Chapter 17, Problem 17.84P

(a)

Interpretation Introduction

Interpretation:

A source of ethanol is the reaction catalysed by ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ of steam with ethylene derived from oil.

  $\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{4(g)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(g)}}\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(g)}}\\ \\ {\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= -47}\text{.8 kJ}{\text{K}}_{\text{c}}{\text{= 9×10}}^{\text{3}}\text{ at 600 K}\end{array}$

At equilibrium, ${\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\text{= 200}\text{.atm}$ and ${\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{= 400}\text{.atm}$.  ${\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}$ has to be calculated.

Concept Introduction:

Equilibrium constant:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

  ${\text{aA}}_{\text{(g)}}{\text{+ bB}}_{\text{(g)}}\rightleftharpoons {\text{cC}}_{\text{(g)}}{\text{+ dD}}_{\text{(g)}}$

The expression of ${\text{K}}_{\text{p}}$ can be given as

  ${\text{K}}_{\text{p}}\text{ = }\frac{{{\text{(P}}_{\text{C}}\text{)}}^{\text{c}}{{\text{(P}}_{\text{D}}\text{)}}^{\text{d}}}{{{\text{(P}}_{\text{A}}\text{)}}^{\text{a}}{{\text{(P}}_{\text{B}}\text{)}}^{\text{b}}}$

Explanation of Solution

From the given ${\text{K}}_{\text{c}}$ value, ${\text{K}}_{\text{p}}$ is calculated.

  $\begin{array}{l}{\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\text{Δn}}\\ \\ \text{Δn = moles of gaseous products - moles of gaseous reactants}\\ \\ \text{ = 1-2 = -1}\\ \\ {\text{K}}_{\text{p}}{\text{= (9×10}}^{\text{3}}\text{)[(0}\text{.0821 L}\text{.atm/mol}\text{.K)(600}{\text{.K)]}}^{\text{-1}}\\ \\ {\text{K}}_{\text{p}}\text{= 1}{\text{.82704×10}}^{\text{2}}\end{array}$

The values are substituted in equilibrium constant expression.

  $\begin{array}{l}{\text{K}}_{\text{p}}\text{ = }\frac{{\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{OH}}}{{\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}}\\ \\ \text{1}{\text{.8270×10}}^{\text{2}}\text{ = }\frac{\text{200}}{{\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}\text{(400)}}\\ \\ {\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}\text{= 2}{\text{.7367×10}}^{\text{-3}}{\text{ = 3×10}}^{\text{-3}}\text{atm}\end{array}$

${\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}$ is ${\text{3×10}}^{\text{-3}}\text{ atm}$.

(b)

Interpretation Introduction

Interpretation:

The highest yield of ethanol obtained at either high or low $\text{P}$ or at high or low $\text{T}$ has to be given.

Concept Introduction:

Le Chatelier’s principle:

Le Chatelier’s principle states that the changes in the temperature, pressure, volume and concentration of the system results in the change in system to attain new equilibrium.  It is used to understand the conditions of a reaction which favours increased product formation.

Change in equilibrium due to pressure changes:

On increase in the system pressure, the equilibrium shifts towards fewer moles of gas, because, for gases,

  $\text{increase in pressure = decrease in volume}\text{.}$

On decrease in the system pressure, the equilibrium shifts towards more moles of gas, because, for gases,

  $\text{decrease in pressure = increase in volume}\text{.}$

Change in equilibrium due to temperature changes:

If the temperature is increased for the system, the equilibrium shifts away from the heat because of the reaction needs extra heat to use.

If the temperature is decreased for the system, the equilibrium shifts towards the heat because the heat needs to be produced to make up for the loss.

Explanation of Solution

The given reaction is

  $\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{4(g)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(g)}}\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(g)}}\\ \\ {\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= -47}\text{.8 kJ}{\text{K}}_{\text{c}}{\text{= 9×10}}^{\text{3}}\text{ at 600 K}\end{array}$

As the ${\text{ΔH}}_{\text{r×n}}^{\text{°}}$ is negative, the reaction is exothermic that is heat is released from the reaction.

  $\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{4(g)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(g)}}\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(g)}}\text{+ heat}\\ \end{array}$

The reaction proceeds towards products at low temperature.  Heat is removed from the reaction and more yield of ethanol is obtained.

In reactants, two moles of gaseous molecules are present and in products, one mole of gaseous molecules are present.  The high pressure favours the formation of ethanol as fewer moles are present.

(c)

Interpretation Introduction

Interpretation:

A source of ethanol is the reaction catalysed by ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ of steam with ethylene derived from oil.

  $\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{4(g)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(g)}}\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(g)}}\\ \\ {\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= -47}\text{.8 kJ}{\text{K}}_{\text{c}}{\text{= 9×10}}^{\text{3}}\text{ at 600 K}\end{array}$

${\text{K}}_{\text{c}}$ of the reaction at $\text{450}\text{.K}$ has to be calculated.

Explanation of Solution

${\text{K}}_{\text{c}}$ of the reaction at $\text{450}\text{.K}$ can be calculated as

Given,

  $\begin{array}{l}{\text{K}}_{\text{1}}{\text{= 9×10}}^{\text{3}}{\text{T}}_{\text{1}}\text{= 600}\text{.K}\\ \\ {\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= (-47}\text{.8 kJ)}\left(\frac{{\text{10}}^{\text{3}}\text{ J}}{\text{1 kJ}}\right)\\ \\ \text{= -4}{\text{.78×10}}^{\text{4}}\text{ J}\end{array}$

${\text{K}}_{\text{c}}$ at $\text{450}\text{.K}$ is

  ${\text{K}}_{\text{2}}\text{=?}{\text{T}}_{\text{2}}\text{=450}\text{.K}\text{R=8}\text{.314 J/mol}\text{.K}$

  $\begin{array}{l}\text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}\text{ = -}\frac{{\text{ΔH}}_{\text{r×n}}^{\text{°}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)\\ \\ \text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{9×10}}^{\text{3}}}\text{ = -}\frac{\text{-4}{\text{.78×10}}^{\text{4}}\text{ J}}{\text{8}\text{.314 J/mol}\text{.K}}\left(\frac{\text{1}}{\text{450}\text{.K}}\text{-}\frac{\text{1}}{\text{600}\text{.K}}\right)\\ \\ \text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{9×10}}^{\text{3}}}\text{ = 3}\text{.1940769}\\ \\ \frac{{\text{K}}_{\text{2}}}{{\text{9×10}}^{\text{3}}}\text{ = 24}\text{.38765}\\ \\ {\text{K}}_{\text{2}}{\text{ = (9×10}}^{\text{3}}\text{)(24}\text{.38765)}\\ \\ {\text{K}}_{\text{2}}\text{ = 2}{\text{.1949×10}}^{\text{5}}\end{array}$

The value of ${\text{K}}_{\text{c}}$ at $\text{450}\text{.K}$ is $\text{2}{\text{.1949×10}}^{\text{5}}$.

(d)

Interpretation Introduction

Interpretation:

In ${\text{NH}}_{\text{3}}$ manufacture, the yield is increased by condensing to a liquid and removing it.  The condensation of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ have the same effect in ethanol production or not has to be explained.

Explanation of Solution

On condensation, ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ yield is not increased.  It is because, the boiling point of ethanol ($\text{78}{\text{.5}}^{\text{°}}\text{C}$) is lower than the boiling point of water (${\text{100}}^{\text{°}}\text{C}$).  On condensation, water molecules are removed.  So, moles of gaseous molecules will get decreased from each side of the reaction.  Hence, the equilibrium is unaffected.