Chapter 17, Problem 17.76P

(a)

Interpretation Introduction

Interpretation:

The given reaction is

  ${\text{SO}}_{\text{2}}\text{(g) + }\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g) }\rightleftharpoons {\text{ SO}}_{\text{3}}\text{(g)}{\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= -99}\text{.2kJ}$

The qualitative combination of $\text{T and P}$ that maximizes the yield of ${\text{SO}}_{\text{3}}$ 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.

(b)

Interpretation Introduction

Interpretation:

The given reaction is

  ${\text{SO}}_{\text{2}}\text{(g) + }\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g) }\rightleftharpoons {\text{ SO}}_{\text{3}}\text{(g)}{\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= -99}\text{.2kJ}$

The effect of $Q\text{ and K}$ has to be given on addition of ${\text{O}}_{\text{2}}$.

Concept Introduction:

Equilibrium constant:

The relationship between the amount of reactants and products that are present at equilibrium in a reversible reaction at a given temperature is said to be equilibrium constant.  It is denoted by $\text{K}$.

Reaction quotient:

The ratio of concentration terms for a reaction is known as reaction quotient.  It is denoted by $Q$.  For a reversible reaction of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ to NO}}_{\text{2}}$, the reaction quotient is represented as

  $\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}\left(\text{g}\right)}\rightleftharpoons {\text{2NO}}_{\text{2(g)}}\\ \\ Q\text{ = }\frac{{\left({\text{NO}}_{\text{2}}\right)}^{\text{2}}}{\left({\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right)}\end{array}$

At equilibrium, $Q\text{ = K}$.

(c)

Interpretation Introduction

Interpretation:

The given reaction is

  ${\text{SO}}_{\text{2}}\text{(g) + }\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g) }\rightleftharpoons {\text{ SO}}_{\text{3}}\text{(g)}{\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= -99}\text{.2kJ}$

The reason for the use of catalysis for the given reaction has to be explained.

Concept Introduction:

Catalysis:

To change or alter the rate of reaction, a catalyst is used.  The process is said to be catalysis.  The catalyst is not consumed in the reaction.