Chapter 4, Problem 4E.7E

(a)

Interpretation Introduction

Interpretation:

The reaction enthalpy for the given reaction has to be determined.

Concept Introduction:

The bond enthalpy is the physical quantity use to measure the strength of chemical bond.  The reaction enthalpy of a reaction in terms of the bond enthalpies is the difference between the standard bond enthalpies reactant and that of the product.  The breaking of bond is always the endothermic process however the formation of bond is the exothermic process.

Answer to Problem 4E.7E

The reaction enthalpy for the given reaction is $\underset{\_}{-202kJ}$.

Explanation of Solution

The given chemical reaction for which reaction enthalpy has to be calculated is shown below.

  ${\text{N}}_2\left(\text{g}\right)+3{\text{F}}_2\left(\text{g}\right)\to 2{\text{NF}}_3\left(\text{g}\right)$

The total bond enthalpy of reactants of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}=\left(1\text{mol}\right)\Delta H_{\text{B}}\left({\text{N}}_{\text{2}}\right)+\left(3\text{mol}\right)\Delta H_{\text{B}}\left(\text{F}-\text{F}\right)$        (1)

Where,

• ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ is total bond enthalpy of reactants.
• $\Delta H_{\text{B}}\left({\text{N}}_{\text{2}}\right)$ is bond enthalpy of ${\text{N}}_{\text{2}}$.
• $\Delta H_{\text{B}}\left(\text{F}-\text{F}\right)$ is bond enthalpy of $\text{F}-\text{F}$ bond.

The value of $\Delta H_{\text{B}}\left({\text{N}}_{\text{2}}\right)$ is $944\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{F}-\text{F}\right)$ is $158\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta H_{\text{B}}\left({\text{N}}_{\text{2}}\right)$ and $\Delta H_{\text{B}}\left(\text{F}-\text{F}\right)$ in equation (1).

  $\begin{array}{c}{\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}=\left(1\text{mol}\right)\times \left(944\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(3\text{mol}\right)\times \left(158\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =944\text{kJ}+474\text{kJ}\\ =\text{1418}\text{kJ}\end{array}$

The total bond enthalpy of products of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}=\left(6\text{mol}\right)\Delta H_{\text{B}}\left(\text{N}-\text{F}\right)$        (2)

Where,

• ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ is total bond enthalpy of product.
• $\Delta H_{\text{B}}\left(\text{N}-\text{F}\right)$ is bond enthalpy of $\text{N}-\text{F}$ bond.

The value of $\Delta H_{\text{B}}\left(\text{N}-\text{F}\right)$ is $270\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta H_{\text{B}}\left(\text{N}-\text{F}\right)$ in equation (2).

  $\begin{array}{c}{\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}=\left(6\text{mol}\right)\times \left(270\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =1620\text{kJ}\end{array}$

The reaction enthalpy of the given reaction is calculated by the relation shown below.

  $\Delta H_{\text{B}}°={\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}-{\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$        (3)

Where,

• $\Delta H_{\text{B}}°$ is reaction enthalpy.
• ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ is total bond enthalpy of reactants.
• ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ is total bond enthalpy of product.

The value of ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ is $\text{1418}\text{kJ}$.

The value of ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ is $1620\text{kJ}$.

Substitute the value of ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ and ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ in equation (3).

    $\begin{array}{c}\Delta H_{\text{B}}°=\text{1418}\text{kJ}-1620\text{kJ}\\ =-202\text{kJ}\end{array}$

The obtained reaction enthalpy of the given reaction is negative, which indicates that the reaction is exothermic in nature.

Thus, the reaction enthalpy for the given reaction is $\underset{\_}{-202kJ}$.

(b)

Interpretation Introduction

Interpretation:

The reaction enthalpy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4E.7E

The reaction enthalpy for the given reaction is $\underset{\_}{-45kJ}$.

Explanation of Solution

The given chemical reaction for which reaction enthalpy has to be calculated is shown below.

