An important industrial source of ethanol is the reaction, catalyzed by H₃PO₄, of steam with ethylene derived from oil:

$$\begin{gathered} {\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\left(g\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{\rightleftharpoons }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\left(g\right) \\ {\mathbf{\Delta H}}_{\mathbf{r}\mathbf{\times }\mathbf{n}}^{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{47}\mathbf{.}\mathbf{8}{\mathbf{KJK}}_{\mathbf{c}}\mathbf{=}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{at}\mathbf{}\mathbf{600}\mathbf{.}\mathbf{k} \end{gathered}$$

(a) At equilibrium,${\mathbf{P}}_{{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}}\mathbf{=}\mathbf{200}\mathbf{.}\mathbf{atm}$and${\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{400}\mathbf{.}\mathbf{atm}$Calculate${\mathbf{P}}_{{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}}$(b) Is the highest yield of ethanol obtained at high or low P? High or low T? (c) Calculate Kc at 450. K. (d) In NH₃ manufacture, the yield is increased by condensing the NH₃ to a liquid and removing it. Would condensing the C₂H₅OH have the same effect in ethanol production? Explain.

Given balance chemical equation is shown below:

${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{g}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(\text{g}\right)$

Given data:${\text{K}}_{\text{c}}{\text{= 9×10}}^{\text{3}}$

${\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\text{= 200.atm}$

${\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{= 400.atm}$

To find the partial pressure of C₂H₄, equation in terms of Kp is defined as,

${\mathbf{K}}_{\mathbf{P}}\mathbf{=}\frac{\left[{\mathbf{P}}_{{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}}\right]}{\left[{\mathbf{P}}_{{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}}\right]\left[{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\right]}$ (1)

We need to find the value of kp using Kc given above:

${\mathbf{K}}_{\mathbf{P}}\mathbf{=}{\mathbf{K}}_{\mathbf{c}}{\left(\mathbf{RT}\right)}^{\mathbf{\Delta n}}$ (2)

Were,$∆n=1-2=-1$

, T=600.K, R=0.0821

Then,${\text{K}}_{\text{P}}{\text{= 9×10}}^{\text{3}}\text{×}{\left(\text{0.082×1600}\right)}^{\text{- 1}}$

$\text{= 182.70 atm}$

Substituting the value of Kp in equation 1 we get,

$\text{182.70 atm =}\frac{\text{200.atm}}{\left[{\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}\right]{\displaystyle \text{×}}\text{400.atm}}$

$\left[{\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}\right]\text{=}\frac{\text{200.atm}}{\text{182.70 atm×400.atm}}$

$=2.7\times {10}^{3}atm$

Hence partial pressure of C₂H₄ is$2.7\times {10}^{3}atm$

In a reaction an increase in pressure decreases the volume to form ethanol.Whenever a change in number of gaseous molecules occurs in a reaction, if the pressure increases (volume decreases) then the equilibrium shifts towards right side to reduce the effect of pressure.

Kc can be solved using van’t Hoff equation given below:

$\mathbf{In}\frac{{\mathbf{k}}_{\mathbf{2}}}{{\mathbf{k}}_{\mathbf{1}}}\mathbf{=}\mathbf{-}\frac{{\mathbf{\Delta H}}_{\mathbf{r}\mathbf{\times }\mathbf{n}}^{\mathbf{0}}}{\mathbf{R}}\left(\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}\right)$

Given values are${\text{∆H}}_{\text{r×n}}^{\text{0}}\text{= - 47.8KJ}$

$$\begin{gathered} {\text{K}}_{\text{c2}}{\text{= 9×10}}^{\text{3}} \\ {\text{T}}_2\text{= 600.k} \end{gathered}$$

Substituting above values in the given equation we get,

$\text{In}\left(\frac{{\text{9×10}}^{\text{3}}}{{\text{k}}_{\text{c1}}}\right)\text{= -}\frac{\text{- 47.8KJ}}{\text{8.314J/mol.K}}\left(\frac{\text{1}}{\text{600k}}\text{-}\frac{\text{1}}{\text{450k}}\right)$

$\text{In}\left(\frac{{\text{9×10}}^{\text{3}}}{{\text{k}}_{\text{c1}}}\right)\text{= -}\frac{{\text{- 47.8×10}}^{\text{3}}\text{J}}{\text{8.314J/mol.K}}\left(\text{- 0.00056}\right)$

role="math" localid="1663324323678" $\text{In}\left(\frac{{\text{9×10}}^{\text{3}}}{{\text{k}}_{\text{c1}}}\right)\text{= - 3.2196}$

${\text{k}}_{\text{c1}}\text{= 9.03}$

In Haber process to increase the yield of ammonia, it undergoes condensation process. This is because boiling point of reactants are less relative to ammonia. So, the gaseous reactants remains as it is, and they return to reaction chamber to continue the process. Likewise, ethanol can also increase the yield using this same procedure. Removing the ethanol from the product mixture and recycling the reactants, it is possible to attain 95% ethanol.