Chapter 14, Problem 14.30P

a)

Interpretation Introduction

Interpretation:

The value of $\Delta H{°}_{\text{rxn}}$ for following reaction has to be determined. Also, whether reaction is endothermic or exothermic has to be identified.

  $2{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(l\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{O}}_2\left(g\right)$

Concept Introduction:

Thermodynamics is a study of energy transfers that can be done by either heat or work. The energy transferred through work involves force. When work is positive then system gains energy while when work is negative then system loses energy. Heat is not a state function and therefore change in enthalpy of reaction $\left(\Delta H{°}_{\text{rxn}}\right)$ has been introduced. The expression to calculate $\Delta H{°}_{\text{rxn}}$ is as follows:

  $\Delta H_{\text{rxn}}=H_{\text{prod}}-H_{\text{react}}$

The useful property of $\Delta H{°}_{\text{rxn}}$ is that it is an additive property.

Answer to Problem 14.30P

The value of $\Delta H{°}_{\text{rxn}}$ for given reaction is $-196{\text{ kJ·mol}}^{\text{–1}}$ and it is exothermic reaction.

Explanation of Solution

Given chemical equation is as follows:

  $2{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(l\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{O}}_2\left(g\right)$        (1)

The expression of $\Delta H{°}_{\text{rxn}}$ for chemical equation (1) is as follows:

  $\Delta H{°}_{\text{rxn}}=\left[\left(2\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)+\left(1\right)\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\right]-\left[\left(2\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right)\right]$        (2)

Substitute $-285.8{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $0{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$, and $-187.8{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right)$ in equation (2).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\left(2\right)\left(-285.8{\text{ kJ·mol}}^{\text{–1}}\right)+\left(1\right)\left(0{\text{ kJ·mol}}^{\text{–1}}\right)\right]-\left[\left(2\right)\left(-187.8{\text{ kJ·mol}}^{\text{–1}}\right)\right]\\ =-196{\text{ kJ·mol}}^{\text{–1}}\end{array}$

Since $\Delta H{°}_{\text{rxn}}$ has negative sign thus reaction is exothermic.

b)

Interpretation Introduction

Interpretation:

The value of $\Delta H{°}_{\text{rxn}}$ for following reaction has to be determined. Also, whether reaction is endothermic or exothermic has to be identified.

  $\text{MgO}\left(s\right)+{\text{CO}}_2\left(g\right)\to {\text{MgCO}}_{\text{3}}\left(s\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 14.30P

The value of $\Delta H{°}_{\text{rxn}}$ for given reaction is $-100.7{\text{ kJ·mol}}^{\text{–1}}$ and it is exothermic reaction.

Explanation of Solution

Given chemical equation is as follows:

  $\text{MgO}\left(s\right)+{\text{CO}}_2\left(g\right)\to {\text{MgCO}}_{\text{3}}\left(s\right)$        (3)

The expression of $\Delta H{°}_{\text{rxn}}$ for chemical equation (3) is as follows:

  $\Delta H{°}_{\text{rxn}}=\left[\left(1\right)\Delta H{°}_{\text{f}}\left({\text{MgCO}}_{\text{3}}\right)\right]-\left[\begin{array}{l}\left(1\right)\Delta H{°}_{\text{f}}\left(\text{MgO}\right)\\ +\left(1\right)\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)\end{array}\right]$        (4)

Substitute $-1095.8{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{MgCO}}_{\text{3}}\right)$, $-601.6{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left(\text{MgO}\right)$, and $-393.5{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)$ in equation (4).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\left(1\right)\left(-1095.8{\text{ kJ·mol}}^{\text{–1}}\right)\right]-\left[\begin{array}{l}\left(1\right)\left(-601.6{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(1\right)\left(-393.5{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]\\ =-100.7{\text{ kJ·mol}}^{\text{–1}}\end{array}$

Since $\Delta H{°}_{\text{rxn}}$ has negative sign thus reaction is exothermic.

c)

Interpretation Introduction

Interpretation:

The value of $\Delta H{°}_{\text{rxn}}$ for following reaction has to be determined. Also, whether reaction is endothermic or exothermic has to be identified.

  ${\text{4NH}}_{\text{3}}\left(g\right)+5{\text{O}}_2\left(g\right)\to 4\text{NO}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 14.30P

The value of $\Delta H{°}_{\text{rxn}}$ for given reaction is $-1175.9{\text{ kJ·mol}}^{\text{–1}}$ and it is exothermic reaction.

Explanation of Solution

Given chemical equation is as follows:

  ${\text{4NH}}_{\text{3}}\left(g\right)+5{\text{O}}_2\left(g\right)\to 4\text{NO}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$        (5)

The expression of $\Delta H{°}_{\text{rxn}}$ for chemical equation (5) is as follows:

  $\Delta H{°}_{\text{rxn}}=\left[\left(4\right)\Delta H{°}_{\text{f}}\left(\text{NO}\right)+\left(6\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\right]-\left[\begin{array}{l}\left(4\right)\Delta H{°}_{\text{f}}\left({\text{NH}}_{\text{3}}\right)\\ +\left(5\right)\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\end{array}\right]$        (6)

Substitute $+91.3{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left(\text{NO}\right)$, $-241.8{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $-45.9{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{NH}}_{\text{3}}\right)$, and $0{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$ in equation (6).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\left(1\right)\left(+91.3{\text{ kJ·mol}}^{\text{–1}}\right)+\left(6\right)\left(-241.8{\text{ kJ·mol}}^{\text{–1}}\right)\right]-\left[\begin{array}{l}\left(4\right)\left(-45.9{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(5\right)\left(0{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]\\ =-1175.9{\text{ kJ·mol}}^{\text{–1}}\end{array}$

Since $\Delta H{°}_{\text{rxn}}$ has negative sign, thus reaction is exothermic.