Chapter 9, Problem 73E

(a)

Interpretation Introduction

Interpretation: The $\text{ΔH}$ value for the given reaction with the help of ${\text{ΔH}}_{\text{f}}^{\text{0}}$ values needs to be determined.

  

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 $\text{ΔH}$ formation of product and reactant.

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

Answer to Problem 73E

  $\text{ΔH= - 940 kJ}$

Explanation of Solution

According to given ball and stick model, the balance chemical reaction must be:

  

  ${\text{2 NH}}_{\text{3(g) }}{\text{ + 3 O}}_{\text{2(g)}}{\text{+ 2CH}}_{\text{4(g)}}^{}\stackrel{}{\to }{\text{2HCN}}_{\text{(g)}}{\text{+6 H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

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

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

  $\begin{array}{l}{\text{2 NH}}_{\text{3(g) }}{\text{ + 3 O}}_{\text{2(g)}}{\text{+2CH}}_{\text{4(g)}}^{}\stackrel{}{\to }{\text{2HCN}}_{\text{(g)}}{\text{+6 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ \text{ΔH= }\sum 2{\text{×ΔH}}_{\text{HCN(g)}}{\text{+6×ΔH}}_{\text{H2O(g)}}\text{ - }\sum {\text{2×ΔH}}_{\text{NH3(g)}}\text{+3}\times {\text{ΔH}}_{\text{O2(g)}}+{\text{2×ΔH}}_{\text{CH4(g)}}\\ \text{ΔH= }\sum 4\text{×135}\text{.1 kJ+6×(-242kJ) - }\sum \text{2×(-46kJ) +2 (-75kJ) + 2 (0)}\\ \text{ΔH= - 940 kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation: The $\text{ΔH}$ value for the given reaction needs to be determined with the help of ${\text{ΔH}}_{\text{f}}^{\text{0}}$ values.

  ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{ + 3 H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}\stackrel{}{\to }{\text{3CaSO}}_{\text{4}}{}_{\text{(s)}}{\text{+ 2H}}_{\text{3}}{\text{PO}}_{\text{4(l)}}$

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 73E

  $\text{ΔH= - 265 kJ}$

Explanation of Solution

Given Information:

  ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{ + 3 H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}\stackrel{}{\to }{\text{3CaSO}}_{\text{4}}{}_{\text{(s)}}{\text{+ 2H}}_{\text{3}}{\text{PO}}_{\text{4(l)}}$

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

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

  $\begin{array}{l}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}{\text{ + 3 H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}\stackrel{}{\to }{\text{3CaSO}}_{\text{4}}{}_{\text{(s)}}{\text{+ 2H}}_{\text{3}}{\text{PO}}_{\text{4(l)}}\\ \text{ΔH= }\sum 2{\text{×ΔH}}_{\text{H3PO4(l)}}{\text{+3×ΔH}}_{\text{CaSO4(s)}}\text{ - }\sum {\text{ΔH}}_{\text{ca3(PO3)2(s)}}\text{+3}\times {\text{ΔH}}_{\text{H2SO4(l)}}\\ \text{ΔH= }\sum 2\text{×(-1267kJ)+3×(-1433kJ) - }\sum \text{(-4126kJ) +3}\times \text{(-814kJ)}\\ \text{ΔH= - 265 kJ}\end{array}$

(c)

Interpretation Introduction

Interpretation: The $\text{ΔH}$ value for the given reaction with the help of ${\text{ΔH}}_{\text{f}}^{\text{0}}$ values needs to be determined.

  ${\text{NH}}_{\text{3}}{}_{\text{(g)}}{\text{ + HCl}}_{\text{(g)}}\stackrel{}{\to }{\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$

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 73E

  $\text{ΔH= - 176 kJ}$

Explanation of Solution

Given:

  ${\text{NH}}_{\text{3}}{}_{\text{(g)}}{\text{ + HCl}}_{\text{(g)}}\stackrel{}{\to }{\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$

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

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

  $\begin{array}{l}{\text{NH}}_{\text{3}}{}_{\text{(g)}}{\text{ + HCl}}_{\text{(g)}}\stackrel{}{\to }{\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}\\ \text{ΔH= }\sum {\text{ΔH}}_{\text{NH4Cl(s)}}\text{ - }\sum {\text{ΔH}}_{\text{NH3(g)}}{\text{+ΔH}}_{\text{HCl(g)}}\\ \text{ΔH= }\sum 1\text{×(-314 kJ) - }\sum 1\times \text{(-46kJ) +1}\times \text{(-92kJ)}\\ \text{ΔH= - 176 kJ}\end{array}$

(d)

Interpretation Introduction

Interpretation: The value of $\text{ΔH}$ for the given reaction with the help of ${\text{ΔH}}_{\text{f}}^{\text{0}}$ values needs to be determined.

