Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presence of superheated steam.

(a) Find$\mathbf{∆}{\mathit{H}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}\mathbf{,}\mathbf{}\mathbf{∆}{\mathit{G}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{∆}{\mathit{S}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}$ given these data at$\mathbf{298}\mathit{K}$ :

(b)At what temperature is the reaction spontaneous?

(c)What are$\mathbf{∆}{\mathit{G}}_{\mathbf{\gamma }\mathbf{N}\mathbf{n}}^{\mathbf{°}}$ and${\mathbf{600}}^{\mathbf{°}}\mathit{C}$ ?

(d)With$\mathbf{5}\mathbf{.}\mathbf{0}$ parts steam to$\mathbf{1}\mathbf{.}\mathbf{0}$ part ethylbenzene in the reactant mixture and the total pressure kept constant at$\mathbf{1}\mathbf{.}\mathbf{3}\mathit{a}\mathit{t}\mathit{m}$ what is$\mathbf{∆}\mathit{G}$$\mathit{a}\mathit{t}\mathbf{}\mathbf{50}\mathbf{\%}$ conversion, that is, when$\mathbf{50}\mathbf{\%}$ of the ethylbenzene has reacted?

Treatment with hydrogen in the presence of a catalyst, such as nickel, palladium, or platinum, is known as catalytic hydrogenation. The reaction requires catalysts to be usable; non-catalytic hydrogenation occurs only at extremely high temperatures. Hydrogenation breaks down hydrocarbons' double and triple bonds.

a) At first, write the reaction of catalytic dehydrogenation of ethylbenzene:

The entalphy of the reaction at${25}^{°}C\left(298K\right)$ can be calculated using Appendix B and given values:

$$\begin{gathered} ∆H_{ren}^{°}=∆H_{product}^{°}-∆H_{reac\tan t}^{°} \\ ∆H_{Txn}^{°}=\left[1∆H^{°}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{- CH = CH}}_{\text{2}}\text{(l)}\right)+1∆H\left(H_2\left(g\right)\right)\right]-\left[1H^{°}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{- CH}}_{\text{2}}{\text{CH}}_{\text{3}}\text{(l)}\right)\right] \\ ∆H_{rxn}^{°}=\left[1mol\times \left(103.80\frac{kJ}{mol}\right)+1mol\times \left(0.00\frac{kJ}{mol}\right)\right]-\left[1mol\times \left(-12.50\frac{kJ}{mol}\right)\right] \\ ∆H_{rxn}^{·}=116.30\frac{kJ}{mol} \end{gathered}$$

The enthalpy change of the reaction at $298Kis116.30\frac{kJ}{mol}$

Similarly, the entropy of the reaction at ${25}^{°}Cat298K$ can be calculated using Appendix B:

$$\begin{gathered} ∆H_{ren}^{°}=∆H_{products}^{°}-∆H_{reac\tan t}^{°} \\ ∆S_{rxn}^{°}=∆S_{product}^{°}-∆S_{reac\tan t}^{°}∆S_{rxn}^{°} \\ =\left[1∆S^{°}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{- CH = CH}}_{\text{2}}\text{(l)}\right)+1∆S^{°}\left(H_2\left(g\right)\right)\right]-\left[1∆S^{°}\left(\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{- CH}}_{\text{2}}{\text{CH}}_{\text{3}}\text{(l)}\right)\right)\right] \\ ∆S^{°}=\left[1mol\times \left(238.00\frac{J}{mol\times K}\right)+1mol\times \left(130.60\frac{J}{mol\times K}\right)\right]-\left(255.00\frac{J}{mol\times K}\right) \\ ∆S_{rxn}^{°}=113.60\frac{J}{mol\times K} \end{gathered}$$

The entropy change of the reaction at ${25}^{°}Cat298K$ is$113.60\frac{J}{mol\times K}$ .

Finally, $∆G_{rxn}^{°}$ can be calculated:

$$\begin{gathered} ∆G_{rxn}^{°}=∆H_{rxn}^{°}-T∆S_{rxn}^{°} \\ ∆G_{rxn}^{°}=116.30\frac{kJ}{mol}-298K\times \left(113.60\times {10}^{-3}\frac{kJ}{mol\times K}\right) \\ ∆G_{rxn}^{°}=84.4472\frac{kJ}{mol} \end{gathered}$$

b) To calculate the temperature for the spontaneous reaction, when$∆G_{rxn}^{°}<0$ .

$$\begin{gathered} ∆G_{rxn}^{°}=∆H_{rxn}^{°}-T\times ∆S_{rxn}^{°}<0 \\ ∆H_{rxn}^{°}<T\times ∆S_{rxn}^{°} \\ \frac{∆H_{rxn}^{°}}{∆S_{rxn}^{°}}<T \\ T>\frac{116.30{\displaystyle \frac{kJ}{mol}}}{113.60\times {10}^{-3}{\displaystyle \frac{kJ}{mol\times K}}} \\ T>1023.7676K \end{gathered}$$

Thus, at the temperatures above $1023.77K$ the reaction is spontaneous.

c) Let us calculate$∆G_{rxn}at{600}^{°}C\left(873K\right)$ :

$$\begin{gathered} ∆G_{rxn}=∆H_{rxn}-T\times ∆S_{rxn}∆G_{rxn} \\ =116.30\frac{kJ}{mol}-873K\times 113.60\times {10}^{-3}\frac{kJ}{mol\times K}\$\$ \\ ∆G_{rxn}=17.1272\frac{kJ}{mol} \end{gathered}$$

Gibb's free energy change of the reaction at ${600}^{°}Cis17.13\frac{kJ}{mol}$

The $K$ can be then found as,

$$\begin{gathered} ∆G_{rxn}=-R\times T\times InK \\ InK=\frac{∆G_{rxn}}{-R\times T} \\ K=e^{InK} \end{gathered}$$

Using the calculated values,

$$\begin{gathered} K=\frac{17.1272\times {10}^3{\displaystyle \frac{J}{mol}}}{-8.314\times {\displaystyle \frac{J}{mol\times K}}\times 873K} \\ InK=-2.\text{35973} \\ \text{K = 0.0944} \end{gathered}$$

The equilibrium constant for the reaction at ${600}^{°}Cis0.0944$

d) Looking at the reaction again, if the initial partial pressure of ethylbenzene was$1.3\frac{1}{6}$ atm and the products partial pressure was$0$ , then the final partial pressure of ethylbenzene was $1.3\times \frac{1}{6}\times 0.5$ after the reaction at$50.0\%$ conversion rate, as shown in Table below:

The reaction quotient can be calculated then,

The reaction quotient is $0.01083$

Then, the Gibb's free energy change of the reaction at standard

temperature can be recalculated as:

$$\begin{gathered} ∆G_{rxn}=∆G_{rxn}^{°}+R\times T\times InQ \\ ∆G_{rxn}=17.1272\times {10}^3\frac{J}{mol}+8.314\frac{J}{mol\times K}\times 298K\times In\left(0.1083\right) \\ ∆G_{rxn}=11619.93\frac{J}{mol} \\ =11.62\frac{kJ}{mol} \end{gathered}$$

At the given conditions $298K,∆G_{rxn}^{°}$and $11.62kJ/mol$