Chapter 17, Problem 17.87P

(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

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

(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.

(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.

(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.