  ${\text{CH}}_3{\text{CHCH}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)\to {\text{CH}}_3\text{CH}\left(\text{OH}\right){\text{CH}}_3\left(\text{g}\right)$

The total bond enthalpy of reactants of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}=\left(\begin{array}{l}\left(6\text{mol}\right)\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)+\left(1\text{mol}\right)\Delta H_{\text{B}}\left(\text{C}-\text{C}\right)+\\ \left(1\text{mol}\right)\Delta H_{\text{B}}\left(\text{C}=\text{C}\right)+\left(2\text{mol}\right)\Delta H_{\text{B}}\left(\text{O}-\text{H}\right)\end{array}\right)$        (4)

Where,

• ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ is total bond enthalpy of reactants.
• $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$ is bond enthalpy of $\text{C}-\text{H}$ bond.
• $\Delta H_{\text{B}}\left(\text{C}-\text{C}\right)$ is bond enthalpy of $\text{C}-\text{C}$ bond.
• $\Delta H_{\text{B}}\left(\text{C=C}\right)$ is bond enthalpy of $\text{C=C}$ bond.
• $\Delta H_{\text{B}}\left(\text{O}-\text{H}\right)$ is bond enthalpy of $\text{O}-\text{H}$ bond.

The value of $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$ is $412\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{C}-\text{C}\right)$ is $348\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{C}=\text{C}\right)$ is $612\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{O}-\text{H}\right)$ is $463\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$, $\Delta H_{\text{B}}\left(\text{C}-\text{C}\right)$, $\Delta H_{\text{B}}\left(\text{C}=\text{C}\right)$ and $\Delta H_{\text{B}}\left(\text{O}-\text{H}\right)$ in equation (4).

  $\begin{array}{c}{\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}=\left(\begin{array}{c}\left(6\text{mol}\right)\times \left(412\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1\text{mol}\right)\times \left(348\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1\text{mol}\right)\times \left(612\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(2\text{mol}\right)\times \left(463\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right)\\ =2472\text{kJ}+348\text{kJ}+612\text{kJ}+926\text{kJ}\\ =4358\text{kJ}\end{array}$

The total bond enthalpy of products of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}=\left(\begin{array}{c}\left(7\text{mol}\right)\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)+\left(2\text{mol}\right)\Delta H_{\text{B}}\left(\text{C}-\text{C}\right)\\ +\left(1\text{mol}\right)\Delta H_{\text{B}}\left(\text{C}-\text{O}\right)+\left(1\text{mol}\right)\Delta H_{\text{B}}\left(\text{O}-\text{H}\right)\end{array}\right)$        (5)

Where,

• ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ is total bond enthalpy of product.
• $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$ is bond enthalpy of $\text{C}-\text{H}$ bond.
• $\Delta H_{\text{B}}\left(\text{C}-\text{C}\right)$ is bond enthalpy of $\text{C}-\text{C}$ bond.
• $\Delta H_{\text{B}}\left(\text{C}-\text{O}\right)$ is bond enthalpy of $\text{C}-\text{O}$ bond.
• $\Delta H_{\text{B}}\left(\text{O}-\text{H}\right)$ is bond enthalpy of $\text{O}-\text{H}$ bond.

The value of $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$ is $412\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{C}-\text{C}\right)$ is $348\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{C}-\text{O}\right)$ is $360\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{O}-\text{H}\right)$ is $463\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$, $\Delta H_{\text{B}}\left(\text{C}-\text{C}\right)$, $\Delta H_{\text{B}}\left(\text{C}-\text{O}\right)$ and $\Delta H_{\text{B}}\left(\text{O}-\text{H}\right)$ in equation (5).

  $\begin{array}{c}{\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}=\left(\begin{array}{c}\left(7\text{mol}\right)\times \left(412\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(2\text{mol}\right)\times \left(348\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1\text{mol}\right)\times \left(360\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1\text{mol}\right)\times \left(463\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right)\\ =2884\text{kJ}+696\text{kJ}+360\text{kJ}+463\text{kJ}\\ =\text{4403}\text{kJ}\end{array}$

The value of ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ is $4358\text{kJ}$.

The value of ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ is $\text{4403}\text{kJ}$.

Substitute the value of ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ and ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ in equation (3).

