91Ó°ÊÓ

Gibbs free energy, also known as Gibbs function, Gibbs energy, or free enthalpy, is a term used to measure the greatest amount of work done in a thermodynamic system when temperature and pressure remain constant.

Entropy is a measure of a system's unpredictability or disorder in general.

Entropy is a thermodynamic property that describes how a system behaves in terms of temperature, pressure, entropy, and heat capacity.

(a)

The reaction given is –

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{C}{\text{H}}_{\text{3}}\stackrel{{\text{Pt/Al}}_{\text{2}}{\text{O}}_{\text{3}}}{→}{\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}{\text{+H}}_{\text{2}}$

First, write the formation reaction of${\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}$ –

$3C(s,graphite)3H_2\left(g\right)→{\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}\left(g\right)$

Solve for $∆S_f^{°}$. The values for${\text{S}}^{\text{o}}{\text{(H}}_{\text{2}}\text{(g))}$ and ${\text{S}}^{\text{o}}\text{(C(s,graphite))}$ can be found on Appendix B.

$$\begin{gathered} ∆S_f^{°}\left({\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}\right)=\underset{}{\overset{}{∑n_p}}{S}^{°}\left(product\right)-\underset{}{\overset{}{∑n_r}}{S}^{°}\left(reac\tan t\right) \\ =\left[n{S}^{°}\left({\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}\left(g\right)\right)\right]-\left[n{S}^{°}\left(\text{C(s,graphite)}\right)+n{S}^{°}\left(H_{2\left(g\right)}\right)\right] \\ =\left(\right[1(267.1)]-[3(5.686)+3(130.6)\left]\right)J/mol·K \\ ∆S_f^{°}\left({\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}\right)=-141.8J/mol·K \end{gathered}$$

Therefore, the value is obtained as .$-141.8J/mol·K$

(b)

Solve for$∆G_f^{°}$ . Use the obtained $∆S_f^{°}$from (a).

$$\begin{gathered} ∆G_f^{°}=∆H_f^{°}-T∆S_f^{°} \\ =20.4kJ/mol-\left(298K\right)\left(-141.8kJ/mol·K×\frac{1kJ}{1000J}\right) \\ ∆G_f^{°}=62.65kJ/mol \end{gathered}$$

Therefore, the value is obtained as $62.65kJ/mol$.

(c)

Solve for$∆H_{rxn}^{°}$. Use the values for$∆H_f^{°}$; obtained before, and for${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{CH}}_{\text{2}}$and$H_2$, they are found in Appendix B.

$$\begin{gathered} ∆H_{rxn}^{°}=\underset{}{\overset{}{∑}}{n}_p∆H_f^{°}\left(product\right)-\underset{}{\overset{}{∑}}{n}_r∆H_f^{°}\left(reac\tan t\right) \\ =\left[n∆H_f^{°}\left({\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}\left(g\right)\right)+n∆H_f^{°}\left(H_2\left(g\right)\right)\right]-\left[n∆H_f^{°}\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\left(g\right)\right)\right] \\ =\left(\left[1\left(20.4\right)+1\left(0\right)\right]-\left[1\left(-105\right)\right]\right)kJ \\ ∆H_{rxn}^{°}=125.4kJ \end{gathered}$$

Next, solve for $∆G_{rxn}^{°}$. Use the values for$∆G_f^{°}$ ; obtained before, and for ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{CH}}_{\text{2}}$ and$H_2$ , they are found in Appendix B.

$$\begin{gathered} ∆G_{rxn}^{°}=\underset{}{\overset{}{∑n_p}}∆G_f^{°}\left(product\right)-\underset{}{\overset{}{∑}}{n}_r∆G_f^{°}\left(reac\tan t\right) \\ =\left[n∆G_f^{°}\left({\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}\left(g\right)\right)+n∆G_f^{°}\left(H_2\left(g\right)\right)\right]-\left[n∆G_f^{°}\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\left(g\right)\right)\right] \\ =\left(\left[1\left(62.65\right)+1\left(0\right)\right]-\left[1\left(-24.5\right)\right]\right)kJ \\ ∆G_{rxn}^{°}=87.15kJ \end{gathered}$$

Therefore, the values are obtained as $∆H_{rxn}^{°}=125.4kJ$ and$∆G_{rxn}^{°}=87.15kJ$ .

