Chapter 13, Problem 13E.4BE

(i)

Interpretation Introduction

Interpretation:

The plot of molar entropy of a collection of harmonic oscillators as a function of $T/{\theta }^{\text{V}}$ is to be plotted.  Also, the standard molar entropy of ethyne at $298\text{K}$ is to be calculated.

Concept introduction:

Entropy is a thermodynamic property which shows the randomness or disorder of the atoms within the system.  The randomness of the molecules increases when solid is melted and liquid is converted into the vapor state.  As in vapor state particles move freely because they are at larger distances from each other.

The molar entropy of a collection of oscillations is expressed by the formula shown below.

    $S_{\text{m}}=Nk\left\{\frac{\beta \epsilon }{e^{\beta \epsilon }-1}-\ln \left(1-e^{-\beta \epsilon }\right)\right\}$

Answer to Problem 13E.4BE

The plot of molar entropy of a collection of harmonic oscillators as a function of $T/{\theta }^{\text{V}}$ is shown below.

The standard molar entropy of ethyne at $298\text{K}$ is $\underset{\_}{5.850J/mol\cdot K}$.

Explanation of Solution

The molar entropy of a collection of oscillations is expressed by the formula shown below.

    $S_{\text{m}}=Nk\left\{\frac{\beta \epsilon }{e^{\beta \epsilon }-1}-\ln \left(1-e^{-\beta \epsilon }\right)\right\}$        (1)

The value of $\beta \epsilon$ is calculated by the formula shown below.

    $\beta \epsilon =\frac{hc\overline{\nu }}{kT}$        (2)

Where,

• $h$ is the Planck constant $\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)$.
• $c$ is the speed of light $\left(2.998\times {10}^8\text{m}/\text{s}=2.998\times {10}^{10}\text{cm}/\text{s}\right)$.
• $\overline{\nu }$ is the wavenumber.
• $k$ is the Boltzmann constant with value $1.38\times {10}^{-23}\text{J}/\text{K}$.

Also, the term ${\theta }^{\text{V}}$ is expressed as shown below.

  $\frac{hc\overline{\nu }}{k}={\theta }^{\text{V}}$        (3)

Therefore, the value of $\beta \epsilon$ is expressed as shown below.

    $\beta \epsilon =\frac{{\theta }^{\text{V}}}{T}$        (4)

Where,

• ${\theta }^{\text{V}}$ is the vibrational temperature.

Therefore, the molar entropy of a collection of oscillations is expressed by the formula shown below.

    $\begin{array}{l}{S}_{\text{m}}=\frac{R\left({\theta }^{\text{V}}/T\right)}{e^{{\theta }^{\text{V}}/T}-1}-R\ln \left(1-e^{{-}^{{\theta }^{\text{V}}/T}}\right)\\ \frac{S_{\text{m}}}{R}=\frac{{\theta }^{\text{V}}/T}{e^{{\theta }^{\text{V}}/T}-1}-\ln \left(1-e^{{-}^{{\theta }^{\text{V}}/T}}\right)\end{array}$        (4)

First, the values of ${\theta }^{\text{V}}$ for different values of wavenumber for the given data are to be calculated by using equation (3) as shown below.

For $\overline{\nu }=612{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 612{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{1.215\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =880.4\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{880.4\text{K}}{298\text{K}}\\ =2.96\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{298\text{K}}{880.4\text{K}}\\ =0.338\end{array}$

For $\overline{\nu }=729{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 729{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{1.448\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =1049.275\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{1049.275\text{K}}{298\text{K}}\\ =3.52\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{298\text{K}}{1049.275\text{K}}\\ =0.284\end{array}$

For $\overline{\nu }=1974{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 1974{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{3.92\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =2840.57\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{2840.57\text{K}}{298\text{K}}\\ =9.53\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{298\text{K}}{2840.57\text{K}}\\ =0.105\end{array}$

For $\overline{\nu }=3287{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 3287{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{6.530\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =4731.88\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{4731.88\text{K}}{298\text{K}}\\ =15.88\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{298\text{K}}{4731.88\text{K}}\\ =0.063\end{array}$

For $\overline{\nu }=3374{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 3374{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{6.702\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =4856.52\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{4856.52\text{K}}{298\text{K}}\\ =16.30\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{298\text{K}}{4856.52\text{K}}\\ =0.0614\end{array}$

Now, the value of $S_{\text{m}}/R$ is calculated with the help of equation (4) by substituting all the calculated values, as shown below.

