Q98CP - Chemistry: Molecular Nature Of Matter And Change

Question

Propylene (propene;${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH = CH}}}_{\mathbf{\text{2}}}$  ) is used to produce polypropylene and many other chemicals. Although most is obtained from the cracking of petroleum, about $\mathbf{2}\mathbf{\%}$   is produced by catalytic dehydrogenation of propane ( ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{3}}}$): 

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

Because this reaction is endothermic, heaters are placed between the reactor vessels to maintain the required temperature. 

(a) If the molar entropy${\mathit{S}}^{\mathbf{°}}$,  , of propylene is$\mathbf{\text{267.1 J/molK}}$  , find its entropy of formation, ${\mathbf{\text{S}}}_{\mathbf{\text{f}}}^{\mathbf{\text{o}}}$ .

(b) Find  $\mathbf{∆}{\mathit{G}}_{\mathbf{f}}^{\mathbf{°}}$ of propylene ($\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{°}}$ for propylene$\mathbf{\text{ = 20.4 kJ/mol}}$ ).

(c) Calculate $\mathbf{∆}{\mathit{H}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}$  and$\mathbf{∆}{\mathit{G}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}$   for the dehydrogenation.

(d) What is the theoretical yield of propylene at $\mathbf{580}\mathbf{°}\mathit{C}$  if the initial pressure of propane is $\mathbf{1}\mathbf{.}\mathbf{00}\mathit{a}\mathit{t}\mathit{m}$ ? 

(e) Would the yield change if the reactor walls were permeable to${\mathit{H}}_{\mathbf{2}}$  ? Explain. 

(f) At what temperature is the dehydrogenation spontaneous, with all substances in the standard state?

 

Step-by-Step Solution

Answer

$H_2$(a) The entropy of formation  $∆S_f^{°}$  is$∆S_f^{°}=-141.8J/mol·K$  .

(b) The value for $∆G_f^{°}$  of propylene is$∆G_f^{°}=62.65kJ/mol$  .

(c) The value for $∆H_{rxn}^{°}$  and$∆G_{rxn}^{°}$   for dehydrogenation is $∆H_{rxn}^{°}=-125.4kJ$  and  $∆G_{rxn}^{°}=87.15kJ$ respectively.

(d) The theoretical yield of propylene is $26.9\%$ .

(e) Yes, the yield of propylene would change and it will increase if the reactor walls were permeable to  .

(f) The temperature at which the dehydrogenation is spontaneous, with all substances in the standard state is$T=978K$  .

Step 1: Concept Introduction

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.

 

Step 2: Entropy of Formation

(a)

The reaction given is –

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

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

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

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.

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

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

Step 3: Gibbs Energy for Propylene

(b)

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

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

 

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

Step 4: Calculation for ∆ H r x n °   a n d   ∆ G r x n °

(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.

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

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.

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

 

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

Step 5: Theoretical Yield of Propylene

(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.

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

Solve for  $∆G_{rxn}^{°}$ –

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

 

Now, solve for$K$   –

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

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 –

 $$\begin{gathered} Theoretical Yield=\frac{0.269atm}{1.00atm}\times 100\% \\ =26.9\% \end{gathered}$$

 

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

Step 6: Change in the yield

(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.

Step 7: Calculation for Temperature

(f)

Solve for $T$  where $∆G^{°}=0$ –

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

 

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