    $\begin{array}{c}\Delta H_{\text{B}}°=4358\text{kJ}-\text{4403}\text{kJ}\\ =-45\text{kJ}\end{array}$

The obtained reaction enthalpy of the given reaction is negative, which indicates that the reaction is exothermic in nature.

Thus, the reaction enthalpy for the given reaction is $\underset{\_}{-45kJ}$.

(c)

Interpretation Introduction

Interpretation:

The reaction enthalpy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4E.7E

The reaction enthalpy for the given reaction is $\underset{\_}{-115kJ}$.

Explanation of Solution

The given chemical reaction for which reaction enthalpy has to be calculated is shown below.

  ${\text{CH}}_4\left(\text{g}\right)+{\text{Cl}}_2\left(\text{g}\right)\to {\text{CH}}_3\text{Cl}\left(\text{g}\right)+\text{HCl}\left(\text{g}\right)$

The total bond enthalpy of reactants of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}=\left(4\text{mol}\right)\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)+\left(1\text{mol}\right)\Delta H_{\text{B}}\left(\text{Cl}-\text{Cl}\right)$        (6)

Where,

• ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ is total bond enthalpy of reactants.
• $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$ is bond enthalpy of $\text{C}-\text{H}$ bond.
• $\Delta H_{\text{B}}\left(\text{Cl}-\text{Cl}\right)$ is bond enthalpy of $\text{Cl}-\text{Cl}$ bond.

The value of $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$ is $412\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{Cl}-\text{Cl}\right)$ is $242\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$ and $\Delta H_{\text{B}}\left(\text{Cl}-\text{Cl}\right)$ in equation (6).

  $\begin{array}{c}{\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}=\left(4\text{mol}\right)\times \left(412\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1\text{mol}\right)\times \left(242\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =1648\text{kJ}+242\text{kJ}\\ =\text{1890}\text{kJ}\end{array}$

The total bond enthalpy of products of the given reaction is calculated by the relation shown below.

    ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}=\left(\begin{array}{l}\left(3\text{mol}\right)\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)+\left(1\text{mol}\right)\Delta H_{\text{B}}\left(\text{C}-\text{Cl}\right)\\ +\left(1\text{mol}\right)\Delta H_{\text{B}}\left(\text{H}-\text{Cl}\right)\end{array}\right)$        (7)

Where,

• ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ is total bond enthalpy of product.
• $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$ is bond enthalpy of $\text{C}-\text{H}$ bond.
• $\Delta H_{\text{B}}\left(\text{C}-\text{Cl}\right)$ is bond enthalpy of $\text{C}-\text{Cl}$ bond.
• $\Delta H_{\text{B}}\left(\text{H}-\text{Cl}\right)$ is bond enthalpy of $\text{H}-\text{Cl}$ bond.

The value of $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$ is $412\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{C}-\text{Cl}\right)$ is $338\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta H_{\text{B}}\left(\text{H}-\text{Cl}\right)$ is $431\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta H_{\text{B}}\left(\text{C}-\text{H}\right)$, $\Delta H_{\text{B}}\left(\text{C}-\text{Cl}\right)$ and $\Delta H_{\text{B}}\left(\text{H}-\text{Cl}\right)$ in equation (7).

  $\begin{array}{c}{\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}=\left(\begin{array}{c}\left(3\text{mol}\right)\times \left(412\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1\text{mol}\right)\times \left(338\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1\text{mol}\right)\times \left(431\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right)\\ =1236\text{kJ}+338\text{kJ}+431\text{kJ}\\ =2005\text{kJ}\end{array}$

The value of ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ is $\text{1890}\text{kJ}$.

The value of ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ is $2005\text{kJ}$.

Substitute the value of ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{reactants}\right)}$ and ${\displaystyle \sum \Delta H_{\text{B}}°\left(\text{products}\right)}$ in equation (3).

    $\begin{array}{c}\Delta H_{\text{B}}°=\text{1890}\text{kJ}-\text{2005}\text{kJ}\\ =-\text{115}\text{kJ}\end{array}$

The obtained reaction enthalpy of the given reaction is negative, which indicates that the reaction is exothermic in nature.

Thus, the reaction enthalpy for the given reaction is $\underset{\_}{-115kJ}$.