For $\overline{\nu }=612{\text{cm}}^{-1}$,

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{2.96}{e^{2.96}-1}-\ln \left(1-e^{-2.96}\right)\\ =\frac{2.96}{19.3-1}-\ln \left(1-0.051\right)\\ =\frac{2.96}{18.3}-\left(-0.052\right)\\ =0.213\end{array}$

For $\overline{\nu }=729{\text{cm}}^{-1}$,

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{3.52}{e^{3.52}-1}-\ln \left(1-e^{-3.52}\right)\\ =\frac{3.52}{33.78-1}-\ln \left(1-0.03\right)\\ =\frac{3.52}{32.78}-\left(-0.030\right)\\ =0.137\end{array}$

For $\overline{\nu }=1974{\text{cm}}^{-1}$.

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{9.53}{e^{9.53}-1}-\ln \left(1-e^{-9.53}\right)\\ =\frac{9.53}{13766.6-1}-\ln \left(1-7.264\times {10}^{-5}\right)\\ =\frac{9.53}{13765.6}-\left(-1.0\times {10}^{-3}\right)\\ =1.7\times {10}^{-3}\end{array}$

For $\overline{\nu }=3287{\text{cm}}^{-1}$.

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{15.88}{e^{15.88}-1}-\ln \left(1-e^{-15.88}\right)\\ =\frac{15.88}{7881273.0-1}-\ln \left(1-1.268\times {10}^{-7}\right)\\ =\frac{15.88}{7881272}-\left(-1.0\times {10}^{-3}\right)\\ =1.002\times {10}^{-3}\end{array}$

For $\overline{\nu }=3374{\text{cm}}^{-1}$.

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{16.30}{e^{16.30}-1}-\ln \left(1-e^{-16.30}\right)\\ =\frac{16.30}{11994994.55-1}-\ln \left(1-8.34\times {10}^{-8}\right)\\ =\frac{16.30}{11994993.55}-\left(-1.0\times {10}^{-3}\right)\\ =1.0013\times {10}^{-3}\end{array}$

Therefore, the values of $\overline{\nu }\left({\text{cm}}^{-1}\right)$, ${\theta }^{\text{V}}/T$ and $C_{V,m}/R$ at $298\text{K}$ are shown below in the Table.

$\overline{\nu }\left({\text{cm}}^{-1}\right)$	${\theta }^{\text{V}}/T$	$T/{\theta }^{\text{V}}$	$S_{\text{m}}/R$
$612$	$2.96$	$0.338$	$0.213$
$612$	$2.96$	$0.338$	$0.213$
$729$	$3.52$	$0.284$	$0.137$
$729$	$3.52$	$0.284$	$0.137$
$1974$	$9.53$	$0.105$	$1.7\times {10}^{-3}$
$3287$	$15.88$	$0.063$	$1.002\times {10}^{-3}$
$3374$	$16.30$	$0.061$	$1.0013\times {10}^{-3}$

Now, the plot of molar entropy of a collection of harmonic oscillators as a function of $T/{\theta }^{\text{V}}$ is shown below.

Figure 1

The standard molar entropy of ethyne is calculated by the formula shown below.

  $S_{\text{m}}/R$        (5)

The value of gas constant is $8.\text{314 }\text{J}/\text{mol}\cdot \text{K}$.

The sum of all the values of $S_{\text{m}}/R$ at different wavenumbers is calculated as shown below.

    $\begin{array}{c}{S}_{\text{m}}/R=0.213+0.213+0.137+0.137+1.7\times {10}^{-3}+1.002\times {10}^{-3}+1.0013\times {10}^{-3}\\ =0.7037\end{array}$

Therefore, the standard molar entropy of ethyne is calculated as shown below.

    $\begin{array}{c}{S}_{\text{m}}=0.7037\times R\\ =0.7037\times 8.\text{314 }\text{J}/\text{mol}\cdot \text{K}\\ =\underset{\_}{5.850J/mol\cdot K}\end{array}$

The standard molar entropy of ethyne at $298\text{K}$ is $\underset{\_}{5.850J/mol\cdot K}$.