(d)

First, solve for$K$and solve$P_{{\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}}$from the$K$expression. Solve for$∆S_{rxn}^{°}$first. Use the values for $S_f^{°}$ solved before, and for${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{CH}}_{\text{2}}$and$H_2$, they are found in Appendix B.

$$\begin{gathered} ∆S_{rxn}^{°}=\underset{}{\overset{}{∑}}{n}_p{S}_f^{°}\left(product\right)-\underset{}{\overset{}{∑}}{n}_r{S}_f^{°}\left(reac\tan t\right) \\ =\left[n{S}_f^{°}\left({\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}\left(g\right)\right)+n{S}_f^{°}\left(H_2\left(g\right)\right)\right]-\left[n{S}_f^{°}\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\left(g\right)\right)\right] \\ =\left(\left[1\left(267.1\right)+1\left(130.6\right)\right]-\left[1\left(269.9\right)\right]\right)J/K \\ ∆S_{rxn}^{°}=127.8J/K \end{gathered}$$

Solve for $∆G_{rxn}^{°}$–

$$\begin{gathered} ∆G_{rxn}^{°}=∆H_{rxn}^{°}-T∆S_{rxn}^{°} \\ =\left(1125.4kJ/mol\right)-\left(853K\right)\left(127.8J/mol·K×\frac{1kJ}{1000J}\right) \\ ∆G_{rxn}^{°}=16.4kJ/mol \end{gathered}$$

Now, solve for$K$–

$$\begin{gathered} ∆G_f^{°}=-RTInk \\ InK=\frac{∆G_{rxn}^{°}}{-RT} \\ =\frac{16.4kJ/mol×{\displaystyle \frac{1000J}{1kJ}}}{-\left(8.314J/mol·k\right)\left(853k\right)} \\ \text{lnK = - 2.3} \\ {\text{K =e}}^{\text{- 2.3}} \\ \text{K = 0.0992} \end{gathered}$$

Write the$K$expression.

$\text{K =}\frac{{\text{P}}_{{\text{CH}}_{\text{3}}\text{CH}}{\text{= CH}}_{\text{2}}·{\text{P}}_{{\text{H}}_{\text{2}}}}{{\text{P}}_{{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}}}$

It is known that${\text{P}}_{{\text{CH}}_{\text{3}}{\text{CH = CH}}_{\text{2}}}{\text{=P}}_{{\text{H}}_{\text{2}}}$ and we know that their change will be equivalent to the change subtracted from the initial pressure of$P_{{\text{CH}}_{\text{3}}{\text{CH}}_2{\text{CH}}_3}$ . Denote changes as $x$ –

$K=\frac{\left(x\right)\left(x\right)}{1.00atm}$

Solve for$x$–

$$\begin{gathered} K=\frac{\left(x\right)\left(x\right)}{1.00atm-x} \\ \left(K\right)\left(1.00-x\right)=x^2 \\ {x}^2+\left(0.0992\right)x-0.0992=0 \\ x=0.269atm \end{gathered}$$

Solve for the theoretical yield –

$$\begin{gathered} TheoreticalYield=\frac{0.269atm}{1.00atm}×100\% \\ =26.9\% \end{gathered}$$

Therefore, the value is obtained as $26.9\%$.

(e)

If the reaction chamber is permeable to $H_2$, $H_2$will now try to escape the chamber because the concentration of $H_2$outside the chamber is lower. The concentration of $H_2$will now decrease. According to Le Chatelier's principle, since the concentration of a product decreases, it will shift the reaction to the right producing more propylene.

Therefore, the yield will change.

(f)

Solve for$T$where$∆G^{°}=0$–

$$\begin{gathered} ∆G^{°}=∆H^{°}-T∆S^{°} \\ 0=∆H^{°}-T∆S^{°} \\ T∆S^{°}=∆H^{°} \\ T=\frac{∆H^{°}}{∆S^{°}} \\ =\frac{125.4kJ×{\displaystyle \frac{1000J}{1kJ}}}{127.8J/K} \\ T=978K \end{gathered}$$

Therefore, the value is obtained as $T=978K$.