Chapter 12, Problem 84QRT

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for each reaction has to be calculated.

  $\begin{array}{l}{\text{S}}_{\text{(s)}}{\text{+O}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{SO}}_{\text{2(g)}}\\ {\text{2SO}}_{\text{2(g)}}{\text{+O}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{2SO}}_{\text{3(g)}}\\ {\text{SO}}_{\text{3(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}\end{array}$

Concept Introduction:

Hess's Law:

The enthalpy change of given reaction is calculated by subtraction of sum of enthalpy of formation reactants from sum of enthalpy of formation reactant products.

  ${\text{ΔH}}_{\text{rex}}\text{=}{\displaystyle \sum {\text{ΔH}}_{\text{product}}}\text{-}{\displaystyle \sum {\text{ΔH}}_{\text{reactant}}}$

(b)

Interpretation Introduction

Interpretation:

The exothermic and endothermic reaction has to be given for below given reactions.

  $\begin{array}{l}{\text{S}}_{\text{(s)}}{\text{+O}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{SO}}_{\text{2(g)}}\\ {\text{2SO}}_{\text{2(g)}}{\text{+O}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{2SO}}_{\text{3(g)}}\\ {\text{SO}}_{\text{3(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}\end{array}$

Concept Introduction:

In the chemical reaction, negative value of enthalpy means exothermic reaction and positive value means endothermic reaction.

(c)

Interpretation Introduction

Interpretation:

The reaction, which contains greater entropy, has to be given from below given reactions.

  $\begin{array}{l}{\text{S}}_{\text{(s)}}{\text{+O}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{SO}}_{\text{2(g)}}\\ {\text{2SO}}_{\text{2(g)}}{\text{+O}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{2SO}}_{\text{3(g)}}\\ {\text{SO}}_{\text{3(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}\end{array}$

Concept Introduction:

If number of mole increases result the increasing entropy value and decreases result the decreasing entropy of reaction.

(d)

Interpretation Introduction

Interpretation:

The reaction, which is favor at low temperature, has to be given from below given reactions.

  $\begin{array}{l}{\text{S}}_{\text{(s)}}{\text{+O}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{SO}}_{\text{2(g)}}\\ {\text{2SO}}_{\text{2(g)}}{\text{+O}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{2SO}}_{\text{3(g)}}\\ {\text{SO}}_{\text{3(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}\end{array}$

Concept Introduction:

If number of mole increases result the increasing entropy value and decreases result the decreasing entropy of reaction.