(ii)

Interpretation Introduction

Interpretation:

The plot of molar entropy of a collection of harmonic oscillators as a function of $T/{\theta }^{\text{V}}$ is to be plotted.  Also, the standard molar entropy of ethyne at $500\text{K}$ is to be calculated.

Concept introduction:

Entropy is a thermodynamic property which shows the randomness or disorder of the atoms within the system.  The randomness of the molecules increases when solid is melted and liquid is converted into the vapor state.  As in vapor state particles move freely because they are at larger distances from each other.

The molar entropy of a collection of oscillations is expressed by the formula shown below.

    $S_{\text{m}}=Nk\left\{\frac{\beta \epsilon }{e^{\beta \epsilon }-1}-\ln \left(1-e^{-\beta \epsilon }\right)\right\}$

Answer to Problem 13E.4BE

The plot of molar entropy of a collection of harmonic oscillators as a function of $T/{\theta }^{\text{V}}$ is shown below.

The standard molar entropy of ethyne at $500\text{K}$ is $\underset{\_}{16.47J/mol\cdot K}$.

Explanation of Solution

The molar entropy of a collection of oscillations is expressed by the formula shown below.

    $S_{\text{m}}=Nk\left\{\frac{\beta \epsilon }{e^{\beta \epsilon }-1}-\ln \left(1-e^{-\beta \epsilon }\right)\right\}$        (1)

The value of $\beta \epsilon$ is calculated by the formula shown below.

    $\beta \epsilon =\frac{hc\overline{\nu }}{kT}$        (2)

Where,

• $h$ is the Planck constant $\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)$.
• $c$ is the speed of light $\left(2.998\times {10}^8\text{m}/\text{s}=2.998\times {10}^{10}\text{cm}/\text{s}\right)$.
• $\overline{\nu }$ is the wavenumber.
• $k$ is the Boltzmann constant with value $1.38\times {10}^{-23}\text{J}/\text{K}$.

Also, the term ${\theta }^{\text{V}}$ is expressed as shown below.

  $\frac{hc\overline{\nu }}{k}={\theta }^{\text{V}}$        (3)

Therefore, the value of $\beta \epsilon$ is expressed as shown below.

    $\beta \epsilon =\frac{{\theta }^{\text{V}}}{T}$        (4)

Where,

• ${\theta }^{\text{V}}$ is the vibrational temperature.

Therefore, the molar entropy of a collection of oscillations is expressed by the formula shown below.

    $\begin{array}{l}{S}_{\text{m}}=\frac{R\left({\theta }^{\text{V}}/T\right)}{e^{{\theta }^{\text{V}}/T}-1}-R\ln \left(1-e^{{-}^{{\theta }^{\text{V}}/T}}\right)\\ \frac{S_{\text{m}}}{R}=\frac{{\theta }^{\text{V}}/T}{e^{{\theta }^{\text{V}}/T}-1}-\ln \left(1-e^{{-}^{{\theta }^{\text{V}}/T}}\right)\end{array}$        (4)

First, the values of ${\theta }^{\text{V}}$ for different values of wavenumber for the given data are to be calculated by using equation (3) as shown below.

For $\overline{\nu }=612{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 612{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{1.215\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =880.4\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{880.4\text{K}}{500\text{K}}\\ =1.76\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{500\text{K}}{880.4\text{K}}\\ =0.568\end{array}$

For $\overline{\nu }=729{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 729{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{1.448\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =1049.275\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{1049.275\text{K}}{500\text{K}}\\ =2.09\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{500\text{K}}{1049.275\text{K}}\\ =0.4765\end{array}$

For $\overline{\nu }=1974{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 1974{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{3.92\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =2840.57\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{2840.57\text{K}}{500\text{K}}\\ =5.68\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{500\text{K}}{2840.57\text{K}}\\ =0.176\end{array}$

For $\overline{\nu }=3287{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 3287{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{6.530\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =4731.88\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{4731.88\text{K}}{500\text{K}}\\ =9.46\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{500\text{K}}{4731.88\text{K}}\\ =0.106\end{array}$

For $\overline{\nu }=3374{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 3374{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{6.702\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =4856.52\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{4856.52\text{K}}{500\text{K}}\\ =9.71\end{array}$

Also, the value of $T/{\theta }^{\text{V}}$ is calculated as shown below.

