An important industrial source of ethanol is the reaction, catalyzed by ${\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}$ , of steam with ethylene derived from oil:

${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{∆}{\mathbf{H}}_{\mathbf{rxn}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{47}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{kJ}\mathbf{}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{c}}\mathbf{=}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{600}\mathbf{K}$

1. At equilibrium, ${\mathbf{P}}_{{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}}\mathbf{=}\mathbf{200}\mathbf{}\mathbf{atm}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{400}\mathbf{}\mathbf{atm}$ Calculate ${\mathbf{P}}_{{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}}$ .
2. Is the highest yield of ethanol obtained at high or low$\mathbf{P}\mathbf{?}$ High or low$\mathbf{T}\mathbf{?}$
3. Calculate ${\mathbf{K}}_{\mathbf{c}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{450}\mathbf{}\mathbf{K}$.
4. In manufacture, the yield is increased by condensing the ${\mathbf{NH}}_{\mathbf{3}}$ to a liquid and removing it. Would condensing the${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}$ have the same effect in ethanol production? Explain.

Ethyl alcohol is also known as grain alcohol, ethanol, and fermentation alcohol. It is an intoxicating component of fermented liquor that is generally formed by yeast fermentation of specific carbohydrates. Ethanol is a colourless, limpid, volatile, combustible, water-miscible liquid with an ether like odour and having a strong, burning taste.

(b) In this problem we are supposed to evaluate the equilibrium reaction that is shown below.

${\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{g}\right)$

where its,

$\begin{array}{rcl}∆{\mathrm{H}}_{\mathrm{rxn}}^{\mathrm{o}}& =& -47.8\mathrm{kJ}\\ & =& -47800\mathrm{J}\end{array}$

And its ${\mathrm{K}}_{\mathrm{c}}=9\times {10}^3\mathrm{at}\mathrm{T}=600\mathrm{K}$.

First, we need to write the expression for the equilibrium constant of the reaction in terms of partial pressures.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=\frac{\mathrm{Product}}{\mathrm{Reactant}} \\ {\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}}{{\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_4}.{\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}} \end{gathered}$$

Now, Calculating ${\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_4}$

In order to calculate ${\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_4}$we need to solve for ${\mathrm{K}}_{\mathrm{p}}$using the given ${\mathrm{K}}_{\mathrm{c}}$.

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{p}}& =& {\mathrm{K}}_{\mathrm{c}}{\left(\mathrm{RT}\right)}^{∆\mathrm{n}}\\ & =& \left(9\times {10}^3\right){\left(0.0821\times 600\right)}^{-1}\\ {\mathrm{K}}_{\mathrm{p}}& =& 182.7\end{array}$

Solve for ${\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_4}$ using the expression for the equilibrium constant in terms of partial pressures.

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{p}}& =& \frac{{\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}}{{\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_4}\times {\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}}\\ {\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_4}& =& \frac{{\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}}{{\mathrm{K}}_{\mathrm{p}}\times {\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}}\\ & =& \frac{200}{\left(182.7\right)\left(400\right)}\\ {\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_4}& =& 3\times {10}^{-3}\mathrm{atm}\end{array}$

Therefore, we get ${\mathrm{P}}_{{\mathrm{C}}_2{\mathrm{H}}_4}=3\times {10}^{-3}\mathrm{atm}$

On the product side, there is ethanol $\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)$. On the reactant side, there are$2\mathrm{mol}$ of gas, while on the product side, there are $1\mathrm{mol}$ of gas. Because the product side contains fewer moles, a higher pressure will favour ethanol production.

A higher pressure indicates that the gas molecules have a restricted amount of room. To cope with the change, the equilibrium will shift to the side with fewer molecules, balancing the pressure.

The reaction is exothermic because the $∆{\mathrm{H}}_{\mathrm{rxn}}^{\mathrm{o}}$is negative. When heat is expelled, exothermic reactions occur

${\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{g}\right)+\mathrm{heat}$

We need to lower the reaction's heat to make more ethanol. To compensate for the "lost" heat, this will favour the product side. As a result, at lower temperatures, ethanol production will be higher.

(c) We can use the Van't Hoff equation to solve for the ${\mathrm{K}}_{\mathrm{c}}$at a different temperature. We denote the given (b) and K₁ and the K_C to be solved as K₂ .

$\begin{array}{rcl}\ln \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}& =& \frac{-∆{\mathrm{H}}^{\mathrm{o}}}{\mathrm{R}}\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\\ & =& -\frac{-47800}{8.314}\frac{1}{450}-\frac{1}{600}\\ \ln \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}& =& 3.194\\ \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}& =& {\mathrm{e}}^{3.194}\\ {\mathrm{K}}_2& =& {\mathrm{K}}_1\times {\mathrm{e}}^{3.194}\\ & =& \left(9\times {10}^3\right)\left({\mathrm{e}}^{3.194}\right)\\ {\mathrm{K}}_2& =& 2\times {10}^5\end{array}$$\begin{array}{rcl}\ln \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}& =& \frac{-∆{\mathrm{H}}^{\mathrm{o}}}{\mathrm{R}}\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\\ & =& -\frac{-47800}{8.314}\frac{1}{450}-\frac{1}{600}\\ \ln \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}& =& 3.194\\ \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}& =& {\mathrm{e}}^{3.194}\\ {\mathrm{K}}_2& =& {\mathrm{K}}_1\times {\mathrm{e}}^{3.194}\\ & =& \left(9\times {10}^3\right)\left({\mathrm{e}}^{3.194}\right)\\ {\mathrm{K}}_2& =& 9\times {10}^5\end{array}$

We have ethene gas, water vapour, and gaseous ethanol in the given reaction. It might work if we convert ethanol to a pure liquid by condensing it because the gaseous ethanol will no longer be present, causing the reaction to proceed to the right. Ethanol, on the other hand, has a lower boiling point than water. When we condense ethanol, the water condenses first, reducing the amount of our reactants.

As a result, condensing ethanol will not work since water has a higher boiling point than ethanol and will condense first.