Chapter 9, Problem 74E

(a)

Interpretation Introduction

Interpretation: Interpret the $\text{ΔH}$ value for the given reaction with the help of ${\text{ΔH}}_{\text{f}}^{\text{0}}$ values related to the production of nitric acid for the Oswald process.

  $\begin{array}{l}{\text{4 NH}}_{\text{3(g) }}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{4 NO}}_{\text{(g)}}{\text{+6 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ {\text{2 NO}}_{\text{(g)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2 NO}}_{\text{2(g)}}\\ {\text{3 NO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{2 HNO}}_{\text{3}}{}_{\text{(aq)}}{\text{+ NO}}_{\text{(g)}}\end{array}$

Concept Introduction:At constant volume the change in heat for a system to change the internal energy is represented as ΔE or q_V. At constant pressure the change in heat for a system to change the enthalpy is represented as ΔH or q_p.

The $\text{ΔH}$ for a chemical reaction can be calculated with the help of difference between $\text{ΔH}$ of product and reactant.

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

Answer to Problem 74E

  $\begin{array}{l}{\text{4 NH}}_{\text{3(g) }}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{4 NO}}_{\text{(g)}}{\text{+6 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\text{ ΔH=-908kJ}\\ {\text{2 NO}}_{\text{(g)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2 NO}}_{\text{2(g)}}\text{ ΔH=-112 kJ}\\ {\text{3 NO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{2 HNO}}_{\text{3}}{}_{\text{(aq)}}{\text{+ NO}}_{\text{(g)}}\text{ΔH=-140 kJ}\end{array}$

Explanation of Solution

  $\begin{array}{l}{\text{4 NH}}_{\text{3(g) }}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{4 NO}}_{\text{(g)}}{\text{+6 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ {\text{2 NO}}_{\text{(g)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2 NO}}_{\text{2(g)}}\\ {\text{3 NO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{2 HNO}}_{\text{3}}{}_{\text{(aq)}}{\text{+ NO}}_{\text{(g)}}\end{array}$

Calculate $\text{ΔH}$ for each of the reaction:

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

  $\begin{array}{l}{\text{4 NH}}_{\text{3(g) }}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{4 NO}}_{\text{(g)}}{\text{+6 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ \text{ΔH= }\sum 4{\text{×ΔH}}_{\text{NO(g)}}{\text{+6×ΔH}}_{\text{H2O(g)}}\text{ - }\sum {\text{4×ΔH}}_{\text{NH3(s)}}\text{+5}\times {\text{ΔH}}_{\text{O2(g)}}\\ \text{ΔH= }\sum 4\text{×90 kJ+6×(-242kJ) - }\sum \text{4×(-46kJ) +5}\times \text{0}\\ \text{ΔH= - 908 kJ}\end{array}$

  $\begin{array}{l}{\text{2 NO}}_{\text{(g)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2 NO}}_{\text{2(g)}}\\ \text{ΔH= }\sum 2{\text{×ΔH}}_{\text{NO2(g)}}\text{ - }\sum {\text{2×ΔH}}_{\text{NO(g)}}{\text{+ ΔH}}_{\text{O2(g)}}\\ \text{ΔH= }\sum 2\text{×34 kJ - }\sum 2\text{×(90 kJ) + 0}\\ \text{ΔH= - 112 kJ}\end{array}$

  $\begin{array}{l}{\text{3 NO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{2 HNO}}_{\text{3}}{}_{\text{(aq)}}{\text{+ NO}}_{\text{(g)}}\\ \text{ΔH= }\sum 2{\text{×ΔH}}_{\text{HNO3(aq)}}{\text{+ ΔH}}_{\text{NO(g)}}\text{ - }\sum {\text{3×ΔH}}_{\text{NO2(g)}}{\text{+ ΔH}}_{\text{H2O(l)}}\\ \text{ΔH= }\sum 2\text{×(-207kJ)+ 90 kJ - }\sum \text{3×34 kJ+ (-286kJ)}\\ \text{ΔH= - 140 kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:Interpret the overall equation for the production of nitric acid by Ostwald process by the combination of given reaction and also interpret the reaction as exothermic and endothermic.

  $\begin{array}{l}{\text{4 NH}}_{\text{3(g) }}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{4 NO}}_{\text{(g)}}{\text{+6 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ {\text{2 NO}}_{\text{(g)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2 NO}}_{\text{2(g)}}\\ {\text{3 NO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{2 HNO}}_{\text{3}}{}_{\text{(aq)}}{\text{+ NO}}_{\text{(g)}}\end{array}$

Concept Introduction:At constant volume the change in heat for a system to change the internal energy is represented as ΔE or q_V. At constant pressure the change in heat for a system to change the enthalpy is represented as ΔH or q_p.

The $\text{ΔH}$ for a chemical reaction can be calculated with the help of difference between $\text{ΔH}$ of product and reactant.

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

Answer to Problem 74E

  $12{\text{NH}}_{\text{3(g) }}+21{\text{O}}_{\text{2(g)}}\stackrel{}{\to }{\text{8 HNO}}_{\text{3}}{}_{\text{(aq)}}+4{\text{ NO}}_{\text{(g)}}{\text{+14 H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Exothermic reaction

Explanation of Solution

Given:

  $\begin{array}{l}{\text{4 NH}}_{\text{3(g) }}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{4 NO}}_{\text{(g)}}{\text{+6 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\text{ ΔH=-908kJ}\\ {\text{2 NO}}_{\text{(g)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2 NO}}_{\text{2(g)}}\text{ ΔH=-112 kJ}\\ {\text{3 NO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{2 HNO}}_{\text{3}}{}_{\text{(aq)}}{\text{+ NO}}_{\text{(g)}}\text{ΔH=-140 kJ}\end{array}$

From the given reactions, the Ostwald reaction for the production of nitric acid:

  $\begin{array}{l}{\text{12 NH}}_{\text{3(g) }}{\text{ + 15 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 12 NO}}_{\text{(g)}}{\text{+ 18 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\text{ ΔH= 3 }\times \text{-908kJ=-2724kJ}\\ {\text{12 NO}}_{\text{(g)}}{\text{ + 6 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{12 NO}}_{\text{2(g)}}\text{ ΔH= 6 }\times \text{-112 kJ=-672 kJ}\\ {\text{12 NO}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{8 HNO}}_{\text{3}}{}_{\text{(aq)}}{\text{+ 4 NO}}_{\text{(g)}}\text{ΔH= 4}\times \text{-140 kJ=-560kJ}\\ {\text{4 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\stackrel{}{\to }{\text{4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ 12{\text{NH}}_{\text{3(g) }}+21{\text{O}}_{\text{2(g)}}\stackrel{}{\to }{\text{8 HNO}}_{\text{3}}{}_{\text{(aq)}}+4{\text{ NO}}_{\text{(g)}}{\text{+14 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\text{ΔH}=-2724-672-5\text{60 =-3956kJ}\end{array}$

Since each step of the reaction is exothermic therefore overall reaction will be exothermic.