m}}^T}{N_{\text{A}}}=2.561\times {10}^{-2}\times {\left(\text{100}\text{K}\right)}^{{}^{5/2}}{\left(36.033\text{g/mol}\right)}^{3/2}\\ =2.561\times {10}^{-2}\times \left(100000\right)\times \left(216.2971\right)\\ =5.54\times {10}^7\end{array}$

Thus the value of translational partition function for ${\text{C}}_3$ at $\text{100}\text{K}$ is $5.54\times {10}^7$.

The expression for rotational partition function, $q^R$ for ${\text{C}}_3$ which is a nonlinear molecule given below.

    $\begin{array}{c}{q}^R=\frac{1}{\sigma }\frac{kT}{hc}{\left(\frac{\pi }{ABC}\right)}^{1/2}\\ =\frac{1.0270}{\sigma }\times {\left(\frac{T/K}{ABC/c{m}^{-3}}\right)}^{3/2}\mathrm{......................}\left(2\right)\end{array}$

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.

The value of $ABC$ is calculated by the expression given below.

    $ABC={\left(\frac{\hslash }{4\pi c}\right)}^3\frac{1}{I_A{I}_B{I}_C}$

Substitute the values of $\hslash$ as $1.055\times {10}^{-34}\text{J}\text{s}$, $c$ $2.998\times {10}^{10}\text{cm/s}$ and moment of inertia for ${\text{C}}_3$ are $I_A=39.340m_u\stackrel{\ast }{{\text{A}}^2},I_B=39.032m_u\stackrel{\ast }{{\text{A}}^2}$ and $I_C=0.3082m_u\stackrel{\ast }{{\text{A}}^2}$ in the above equation.

    $\begin{array}{c}ABC={\left(\frac{1.055\times {10}^{-34}\text{J}\text{s}}{4\times 3.14\times 2.998\times {10}^{10}\text{cm/s}}\right)}^3\frac{1}{m_u\left(39.340\stackrel{\ast }{{\text{A}}^2}\times 39.032\stackrel{\ast }{{\text{A}}^2}\times 0.3082\stackrel{\ast }{{\text{A}}^2}\right)}\\ ={\left(\frac{1.055\times {10}^{-34}\text{J}\text{s}}{4\times 3.14\times 2.998\times {10}^{10}\text{cm/s}}\right)}^3\times \frac{{10}^{60}{\text{m}}^{-6}}{{\left(1.66054\times {10}^{-27}\text{kg}\right)}^3\times \stackrel{\ast }{{\text{A}}^6}\left(39.340\stackrel{\ast }{{\text{A}}^2}\times 39.032\stackrel{\ast }{{\text{A}}^2}\times 0.3082\stackrel{\ast }{{\text{A}}^2}\right)}\\ =2.1993\times {10}^{-5-102-30}\frac{{\text{kg}}^3{\text{m}}^6{\text{s}}^{-3}}{{\text{cm}}^3{\text{s}}^{-3}}\times \frac{{10}^{60}{\text{m}}^{-6}}{{\left(1.66054\times {10}^{-27}\text{kg}\right)}^3\times \stackrel{\ast }{{\text{A}}^6}\left(39.340\times 39.032\times 0.3082\right)}\\ =10.134{\text{cm}}^{-3}\end{array}$

Substitute the values of $ABC$ as $10.134{\text{cm}}^{-3}$ $T=10\text{0}\text{K}$ and $\sigma$ as $2$ in equation (2).

    $\begin{array}{c}{q}^R=\frac{1.0270}{2}\times \left(\frac{{\left(100\right)}^{3/2}}{{\left(10.134\right)}^{1/2}}\right)\\ =\frac{1.0270}{2}\times \left(\frac{1000}{3.18}\right)\\ =0.5135\times 314.47\\ =161.48\end{array}$

Thus, the value of rotational partition function of ${\text{C}}_3$ at $\text{100}\text{K}$ is $161.48$.

The vibrational degrees of freedom of ${\text{C}}_3$ is given below.

    $q_{}^V=\left(\frac{1}{1-e^{\frac{-hc\tilde{v}}{kT}}}\right)\mathrm{..................}\left(4\right)$

Where,

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

Substitute the value of $h$, $c$, $k$ T and $\overline{\nu }$ as $63.4{\text{cm}}^{-1}$ in equation (4).

    $\begin{array}{c}{q}_1^V={\left(1-\exp \left(\frac{6.626\times {10}^{-34}\text{J}\text{s}\times 2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\times 63.4{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\times 100\text{K}}\right)\right)}^{-1}\\ ={\left(1-\exp \left(91.26\times {10}^{-3}\right)\right)}^{-1}\\ ={\left(1-1.09555\right)}^{-1}\\ =10.465\end{array}$

Thus, the vibrational number for ${\text{C}}_{\text{3}}$ at $\text{100}\text{K}$ is $10.465$.

The degeneracy of electronic ground sate is $q^E=1$ because the lowest-mode of partition function is also one.

Substitute the value of $q^E$, $q^T$, $q^R$, $R$ as $8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$, $T$ and $q^V$ for ${\text{C}}_3$ in equation (1).

    $\begin{array}{c}{G}_{\text{m}}^{\Theta }\left(10\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)=8.3145\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 100\text{K}\ln \left[\left(5.54\times {10}^7\right)\left(161.48\right)\left(10.465\right)\left(1\right)\right]\\ =\underset{\_}{2.10\times 10^{4}Jmo{l}^{-1}}\end{array}$

Therefore, the value of $G_{\text{m}}^{\Theta }\left(10\text{K}\right)-G_{\text{m}}^{\Theta }\left(0\right)$ for ${\text{C}}_3$ at $\text{100}\text{K}$ is $\underset{\_}{2.10\times 10^{4}Jmo{l}^{-1}}$.