    $\begin{array}{c}T/{\theta }^{\text{V}}=\frac{500\text{K}}{4856.52\text{K}}\\ =0.103\end{array}$

Now, the value of $S_{\text{m}}/R$ is calculated with the help of equation (4) by substituting all the calculated values, as shown below.

For $\overline{\nu }=612{\text{cm}}^{-1}$,

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{1.76}{e^{1.76}-1}-\ln \left(1-e^{-1.76}\right)\\ =\frac{1.76}{5.812-1}-\ln \left(1-0.172\right)\\ =\frac{1.76}{4.812}-\left(-0.188\right)\\ =0.553\end{array}$

For $\overline{\nu }=729{\text{cm}}^{-1}$,

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{2.09}{e^{2.09}-1}-\ln \left(1-e^{-2.09}\right)\\ =\frac{2.09}{8.085-1}-\ln \left(1-0.1236\right)\\ =\frac{2.09}{7.085}-\left(-0.1319\right)\\ =0.426\end{array}$

For $\overline{\nu }=1974{\text{cm}}^{-1}$.

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{5.68}{e^{5.68}-1}-\ln \left(1-e^{-5.68}\right)\\ =\frac{5.68}{292.95-1}-\ln \left(1-3.4135\times {10}^{-3}\right)\\ =\frac{5.68}{291.950}-\left(-3.419\times {10}^{-3}\right)\\ =0.0228\end{array}$

For $\overline{\nu }=3287{\text{cm}}^{-1}$.

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{9.46}{e^{9.46}-1}-\ln \left(1-e^{-9.46}\right)\\ =\frac{9.46}{12835.8-1}-\ln \left(1-7.790\times {10}^{-5}\right)\\ =\frac{9.46}{12834.8}-\left(-7.790\times {10}^{-5}\right)\\ =8.14\times {10}^{-4}\end{array}$

For $\overline{\nu }=3374{\text{cm}}^{-1}$.

    $\begin{array}{c}\frac{S_{\text{m}}}{R}=\frac{9.71}{e^{9.71}-1}-\ln \left(1-e^{-9.71}\right)\\ =\frac{9.71}{16481.60-1}-\ln \left(1-6.067\times {10}^{-5}\right)\\ =\frac{9.71}{16480.6}-\left(-6.067\times {10}^{-5}\right)\\ =6.5\times {10}^{-4}\end{array}$

Therefore, the values of $\overline{\nu }\left({\text{cm}}^{-1}\right)$, ${\theta }^{\text{V}}/T$ and $C_{V,m}/R$ at $298\text{K}$ are shown below in the Table.

$\overline{\nu }\left({\text{cm}}^{-1}\right)$	${\theta }^{\text{V}}/T$	$T/{\theta }^{\text{V}}$	$S_{\text{m}}/R$
$612$	$1.76$	$0.568$	$0.553$
$612$	$1.76$	$0.568$	$0.553$
$729$	$2.09$	$0.4765$	$0.426$
$729$	$2.09$	$0.4765$	$0.426$
$1974$	$5.68$	$0.176$	$0.0228$
$3287$	$9.46$	$0.106$	$8.14\times {10}^{-4}$
$3374$	$9.71$	$0.103$	$6.5\times {10}^{-4}$

Now, the plot of molar entropy of a collection of harmonic oscillators as a function of $T/{\theta }^{\text{V}}$ is shown below.

Figure 2

The standard molar entropy of ethyne is calculated by the formula shown below.

  $\text{Standard molar entropy}=S_{\text{m}}/R$        (5)

The value of gas constant is $8.\text{314 }\text{J}/\text{mol}\cdot \text{K}$.

The sum of all the values of $S_{\text{m}}/R$ at different wavenumbers is calculated as shown below.

    $\begin{array}{c}{S}_{\text{m}}/R=0.553+0.553+0.426+0.426+0.0228+8.14\times {10}^{-4}+6.5\times {10}^{-4}\\ =1.982\end{array}$

Therefore, the standard molar entropy of ethyne is calculated as shown below.

    $\begin{array}{c}{S}_{\text{m}}=1.982\times R\\ =1.982\times 8.\text{314 }\text{J}/\text{mol}\cdot \text{K}\\ =\underset{\_}{16.47J/mol\cdot K}\end{array}$

The standard molar entropy of ethyne at $500\text{K}$ is $\underset{\_}{16.47J/mol\cdot K}$.