  

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 $\text{ΔH}$ formation of product and reactant.

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

Answer to Problem 73E

  $\text{ΔH= - 1235 kJ}$

Explanation of Solution

According to given ball and stick model, the balance chemical reaction must be:

  

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l) }}{\text{ + 3 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2CO}}_{\text{2}}{}_{\text{(g)}}{\text{+ 3 H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

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

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

  $\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l) }}{\text{ + 3 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2CO}}_{\text{2}}{}_{\text{(g)}}{\text{+ 3 H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ \text{ΔH= }\sum 2{\text{×ΔH}}_{\text{CO2(g)}}{\text{+3×ΔH}}_{\text{H2O(g)}}\text{ - }\sum {\text{ΔH}}_{\text{C2H5OH(l)}}\text{+3}\times {\text{ΔH}}_{\text{O2(g)}}\\ \text{ΔH= }\sum 4\text{×(-393}\text{.5 kJ)+3×(-242kJ) - }\sum 1\text{×(-278kJ) +3 (0) }\\ \text{ΔH= - 1235 kJ}\end{array}$

(e)

Interpretation Introduction

Interpretation: The $\text{ΔH}$ value for the given reaction with the help of ${\text{ΔH}}_{\text{f}}^{\text{0}}$ values needs to be determined.

  ${\text{SiCl}}_{\text{4(l)}}{\text{ + 2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{SiO}}_2{}_{\text{(s)}}{\text{+ 4 HCl}}_{\text{(aq)}}$

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 $\text{ΔH}$ formation of product and reactant.

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

Answer to Problem 73E

  $\text{ΔH= -320 kJ}$

Explanation of Solution

Given Information:

  ${\text{SiCl}}_{\text{4(l)}}{\text{ + 2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{SiO}}_2{}_{\text{(s)}}{\text{+ 4 HCl}}_{\text{(aq)}}$

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

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

  $\begin{array}{l}{\text{SiCl}}_{\text{4(l)}}{\text{ + 2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{SiO}}_2{}_{\text{(s)}}{\text{+ 4 HCl}}_{\text{(aq)}}\\ \text{ΔH= }\sum {\text{ΔH}}_{\text{SiO2(s)}}{\text{+4×ΔH}}_{\text{HCl(aq)}}\text{ - }\sum {\text{ΔH}}_{\text{SiCl4(l)}}\text{+2}\times {\text{ΔH}}_{\text{H2O(l)}}\\ \text{ΔH= }\sum \text{(-911 kJ)+4×(-167kJ) - }\sum \text{(-687 kJ) +2}\times \text{(-286kJ)}\\ \text{ΔH= -320 kJ}\end{array}$

(f)

Interpretation Introduction

Interpretation: The $\text{ΔH}$ value for the given reaction with the help of ${\text{ΔH}}_{\text{f}}^{\text{0}}$ values needs to be determined.

  ${\text{MgO}}_{\text{(s)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{Mg(OH)}}_2{}_{\text{(s)}}$

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 73E

  $\text{ΔH= -37 kJ}$

Explanation of Solution

Given:

  ${\text{MgO}}_{\text{(s)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{Mg(OH)}}_2{}_{\text{(s)}}$

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

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

  $\begin{array}{l}{\text{MgO}}_{\text{(s)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{Mg(OH)}}_2{}_{\text{(s)}}\\ \text{ΔH= }\sum {\text{ΔH}}_{\text{Mg(OH)2(s)}}\text{ - }\sum {\text{ΔH}}_{\text{MgO(s)}}{\text{+ΔH}}_{\text{H2O(l)}}\\ \text{ΔH= }\sum \text{(-925 kJ)- }\sum \text{(-602 kJ) +(-286kJ)}\\ \text{ΔH= -37 kJ}\end{array}$